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== This room is an usable summery of my notes ==
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<small>Whole my previous notes is visible in the [https://www.encyclopediaofmath.org/index.php?title=User_talk:Musictheory2math&diff=prev&oldid=43134 revision as of 18:42, 13 April 2018] Alireza Badali 21:52, 13 April 2018 (CEST)</small>
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== <big><big>$\mathscr B$</big></big> $theory$ (algebraic topological analytical number theory) ==
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Logarithm function as an inverse of the function $f:\Bbb N\to\Bbb R,\,f(n)=a^n,\,a\in\Bbb R$ has prime numbers properties because in usual definition of prime numbers multiplication operation is a point meantime we have $a^n=a\times a\times ...a,$ $n$ times, hence prime number theorem or its extensions or some other forms is applied in $B$ theory for solving problems on prime numbers exclusively and not all natural numbers.
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=== Goldbach's conjecture ===
  
 
'''Lemma''': For each subinterval $(a,b)$ of $[0.1,1),\,\exists m\in \Bbb N$ that $\forall k\in \Bbb N$ with $k\ge m$ then $\exists t\in (a,b)$ that $t\cdot 10^k\in \Bbb P$.
 
'''Lemma''': For each subinterval $(a,b)$ of $[0.1,1),\,\exists m\in \Bbb N$ that $\forall k\in \Bbb N$ with $k\ge m$ then $\exists t\in (a,b)$ that $t\cdot 10^k\in \Bbb P$.
:[https://math.stackexchange.com/questions/2482941/a-simple-question-about-density-in-the-interval-0-1-1/2483079#2483079 Proof] given by [https://math.stackexchange.com/users/149178/adayah @Adayah] from stackexchange.com: Without loss of generality (by passing to a smaller subinterval) we can assume that $(a, b) = \left( \frac{s}{10^r}, \frac{t}{10^r} \right)$, where $s, t, r$ are positive integers and $s < t$. Let $\alpha = \frac{t}{s}$.  
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:[https://math.stackexchange.com/questions/2482941/a-simple-question-about-density-in-the-interval-0-1-1/2483079#2483079 Proof] given by [https://math.stackexchange.com/users/149178/adayah @Adayah] from stackexchange site: Without loss of generality (by passing to a smaller subinterval) we can assume that $(a, b) = \left( \frac{s}{10^r}, \frac{t}{10^r} \right)$, where $s, t, r$ are positive integers and $s < t$. Let $\alpha = \frac{t}{s}$.  
 
:The statement is now equivalent to saying that there is $m \in \mathbb{N}$ such that for every $k \geqslant m$ there is a prime $p$ with $10^{k-r} \cdot s < p < 10^{k-r} \cdot t$.
 
:The statement is now equivalent to saying that there is $m \in \mathbb{N}$ such that for every $k \geqslant m$ there is a prime $p$ with $10^{k-r} \cdot s < p < 10^{k-r} \cdot t$.
 
:We will prove a stronger statement: there is $m \in \mathbb{N}$ such that for every $n \geqslant m$ there is a prime $p$ such that $n < p < \alpha \cdot n$. By taking a little smaller $\alpha$ we can relax the restriction to $n < p \leqslant \alpha \cdot n$.
 
:We will prove a stronger statement: there is $m \in \mathbb{N}$ such that for every $n \geqslant m$ there is a prime $p$ such that $n < p < \alpha \cdot n$. By taking a little smaller $\alpha$ we can relax the restriction to $n < p \leqslant \alpha \cdot n$.
 
:Now comes the prime number theorem: $$\lim_{n \to \infty} \frac{\pi(n)}{\frac{n}{\log n}} = 1$$
 
:Now comes the prime number theorem: $$\lim_{n \to \infty} \frac{\pi(n)}{\frac{n}{\log n}} = 1$$
 
:where $\pi(n) = \# \{ p \leqslant n : p$ is prime$\}.$ By the above we have $$\frac{\pi(\alpha n)}{\pi(n)} \sim \frac{\frac{\alpha n}{\log(\alpha n)}}{\frac{n}{\log(n)}} = \alpha \cdot \frac{\log n}{\log(\alpha n)} \xrightarrow{n \to \infty} \alpha$$
 
:where $\pi(n) = \# \{ p \leqslant n : p$ is prime$\}.$ By the above we have $$\frac{\pi(\alpha n)}{\pi(n)} \sim \frac{\frac{\alpha n}{\log(\alpha n)}}{\frac{n}{\log(n)}} = \alpha \cdot \frac{\log n}{\log(\alpha n)} \xrightarrow{n \to \infty} \alpha$$
:hence $\displaystyle \lim_{n \to \infty} \frac{\pi(\alpha n)}{\pi(n)} = \alpha$. So there is $m \in \mathbb{N}$ such that $\pi(\alpha n) > \pi(n)$ whenever $n \geqslant m$, which means there is a prime $p$ such that $n < p \leqslant \alpha \cdot n$, and that is what we wanted.♦
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:hence $\displaystyle \lim_{n \to \infty} \frac{\pi(\alpha n)}{\pi(n)} = \alpha$. So there is $m \in \mathbb{N}$ such that $\pi(\alpha n) > \pi(n)$ whenever $n \geqslant m$, which means there is a prime $p$ such that $n < p \leqslant \alpha \cdot n$, and that is what we wanted♦
  
  
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'''Guess''' $1$: $g$ isn't an injective function.
 
'''Guess''' $1$: $g$ isn't an injective function.
  
'''Question''' $1$: Assuming above guess let $[a,a]:=\{a\}$ and $\forall n\in\Bbb N,\, h_n$ is the least subinterval of $[0.01,0.1)$ like $[a,b]$ in terms of size of $b-a$ such that $\{\epsilon\in (0,0.09)\cap (\bigcup _{k\in\Bbb N} r_k(\Bbb N))\mid g(\epsilon)=n\}\subsetneq h_n$ and obviously $g(a)=n=g(b)$ now the question is $\forall n,m\in\Bbb N$ that $m\neq n$ is $h_n\cap h_m=\emptyset$?
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'''Question''' $1$: Assuming guess $1$, let $[a,a]:=\{a\}$ and $\forall n\in\Bbb N,\, h_n$ is the least subinterval of $[0.01,0.1)$ like $[a,b]$ in terms of size of $b-a$ such that $\{\epsilon\in (0,0.09)\cap (\bigcup _{k\in\Bbb N} r_k(\Bbb N))\mid g(\epsilon)=n\}\subsetneq h_n$ and obviously $g(a)=n=g(b)$ now the question is $\forall n,m\in\Bbb N$ that $m\neq n$ is $h_n\cap h_m=\emptyset$?
:[https://math.stackexchange.com/questions/2518063/a-medium-question-about-a-set-related-to-prime-numbers/2526481#2526481 Guidance] given by [https://math.stackexchange.com/users/276986/reuns @reuns] from stackexchange.com:  
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:[https://math.stackexchange.com/questions/2518063/a-medium-question-about-a-set-related-to-prime-numbers/2526481#2526481 Guidance] given by [https://math.stackexchange.com/users/276986/reuns @reuns] from stackexchange site:  
 
:* For $n \in \mathbb{N}$ then $r(n) = 10^{-\lceil \log_{10}(n) \rceil} n$, ie. $r(19) = 0.19$. We look at the image by $r$ of the primes $\mathbb{P}$.
 
:* For $n \in \mathbb{N}$ then $r(n) = 10^{-\lceil \log_{10}(n) \rceil} n$, ie. $r(19) = 0.19$. We look at the image by $r$ of the primes $\mathbb{P}$.
 
:* Let $F((c,d)) = \min \{ p \in \mathbb{P}, r(p) \in (c,d)\}$ and $f((c,d)) = \pi(F(c,d))= \min \{ n, r(p_n) \in (c,d)\}$  ($\pi$ is the prime counting function)
 
:* Let $F((c,d)) = \min \{ p \in \mathbb{P}, r(p) \in (c,d)\}$ and $f((c,d)) = \pi(F(c,d))= \min \{ n, r(p_n) \in (c,d)\}$  ($\pi$ is the prime counting function)
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Suppose $r:\Bbb N\to (0,1)$ is a function given by $r(n)$ is obtained with putting a point at the beginning of $n$ instance $r(34880)=0.34880$ and similarly consider $\forall k\in\Bbb N\cup\{0\},$ $r_k: \Bbb N \to (0,1)$ given by $\forall n\in\Bbb N$, $r_k(n)=10^{-k}\cdot r(n)$ and let $S=r(\Bbb P)$ and $S_1=\bigcup _{k\in\Bbb N\cup\{0\}}r_k(\Bbb P)$.
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Suppose $r:\Bbb N\to (0,1)$ is a function given by $r(n)$ is obtained by putting a point at the beginning of $n$ instance $r(34880)=0.34880$ and similarly consider $\forall k\in\Bbb N,\, w_k:\Bbb N\to (0,1)$ is a function given by $\forall n\in\Bbb N,$ $w_k(n)=10^{1-k}\cdot r(n)$ and let $S=\bigcup _{k\in\Bbb N}w_k(\Bbb P)$.
  
'''Theorem''' $1$: $S$ is dense in the $[0.1,1]$. (proof using above lemma)
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'''Theorem''' $1$: $r(\Bbb P)$ is dense in the interval $[0.1,1]$. (proof using lemma above)
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:Regarding to expression form of Goldbach's conjecture, by using this theorem, I wanted enmesh prime numbers properties (prime number theorem should be used for proving this theorem and there is no way except using prime number theorem to prove this density because there is no deference between a prime $p$ and its image $r(p)$ other than a sign or a mark as a point for instance $59$ & $0.59$.) towards Goldbach hence I planned this method.
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:A corollary: For each natural number like $a=a_1a_2a_3...a_k$ that $a_j$ is $j$_th digit for $j=1,2,3,...,k$, there is a natural number like $b=b_1b_2b_3...b_r$ such that the number $c=a_1a_2a_3...a_kb_1b_2b_3...b_r$ is a prime number.
  
'''Corollary''': For each natural number like $a=a_1a_2a_3...a_k$ that $a_j$ is $j$_th digit for $j=1,2,3,...,k$, there is a natural number like $b=b_1b_2b_3...b_r$ such that the number $c=a_1a_2a_3...a_kb_1b_2b_3...b_r$ is a prime number.
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'''Theorem''' $2$: $S$ is dense in the interval $[0,1]$ and $S\times S$ is dense in the $[0,1]\times [0,1]$.
  
'''Theorem''' $2$: $S_1$ is dense in the interval $[0,1]$ and $S_1\times S_1$ is dense in the $[0,1]\times [0,1]$.
 
  
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'''An algorithm''' that makes new cyclic groups on $\Bbb N$:
  
'''An algorithm''' that makes new cyclic groups on $\Bbb N$:
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Let $\Bbb N$ be that group and at first write integers as a sequence with starting from $0$ and let identity element $e=1$ be corresponding with $0$ and two generating elements $m$ & $n$ be corresponding with $1$ & $-1$ so we have $\Bbb N=\langle m\rangle=\langle n\rangle$ for instance: $$0,1,2,-1,-2,3,4,-3,-4,5,6,-5,-6,7,8,-7,-8,9,10,-9,-10,11,12,-11,-12,...$$ $$1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,...$$
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then regarding to the sequence find a rotation number that for this sequence is $4$ and hence equations should be written with module $4$, then consider $4m-2,4m-1,4m,4m+1$ that the last should be $km+1$ and initial be $km+(2-k)$ otherwise equations won't match with definitions of members inverse, and make a table of products of those $k$ elements but during writing equations pay attention if an equation is right for given numbers it will be right generally for other numbers and of course if integers corresponding with two member don't have same signs then product will be a piecewise-defined function for example $12\star _u 15=6$ or $(4\times 3)\star _u (4\times 4-1)=6$ because $(-5)+8=3$ & $-5\to 12,\,\, 8\to 15,\,\, 3\to 6,$ that implies $(4n)\star _u (4m-1)=4m-4n+2$ if $4m-1\gt 4n$ of course it's better at first members inverse be defined for example since $(-9)+9=0$ & $0\to 1,\,\, -9\to 20,\,\, 9\to 18$ so $20\star _u 18=1$, that implies $(4m)\star _u (4m-2)=1$, and with a little bit addition and multiplication all equations will be obtained simply that for this example is:
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$\begin{cases} m\star _u 1=m\\ (4m)\star _u (4m-2)=1=(4m+1)\star _u (4m-1)\\ (4m-2)\star _u (4n-2)=4m+4n-5\\ (4m-2)\star _u (4n-1)=4m+4n-2\\ (4m-2)\star _u (4n)=\begin{cases} 4m-4n-1 & 4m-2\gt 4n\\ 4n-4m+1 & 4n\gt 4m-2\\ 3 & m=n+1\end{cases}\\ (4m-2)\star _u (4n+1)=\begin{cases} 4m-4n-2 & 4m-2\gt 4n+1\\ 4n-4m+4 & 4n+1\gt 4m-2\end{cases}\\ (4m-1)\star _u (4n-1)=4m+4n-1\\ (4m-1)\star _u (4n)=\begin{cases} 4m-4n+2 & 4m-1\gt 4n\\ 4n-4m & 4n\gt 4m-1\\ 2 & m=n\end{cases}\\ (4m-1)\star _u (4n+1)=\begin{cases} 4m-4n-1 & 4m-1\gt 4n+1\\ 4n-4m+1 & 4n+1\gt 4m-1\\ 3 & m=n+1\end{cases}\\ (4m)\star _u (4n)=4m+4n-3\\ (4m)\star _u (4n+1)=4m+4n\\ (4m+1)\star _u  (4n+1)=4m+4n+1\\ \Bbb N=\langle 2\rangle=\langle 4\rangle\end{cases}$
  
Let $(\Bbb N,\star)$ be that group and we have $\Bbb N=\langle 2\rangle$ & $e=1$ and at first write integers as a sequence with starting from $0$ for instance: $$0,1,2,-1,-2,3,4,-3,-4,5,6,-5,-6,7,8,-7,-8,9,10,-9,-10,11,12,-11,-12,...$$ $$1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,...$$
 
then regarding to the sequence find a rotation number that for this sequence is $4$ and hence equations should be written with module $4$, then consider $4m-2,4m-1,4m,4m+1$ that the last should be $km+1$ and initial be $km+(k-2)$ otherwise equations won't match with members inverse definitions, and make a table of productions of those $k$ elements but during writing equations pay attention if an equation is right for given numbers it will be right generally for other numbers for example $12\star 15=6$ or $(4\times 3)\star (4\times 4-1)=6$ because $(-5)+8=3$ & $-5\to 12,\,\, 8\to 15,\,\, 3\to 6,$ that implies $(4n)\star (4m-1)=4m-4n+2$ if $4m-1\gt 4n$ of course it's better at first members inverse be defined for example since $(-9)+9=0$ & $0\to 1,\,\, -9\to 20,\,\, 9\to 18$ so $20\star 18=1$, that implies $(4m)\star (4m-2)=1$, and with a little addition and subtraction all equations will be obtained simply that for this example is:
 
  
$\begin{cases} m\star _T 1=m\\ (4m)\star _T (4m-2)=1=(4m+1)\star _T (4m-1)\\ (4m)\star _T (4m+2)=3=(4m+1)\star _T (4m+3)\\ (4m-2)\star _T (4n-2)=4m+4n-5\\ (4m-2)\star _T (4n-1)=4m+4n-2\\ (4m-2)\star _T (4n)=\begin{cases} 4m-4n-1 & 4m-2\gt 4n\\ 4n-4m+1 & 4n\gt 4m-2\end{cases}\\ (4m-2)\star _T (4n+1)=\begin{cases} 4m-4n-2 & 4m-2\gt 4n+1\\ 4n-4m+4 & 4n+1\gt 4m-2\end{cases}\\ (4m-1)\star _T (4n-1)=4m+4n-1\\ (4m-1)\star _T (4n)=\begin{cases} 4m-4n+2 & 4m-1\gt 4n\\ 4n-4m & 4n\gt 4m-1\\ 2 & m=n\end{cases}\\ (4m-1)\star _T (4n+1)=\begin{cases} 4m-4n-1 & 4m-1\gt 4n+1\\ 4n-4m+1 & 4n+1\gt 4m-1\end{cases}\\ (4m)\star _T (4n)=4m+4n-3\\ (4m)\star _T (4n+1)=4m+4n\\ (4m+1)\star _T  (4n+1)=4m+4n+1\end{cases}$
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Problem $1$: By using matrices rewrite operation of every group on $\Bbb N$.
  
  
 
Assume $\forall m,n\in\Bbb N$: $\begin{cases} n\star 1=n\\ (2n)\star (2n+1)=1\\ (2n)\star (2m)=2n+2m\\ (2n+1)\star (2m+1)=2n+2m+1\\ (2n)\star (2m+1)=\begin{cases} 2m-2n+1 & 2m+1\gt 2n\\ 2n-2m & 2n\gt 2m+1\end{cases}\end{cases}$
 
Assume $\forall m,n\in\Bbb N$: $\begin{cases} n\star 1=n\\ (2n)\star (2n+1)=1\\ (2n)\star (2m)=2n+2m\\ (2n+1)\star (2m+1)=2n+2m+1\\ (2n)\star (2m+1)=\begin{cases} 2m-2n+1 & 2m+1\gt 2n\\ 2n-2m & 2n\gt 2m+1\end{cases}\end{cases}$
  
and $p_n\star _1p_m=p_{n\star m}$ that $p_n$ is $n$_th prime, obviously $(\Bbb N,\star)$ & $(\Bbb P,\star _1)$ are groups and $\langle 2\rangle =(\Bbb N,\star)\cong (\Bbb Z,+)\cong (\Bbb P,\star _1)=\langle 3\rangle$.
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and $p_n\star _1p_m=p_{n\star m}$ that $p_n$ is $n$_th prime with $e=p_1=2$, obviously $(\Bbb N,\star)$ & $(\Bbb P,\star _1)$ are groups and $\langle 2\rangle =\langle 3\rangle =(\Bbb N,\star)\simeq (\Bbb Z,+)\simeq (\Bbb P,\star _1)=\langle 3\rangle=\langle 5\rangle$.
  
  
Suppose $\forall k\in\Bbb N,\, w_k:\Bbb N\to (0,1)$ is a function given by $\forall n\in\Bbb N,\, w_k(n)=r_{k-1}(n)$.
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'''Theorem''' $3$: $(S,\star _S)$ is a group as: $\forall p,q\in\Bbb P,\,\forall m,n\in\Bbb N,\,\forall w_m(p),w_n(q)\in S,$
  
'''Theorem''' $3$: Let $e=0.2$  & $\forall w_m(p),w_n(q)\in S_1,\, w_m(p)\star _{S_1} w_n(q)=w_{m\star n} (p\star _1 q)$ that $p,q\in\Bbb P,$ $m,n\in\Bbb N$ & $(w_m(p))^{-1}=w_{m^{-1}}(p^{-1})$ that $m\star m^{-1}=1,\, p\star _1 p^{-1}=2$ now $\langle 0.02,0.3\rangle=(S_1,\star _{S_1})\cong\Bbb Z\oplus\Bbb Z$, of course using above algorithm to generate cyclic groups on $\Bbb N$, we can impose another group structure on $\Bbb N$ and consequently on $\Bbb P$ but eventually $S_1$ with an operation analogous above operation $\star _{S_1}$ will be an Abelian group.
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$\begin{cases} e=0.2\\ \\(w_m(p))^{-1}=w_{m^{-1}}(p^{-1}) & m\star m^{-1}=1,\, p\star _1 p^{-1}=2\\ \\w_m(p)\star _S w_n(q)=w_{m\star n} (p\star _1 q)\end{cases}$
:'''Question''' $2$: Under which topology will $S_1$ be a topological group?
 
  
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hence $\langle 0.02,0.3\rangle=(S,\star _S)\simeq\Bbb Z\oplus\Bbb Z$.
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:of course using algorithm above to generate cyclic groups on $\Bbb N$, we can impose another group structure on $\Bbb N$ and consequently on $\Bbb P$ but eventually $S$ with an operation analogous above operation $\star _S$ will be an Abelian group.
  
'''Theorem''' $4$: Let $e=(0.2,0.2)$ & $\forall (w_{m_1}(p_1),w_{m_2}(p_2)),(w_{n_1}(q_1),w_{n_2}(q_2))\in S_1\times S_1,$ $$(w_{m_1}(p_1),w_{m_2}(p_2))\star _{S_1\times S_1} (w_{n_1}(q_1),w_{n_2}(q_2))=(w_{m_1\star n_1} (p_1\star _1 q_1),w_{m_2\star n_2}(p_2\star _1 q_2))$$ that $m_1,n_1,m_2,n_2\in\Bbb N,\, p_1,p_2,q_1,q_2\in\Bbb P$ & $(w_m(p),w_n(q))^{-1}=(w_{m^{-1}}(p^{-1}),w_{n^{-1}}(q^{-1}))$ that $m\star m^{-1}=1=n\star n^{-1}$, $p\star _1p^{-1}=2=q\star _1q^{-1}$ now $$\langle (0.02,0.2),(0.2,0.02),(0.3,0.2),(0.2,0.3)\rangle=(S_1\times S_1,\star _{S_1\times S_1})\cong\Bbb Z\oplus\Bbb Z\oplus\Bbb Z\oplus\Bbb Z,$$ of course using above algorithm to generate cyclic groups on $\Bbb N$, we can impose another group structure on $\Bbb N$ and consequently on $\Bbb P$ but eventually $S_1\times S_1$ with an operation analogous above operation $\star _{S_1\times S_1}$ will be an Abelian group.
 
:'''Question''' $3$: Under which topology will $S_1\times S_1$ be a topological group?
 
  
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'''Theorem''' $4$: $(S\times S,\star _{S\times S})$ is a group as: $\forall m_1,n_1,m_2,n_2\in\Bbb N,\,\forall p_1,p_2,q_1,q_2\in\Bbb P,$ $\forall (w_{m_1}(p_1),w_{m_2}(p_2)),(w_{n_1}(q_1),w_{n_2}(q_2))\in S\times S,$
  
I want make some topologies having '''prime numbers properties''' presentable in collection of '''open sets''', in principle when we image a prime $p$ to real numbers as $w_k(p)$ indeed we accompany prime numbers properties into real numbers which regarding to the expression form of prime number theorem for this aim we should use an important mathematical technique as logarithm function into some planned topologies: '''question''' $4$: Let $M$ be a topological space and $A,B$ are subsets of $M$ with $A\subset B$ and $A$ is dense in $B,$ since $A$ is dense in $B,$ is there some way in which a topology on $B$ may be induced other than the subspace topology? I am also interested in specialisations, for example if $M$ is Hausdorff or Euclidean.
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$\begin{cases} e=(0.2,0.2)\\ \\(w_{m_1}(p_1),w_{m_2}(p_2))^{-1}=(w_{m_1^{-1}}(p_1^{-1}),w_{m_2^{-1}}(p_2^{-1}))\\ \text{such that}\quad m_1\star m_1^{-1}=1=m_2\star m_2^{-1},\, p_1\star _1p_1^{-1}=2=p_2\star _1p_2^{-1}\\ \\(w_{m_1}(p_1),w_{m_2}(p_2))\star _{S\times S} (w_{n_1}(q_1),w_{n_2}(q_2))=(w_{m_1\star n_1} (p_1\star _1 q_1),w_{m_2\star n_2}(p_2\star _1 q_2))\end{cases}$
:Perhaps this technique is useful: $\forall n\in\Bbb N,$ and for each subinterval $(a,b)$ of $[0.1,1),$ that $a\neq b,$
 
  
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hence $\langle (0.02,0.2),(0.2,0.02),(0.3,0.2),(0.2,0.3)\rangle=(S\times S,\star _{S\times S})\simeq\Bbb Z\oplus\Bbb Z\oplus\Bbb Z\oplus\Bbb Z$.
 +
:of course using algorithm above to generate cyclic groups on $\Bbb N$, we can impose another group structure on $\Bbb N$ and consequently on $\Bbb P$ but eventually $S\times S$ with an operation analogous above operation $\star _{S\times S}$ will be an Abelian group.
 +
 +
 +
I want make some topologies having '''prime numbers properties''' presentable in the collection of '''open sets''', in principle when we image a prime $p$ to real numbers as $w_k(p)$ indeed we accompany prime numbers properties among real numbers which regarding to the expression form of prime number theorem for this aim we should use an important mathematical technique as logarithm function into some planned topologies: '''question''' $2$: Let $M$ be a topological space and $A,B$ are subsets of $M$ with $A\subset B$ and $A$ is dense in $B,$ since $A$ is dense in $B,$ is there some way in which a topology on $B$ may be induced other than the subspace topology? I am also interested in specialisations, for example if $M$ is Hausdorff or Euclidean. ($M=\Bbb R,\,B=[0,1],\,A=S$ or $M=\Bbb R^2,$ $B=[0,1]\times[0,1],$ $A=S\times S$)
 +
:Perhaps this technique is useful: an extension of prime number theorem: $\forall n\in\Bbb N,$ and for each subinterval $(a,b)$ of $[0.1,1),$ that $a\neq b,$
 
:$\begin{cases} U_{(a,b)}:=\{n\in\Bbb N\mid a\le r(n)\le b\},\\ \\V_{(a,b)}:=\{p\in\Bbb P\mid a\le r(p)\le b\},\\ \\U_{(a,b),n}:=\{m\in U_{(a,b)}\mid m\le n\},\\ \\V_{(a,b),n}:=\{m\in V_{(a,b)}\mid m\le n\},\\ \\w_{(a,b),n}:=(\#U_{(a,b),n})^{-1}\cdot\#V_{(a,b),n}\cdot\log n,\\ \\w_{(a,b)}:=\lim _{n\to\infty} w_{(a,b),n}\end{cases}$
 
:$\begin{cases} U_{(a,b)}:=\{n\in\Bbb N\mid a\le r(n)\le b\},\\ \\V_{(a,b)}:=\{p\in\Bbb P\mid a\le r(p)\le b\},\\ \\U_{(a,b),n}:=\{m\in U_{(a,b)}\mid m\le n\},\\ \\V_{(a,b),n}:=\{m\in V_{(a,b)}\mid m\le n\},\\ \\w_{(a,b),n}:=(\#U_{(a,b),n})^{-1}\cdot\#V_{(a,b),n}\cdot\log n,\\ \\w_{(a,b)}:=\lim _{n\to\infty} w_{(a,b),n}\end{cases}$
 
::Guess $2$: $\forall (a,b)\subset [0.1,1),\,w_{(a,b)}=0.9^{-1}\cdot (b-a)$.
 
::Guess $2$: $\forall (a,b)\subset [0.1,1),\,w_{(a,b)}=0.9^{-1}\cdot (b-a)$.
:::[https://math.stackexchange.com/questions/2683513/an-extension-of-prime-number-theorem/2683561#2683561 Answer] given by [https://math.stackexchange.com/users/82961/peter $@$Peter]: Imagine a very large number $N$ and consider the range $[10^N,10^{N+1}]$. The natural logarithms of $10^N$ and $10^{N+1}$ only differ by $\ln(10)\approx 2.3$ Hence the reciprocals of the logarithms of all primes in this range virtually coincicde. Because of the approximation $$\int_a^b \frac{1}{\ln(x)}dx$$ for the number of primes in the range $[a,b]$ the number of primes is approximately the length of the interval divided by $\frac{1}{\ln(10^N)}$, so is approximately equally distributed. Hence your conjecture is true.
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:::[https://math.stackexchange.com/questions/2683513/an-extension-of-prime-number-theorem/2683561#2683561 Answer] given by [https://math.stackexchange.com/users/82961/peter $@$Peter] from stackexchange site: Imagine a very large number $N$ and consider the range $[10^N,10^{N+1}]$. The natural logarithms of $10^N$ and $10^{N+1}$ only differ by $\ln(10)\approx 2.3$ Hence the reciprocals of the logarithms of all primes in this range virtually coincicde. Because of the approximation $$\int_a^b \frac{1}{\ln(x)}dx$$ for the number of primes in the range $[a,b]$ the number of primes is approximately the length of the interval divided by $\frac{1}{\ln(10^N)}$, so is approximately equally distributed. Hence your conjecture is true.
 
:::Benfords law seems to contradict this result , but this only applies to sequences producing primes as the Mersenne primes and not if the primes are chosen randomly in the range above.
 
:::Benfords law seems to contradict this result , but this only applies to sequences producing primes as the Mersenne primes and not if the primes are chosen randomly in the range above.
  
  
'''Obviously''' $(S_1,\lt _1)$ is a well-ordering set with order relation $\lt _1$ as: $\forall i,n,k\in\Bbb N$ if $p_n$ be $n$-th prime number, relation $\lt _1$ is defined with: $w_i(p_n)\lt _1w_i(p_{n+k})\lt _1w_{i+1}(p_n)$ or $$0.2\lt _10.3\lt _10.5\lt _10.7\lt _10.11\lt _10.13\lt _10.17\lt _1...0.02\lt _10.03\lt _10.05\lt _10.07\lt _10.011\lt _1$$ $$0.013\lt _10.017\lt _1...0.002\lt _10.003\lt _10.005\lt _10.007\lt _10.0011\lt _10.0013\lt _10.0017\lt _1...$$ and consequently $(S_1\times S_1,\lt _2)$ will be another well-ordering set with dictionary order relation $\lt _2$ as: $\forall (a,b),(c,d)\in S_1\times S_1,\, (a,b)\lt _2 (c,d)$ if $a\lt _1 c,$ or if $a=c$ and $b\lt _1d$, and assume $M=S_1\times S_1$ is a topological space with order topology induced by $(S_1\times S_1,\lt _2)$.
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'''Theorem''' $5$: Let $t_n:\Bbb N\to\Bbb N\setminus\{n\in\Bbb N: 10\mid n\}$ is a surjective strictly monotonically increasing sequence now $\{t_n\}_{n\in\Bbb N}$ is a cyclic group with: $\begin{cases} e=1\\ t_n^{-1}=t_{n^{-1}}\quad\text{that}\quad n\star n^{-1}=1\\ t_n\star _tt_m=t_{n\star m}\end{cases}$
:Question $5$: Does there exist any well known topology on $(0,1)\times (0,1)$ such that $S_1\times S_1$ with subspace topology be homeomorph to the topological space $M$, and is $(S_1\times S_1,\star _{S_1\times S_1})$ a topological group under topology of $M$?
+
 
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that $(\{t_n\}_{n\in\Bbb N},\star _t)=\langle 2\rangle=\langle 3\rangle$ and let $E:=\bigcup _{k\in\Bbb N} w_k(\Bbb N\setminus\{n\in\Bbb N: 10\mid n\})$ so $(E,\star _E)$ is an Abelian group with $\forall m,n\in\Bbb N,$ $\forall a,b\in\Bbb N\setminus\{n\in\Bbb N: 10\mid n\}$: $\,\,\begin{cases} e=0.1\\ w_n(a)^{-1}=w_{n^{-1}}(a^{-1})\quad\text{that}\quad n\star n^{-1}=1,\, a\star _tb=1\\ w_n(a)\star _Ew_m(b)=w_{n\star m}(a\star _tb)\end{cases}$
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 +
that $\langle 0.01,0.2\rangle=E\simeq\Bbb Z\oplus\Bbb Z$
 +
 
  
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'''now''' assume $(S\times S)\oplus E$ is external direct sum of the groups $S\times S$ and $E$ with $e=(0.2,0.2,0.1)$ and $\langle (0.02,0.2,0.1),(0.2,0.02,0.1),(0.3,0.2,0.1),(0.2,0.3,0.1),(0.2,0.2,0.01),(0.2,0.2,0.2)\rangle=$ $(S\times S)\oplus E\simeq\Bbb Z\oplus\Bbb Z\oplus\Bbb Z\oplus\Bbb Z\oplus\Bbb Z\oplus\Bbb Z$.
  
Let $t_n:\Bbb N\to\Bbb N\setminus\{n\in\Bbb N\mid 10\mid n\}$ is a surjective sequence that $\forall n\in\Bbb N,\,\,t_n\lt t_{n+1}$ now $\{t_n\}_{n\in\Bbb N}$ is a cyclic group with: $\begin{cases} e=1\\ t_n^{-1}=t_{n^{-1}}\quad\text{that}\quad n\star n^{-1}=1\\ t_n\star _tt_m=t_{n\star m}\end{cases}$
 
  
that $(\{t_n\}_{n\in\Bbb N},\star _t)=\langle 2\rangle=\langle 3\rangle$ and $E:=\bigcup _{k\in\Bbb N} w_k(\Bbb N\setminus\{n\in\Bbb N\mid 10\mid n\})$ is an Abelian group with $\forall m,n\in\Bbb N\,\,\forall a,b\in\Bbb N\setminus\{n\in\Bbb N\mid 10\mid n\}$: $\begin{cases} e=0.1\\ w_n(a)^{-1}=w_{n^{-1}}(a^{-1})\quad\text{that}\quad n\star n^{-1}=1,\, a\star _tb=1\\ w_n(a)\star _Ew_m(b)=w_{n\star m}(a\star _tb)\end{cases}$
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'''Theorem''' $6$: $(S,\lt _1)$ is a well-ordering set with order relation $\lt _1$ as: $\forall i,n,k\in\Bbb N$ if $p_n$ be $n$-th prime number, relation $\lt _1$ is defined with: $w_i(p_n)\lt _1w_i(p_{n+k})\lt _1w_{i+1}(p_n)$ or $$0.2\lt _10.3\lt _10.5\lt _10.7\lt _10.11\lt _10.13\lt _10.17\lt _1...0.02\lt _10.03\lt _10.05\lt _10.07\lt _10.011\lt _1$$ $$0.013\lt _10.017\lt _1...0.002\lt _10.003\lt _10.005\lt _10.007\lt _10.0011\lt _10.0013\lt _10.0017\lt _1...$$
  
that $\langle 0.01,0.2\rangle=E\cong\Bbb Z\oplus\Bbb Z$
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and $(E,\lt _2)$ is another well ordering set with order relation $\lt _2$ as: $\forall i,n,k\in\Bbb N$ that $10\nmid n,\, 10\nmid n+k,$ $w_i(n)\lt _2w_i(n+k)\lt _2w_{i+1}(n)$ or $$0.1\lt _2 0.2\lt _2 0.3\lt _2 ...0.9\lt _2 0.11\lt _2 0.12\lt _2 ...0.19\lt _2 0.21\lt _2 ...0.01\lt _2 0.02\lt _2 0.03\lt _2 ...0.09$$ $$\lt _2 0.011\lt _2 0.012\lt _2 ...0.019\lt _2 0.021\lt _2 ...0.001\lt _2 0.002\lt _2 0.003\lt _2 ...0.009\lt _2 0.0011\lt _2 ...$$
  
 +
now $M:=S\times S\times E$ is a well-ordering set with order relation $\lt _3$ as: $\forall (a,b,t),(c,d,u)\in S\times S\times E,$ $(a,b,t)\lt _3(c,d,u)$ if $\,\,\begin{cases} t\lt _2u & or\\ t=u,\,\, a+b\lt _2c+d & or\\ t=u,\,\, a+b=c+d,\,\, b\lt _1 d\end{cases}$ ♦
  
now assume $(S_1\times S_1)\oplus E$ is external direct product of the groups $S_1\times S_1$ and $(E,\star _E)$ with $e=(0.2,0.2,0.1)$ and $\langle (0.02,0.2,0.1),(0.2,0.02,0.1),(0.3,0.2,0.1),(0.2,0.3,0.1),(0.2,0.2,0.01),(0.2,0.2,0.2)\rangle=$ $(S_1\times S_1)\oplus E\cong\Bbb Z\oplus\Bbb Z\oplus\Bbb Z\oplus\Bbb Z\oplus\Bbb Z\oplus\Bbb Z$.
 
  
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'''now''' assume $M$ is a topological space ('''Hausdorff space''') induced by order relation $\lt _3$.
  
and $(E,\lt _3)$ is a well ordering set with order relation $\lt _3$ as: $\forall i,n,k\in\Bbb N$ that $10\nmid n,\, 10\nmid n+k,$ $w_i(n)\lt w_i(n+k)\lt w_{i+1}(n)$ or $$0.1\lt _3 0.2\lt _3 0.3\lt _3 ...0.9\lt _3 0.11\lt _3 0.12\lt _3 ...0.19\lt _3 0.21\lt _3 ...0.01\lt _3 0.02\lt _3 0.03\lt _3 ...0.09$$ $$\lt _3 0.011\lt _3 0.012\lt _3 ...0.019\lt _3 0.021\lt _3 ...0.001\lt _3 0.002\lt _3 0.003\lt _3 ...0.009\lt _3 0.0011\lt _3 ...$$ now $E$ will be a topological space induced by $(E,\lt _3)$ and assume $D=M\times E$ is a topological space with produced topology.
 
:Question $6$: Is $(S_1\times S_1)\oplus E$ a topological group under topology of $D$?
 
  
 +
'''Question''' $3$: Is $S$ a topological group under topology induced by order relation $\lt_1$ and is $(S\times S)\oplus E$ a topological group under topology of $M$?
  
'''A Conjecture''': For each even natural number $t$ greater than $4$ and $\forall c,m\in\Bbb N\cup\{0\}$ that $10^c\mid t,\, 10^{1+c}\nmid t$, $A_m=\{(a,b)\mid a,b\in S_1,\, 10^{-1-m}\le a+b\lt 10^{-m}\}$ and if $u$ is the number of digits in $t$ then $\exists (a,b)\in A_c$ such that $t=10^{c+u}\cdot (a+b),\, 10^{c+u}\cdot a,10^{c+u}\cdot b\in\Bbb P\setminus\{2\},\, (a,b,10^{-c-u}\cdot t)\in D$.
+
 
:This conjecture is an equivalent to Goldbach's conjecture.
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'''A new version of Goldbach's conjecture''': For each even natural number $t$ greater than $4$ and $\forall c,m\in\Bbb N\cup\{0\}$ that $10^c\mid t,\, 10^{1+c}\nmid t$, $A_m=\{(a,b)\mid a,b\in S,\, 10^{-1-m}\le a+b\lt 10^{-m}\}$ and if $u$ is the number of digits in $t$ then $\exists (a,b)\in A_c$ such that $t=10^{c+u}\cdot (a+b),\, 10^{c+u}\cdot a,10^{c+u}\cdot b\in\Bbb P\setminus\{2\},\, (a,b,10^{-c-u}\cdot t)\in M$.
 +
:Using homotopy groups Goldbach's conjecture will be proved.
  
 
  Alireza Badali 08:27, 31 March 2018 (CEST)
 
  Alireza Badali 08:27, 31 March 2018 (CEST)
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=== Polignac's conjecture ===
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In previous chapter above I used an important technique by theorem $1$ for presentment of prime numbers properties as density in discussion that using prime number theorem it became applicable, anyway, but now I want perform another method for Twin prime conjecture (Polignac) in principle prime numbers properties are ubiquitous in own natural numbers.
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'''Theorem''' $1$: $(\Bbb N,\star _T)$ is a group with: $\forall m,n\in\Bbb N,$
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$\begin{cases} (12m-10)\star_T(12m-9)=1=(12m-8) \star_T(12m-5)=(12m-7) \star_T(12m-4)=\\ (12m-6) \star_T(12m-1)=(12m-3) \star_T(12m)=(12m-2) \star_T(12m+1)\\ (12m-10) \star_T(12n-10)=12m+12n-19\\ (12m-10) \star_T(12n-9)=\begin{cases} 12m-12n+1 & 12m-10\gt 12n-9\\ 12n-12m-2 & 12n-9\gt 12m-10\end{cases}\\ (12m-10) \star_T(12n-8)=12m+12n-15\\ (12m-10) \star_T(12n-7)=12m+12n-20\\ (12m-10) \star_T(12n-6)=12m+12n-11\\ (12m-10) \star_T(12n-5)=\begin{cases} 12m-12n-3 & 12m-10\gt 12n-5\\ 12n-12m+8 & 12n-5\gt 12m-10\end{cases}\\ (12m-10) \star_T(12n-4)=\begin{cases} 12m-12n-6 & 12m-10\gt 12n-4\\ 12n-12m+3 & 12n-4\gt 12m-10\end{cases}\\ (12m-10) \star_T(12n-3)=12m+12n-18\\ (12m-10) \star_T(12n-2)=\begin{cases} 12m-12n-10 & 12m-10\gt 12n-2\\ 12n-12m+11 & 12n-2\gt 12m-10\end{cases}\\ (12m-10) \star_T(12n-1)=\begin{cases} 12m-12n-7 & 12m-10\gt 12n-1\\ 12n-12m+12 & 12n-1\gt 12m-10\end{cases}\\ (12m-10) \star_T(12n)=\begin{cases} 12m-12n-8 & 12m-10\gt 12n\\ 12n-12m+7 & 12n\gt 12m-10\end{cases}\\ (12m-10) \star_T(12n+1)=12m+12n-10\\ (12m-9) \star_T(12n-9)=12m+12n-16\\ (12m-9) \star_T(12n-8)=\begin{cases} 12m-12n & 12m-9\gt 12n-8\\ 12n-12m+5 & 12n-8\gt 12m-9\end{cases}\\ (12m-9) \star_T(12n-7)=\begin{cases} 12m-12n-1 & 12m-9\gt 12n-7\\ 12n-12m+2 & 12n-7\gt 12m-9\end{cases}\\ (12m-9) \star_T(12n-6)=\begin{cases} 12m-12n-4 & 12m-9\gt 12n-6\\ 12n-12m+9 & 12n-6\gt 12m-9\end{cases}\\ (12m-9) \star_T(12n-5)=12m+12n-12\\ (12m-9) \star_T(12n-4)=12m+12n-17\\ (12m-9) \star_T(12n-3)=\begin{cases} 12m-12n-5 & 12m-9\gt 12n-3\\ 12n-12m+4 & 12n-3\gt 12m-9\end{cases}\\ (12m-9) \star_T(12n-2)=12m+12n-9\\ (12m-9) \star_T(12n-1)=12m+12n-14\\ (12m-9) \star_T(12n)=12m+12n-13\\ (12m-9)\star_T(12n+1)=\begin{cases} 12m-12n-9 & 12m-9\gt 12n+1\\ 12n-12m+6 & 12n+1\gt 12m-9\end{cases}\\ (12m-8) \star_T(12n-8)=12m+12n-11\\ (12m-8) \star_T(12n-7)=12m+12n-18\\ (12m-8) \star_T(12n-6)=12m+12n-7\\ (12m-8) \star_T(12n-5)=\begin{cases} 12m-12n+1 & 12m-8\gt 12n-5\\ 12n-12m-2 & 12n-5\gt 12m-8\end{cases}\\ (12m-8) \star_T(12n-4)=\begin{cases} 12m-12n+2 & 12m-8\gt 12n-4\\ 12n-12m-1 & 12n-4\gt 12m-8\\ 2 & m=n\end{cases}\\ (12m-8) \star_T(12n-3)=12m+12n-10\\ (12m-8) \star_T(12n-2)=\begin{cases} 12m-12n-8 & 12m-8\gt 12n-2\\ 12n-12m+7 & 12n-2\gt 12m-8\end{cases}\\ (12m-8) \star_T(12n-1)=\begin{cases} 12m-12n-3 & 12m-8\gt 12n-1\\ 12n-12m+8 & 12n-1\gt 12m-8\end{cases}\\ (12m-8) \star_T(12n)=\begin{cases} 12m-12n-6 & 12m-8\gt 12n\\ 12n-12m+3 & 12n\gt 12m-8\end{cases}\\ (12m-8) \star_T(12n+1)=12m+12n-8\\ (12m-7) \star_T(12n-7)=12m+12n-15\\ (12m-7) \star_T(12n-6)=12m+12n-10\\ (12m-7) \star_T(12n-5)=\begin{cases} 12m-12n-6 & 12m-7\gt 12n-5\\ 12n-12m+3 & 12n-5\gt 12m-7\end{cases}\\ (12m-7) \star_T(12n-4)=\begin{cases} 12m-12n+1 & 12m-7\gt 12n-4\\ 12n-12m-2 & 12n-4\gt 12m-7\end{cases}\\ (12m-7) \star_T(12n-3)=12m+12n-11\\ (12m-7) \star_T(12n-2)=\begin{cases} 12m-12n-7 & 12m-7\gt 12n-2\\ 12n-12m+12 & 12n-2\gt 12m-7\end{cases}\\ (12m-7) \star_T(12n-1)=\begin{cases} 12m-12n-8 & 12m-7\gt 12n-1\\ 12n-12m+7 & 12n-1\gt 12m-7\end{cases}\\ (12m-7) \star_T(12n)=\begin{cases} 12m-12n-3 & 12m-7\gt 12n\\ 12n-12m+8 & 12n\gt 12m-7\end{cases}\\ (12m-7) \star_T(12n+1)=12m+12n-7\\ (12m-6) \star_T(12n-6)=12m+12n-3\\ (12m-6) \star_T(12n-5)=\begin{cases} 12m-12n+5 & 12m-6\gt 12n-5\\ 12n-12m & 12n-5\gt 12m-6\\ 5 & m=n\end{cases}\\ (12m-6) \star_T(12n-4)=\begin{cases} 12m-12n+4 & 12m-6\gt 12n-4\\ 12n-12m-5 & 12n-4\gt 12m-6\\ 4 & m=n\end{cases}\\ (12m-6) \star_T(12n-3)=12m+12n-8\\ (12m-6) \star_T(12n-2)=\begin{cases} 12m-12n-6 & 12m-6\gt 12n-2\\ 12n-12m+3 & 12n-2\gt 12m-6\end{cases}\\ (12m-6) \star_T(12n-1)=\begin{cases} 12m-12n+1 & 12m-6\gt 12n-1\\ 12n-12m-2 & 12n-1\gt 12m-6\end{cases}\\ (12m-6) \star_T(12n)=\begin{cases} 12m-12n+2 & 12m-6\gt 12n\\ 12n-12m-1 & 12n\gt 12m-6\\ 2 & m=n\end{cases}\\ (12m-6) \star_T(12n+1)=12m+12n-6\\ (12m-5) \star_T(12n-5)=12m+12n-14\\ (12m-5) \star_T(12n-4)=12m+12n-13\\ (12m-5) \star_T(12n-3)=\begin{cases} 12m-12n-1 & 12m-5\gt 12n-3\\ 12n-12m+2 & 12n-3\gt 12m-5\end{cases}\\ (12m-5) \star_T(12n-2)=12m+12n-5\\ (12m-5) \star_T(12n-1)=12m+12n-4\\ (12m-5) \star_T(12n)=12m+12n-9\\ (12m-5) \star_T(12n+1)=\begin{cases} 12m-12n-5 & 12m-5\gt 12n+1\\ 12n-12m+4 & 12n+1\gt 12m-5\end{cases}\\ (12m-4) \star_T(12n-4)=12m+12n-12\\ (12m-4) \star_T(12n-3)=\begin{cases} 12m-12n & 12m-4\gt 12n-3\\ 12n-12m+5 & 12n-3\gt 12m-4\end{cases}\\ (12m-4) \star_T(12n-2)=12m+12n-4\\ (12m-4) \star_T(12n-1)=12m+12n-9\\ (12m-4) \star_T(12n)=12m+12n-14\\ (12m-4) \star_T(12n+1)=\begin{cases} 12m-12n-4 & 12m-4\gt 12n+1\\ 12n-12m+9 & 12n+1\gt 12m-4\end{cases}\\ (12m-3) \star_T(12n-3)=12m+12n-7\\ (12m-3) \star_T(12n-2)=\begin{cases} 12m-12n-3 & 12m-3\gt 12n-2\\ 12n-12m+8 & 12n-2\gt 12m-3\end{cases}\\ (12m-3) \star_T(12n-1)=\begin{cases} 12m-12n-6 & 12m-3\gt 12n-1\\ 12n-12m+3 & 12n-1\gt 12m-3\end{cases}\\ (12m-3) \star_T(12n)=\begin{cases} 12m-12n+1 & 12m-3\gt 12n\\ 12n-12m-2 & 12n\gt 12m-3\end{cases}\\ (12m-3) \star_T(12n+1)=12m+12n-3\\ (12m-2) \star_T(12n-2)=12m+12n-2\\ (12m-2) \star_T(12n-1)=12m+12n-1\\ (12m-2) \star_T(12n)=12m+12n\\ (12m-2) \star_T(12n+1)=\begin{cases} 12m-12n-2 & 12m-2\gt 12n+1\\ 12n-12m+1 & 12n+1\gt 12m-2\end{cases}\\ (12m-1) \star_T(12n-1)=12m+12n\\ (12m-1) \star_T(12n)=12m+12n-5\\ (12m-1) \star_T(12n+1)=\begin{cases} 12m-12n-1 & 12m-1\gt 12n+1\\ 12n-12m+2 & 12n+1\gt 12m-1\end{cases}\\ (12m) \star_T(12n)=12m+12n-4\\ (12m) \star_T(12n+1)=\begin{cases} 12m-12n & 12m\gt 12n+1\\ 12n-12m+5 & 12n+1\gt 12m\end{cases}\\ (12m+1) \star_T(12n+1)=12m+12n+1\end{cases}$
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that $\forall k\in\Bbb N,\,\langle 2\rangle =\langle 3\rangle =\langle (2k+1)\star _T (2k+3)\rangle=(\Bbb N,\star _T)\simeq (\Bbb Z,+)$ and $\langle (2k)\star _T(2k+2)\rangle\neq\Bbb N$ and each prime in $\langle 5\rangle$ is to form of $5+12k$ or $13+12k$, $k\in\Bbb N\cup\{0\}$ and each prime in $\langle 7\rangle$ is to form of $7+12k$ or $13+12k$, $k\in\Bbb N\cup\{0\}$ and $\langle 5\rangle\cap\langle 7\rangle=\langle 13\rangle$ and $\Bbb N=\langle 5\rangle\oplus\langle 7\rangle$ but there isn't any proper subgroup including all primes of the form $11+12k,$ $k\in\Bbb N\cup\{0\}$ (probably I have to make another better).
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:Proof:
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$$0,-1,1,-3,-2,-5,3,2,-4,6,5,4,-6,-7,7,-9,-8,-11,9,8,-10,12,11,10,-12,-13,13,-15,$$ $$1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,$$ $$-14,-17,15,14,-16,18,17,16,-18,-19,19,-21,-20,-23,21,20,-22,24,23,22,-24,...$$ $$29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,...$$
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'''Guess''' $1$: For each group on $\Bbb N$ like $(\Bbb N,\star)$ generated from algorithm above, if $p_i$ be $i$_th prime number and $x_i$ be $i$_th composite number then $\exists m\in\Bbb N,\,\forall n\in\Bbb N$ that $n\ge m$ we have: $2\star3\star5\star7...\star p_n=\prod_{i=1}^{n}p_i\gt\prod _{i=1}^{n}x_i=4\star6\star8\star9...\star x_n$
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'''Guess''' $2$: For each group on $\Bbb N$ like $(\Bbb N,\star)$ generated from algorithm above, we have: $\lim_{n\to\infty}\prod _{n=1}^{\infty}p_n,\lim_{n\to\infty}\prod _{n=1}^{\infty}x_n\in\Bbb N,\,\,(\lim_{n\to\infty}\prod _{n=1}^{\infty}p_n)\star(\lim_{n\to\infty}\prod _{n=1}^{\infty}x_n)=1$.
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now let the group $G$ be external direct sum of three copies of the group $(\Bbb N,\star _T)$, hence $G=\Bbb N\oplus\Bbb N\oplus\Bbb N$.
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'''Theorem''' $2$: $(\Bbb N\times\Bbb N\times\Bbb N,\lt _T)$ is a well ordering set with order relation $\lt _T$ as: $\forall (m_1,n_1,t_1),(m_2,n_2,t_2)\in\Bbb N\times\Bbb N\times\Bbb N,\quad (m_1,n_1,t_1)\lt _T(m_2,n_2,t_2)$ if $\begin{cases} t_1\lt t_2 & or\\ t_1=t_2,\, m_1-n_1\lt m_2-n_2 & or\\ t_1=t_2,\, m_1-n_1=m_2-n_2,\, n_1\lt n_2\end{cases}$
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and suppose $M=\Bbb N\times\Bbb N\times\Bbb N$ is a topological space ('''Hausdorff space''') induced by order relation $\lt _T$.
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'''Question''' $1$: Is $G$ a topological group with topology of $M$?
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'''Now''' regarding to the group $(\Bbb N,\star_T)$, I am planning an algebraic form of prime number theorem towards twin prime conjecture:
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Recall the statement of the prime number theorem: Let $x$ be a positive real number, and let $\pi(x)$ denote the number of primes that are less than or equal to $x$. Then the ratio $\pi(x)\cdot{\log x\over x}$ can be made arbitrarily close to $1$ by taking $x$ sufficiently large.
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Question $2$: Suppose $\pi_1(x)$ is all prime numbers of the form $4k+1$ and less than $x$ and $\pi_2(x)$ is all prime numbers of the form $4k+3$ and less than $x$. Do $\lim_{x\to\infty}\pi_1(x)\cdot{\log x\over x}=0.5=\lim_{x\to\infty}\pi_2(x)\cdot{\log x\over x}\ ?$
 +
:[https://math.stackexchange.com/questions/2769471/another-extension-of-prime-number-theorem/2769494#2769494 Answer] given by [https://math.stackexchange.com/users/174927/milo-brandt $@$Milo Brandt] from stackexchange site: Basically, for any $k$, the primes are equally distributed across the congruence classes $\langle n\rangle$ mod $k$ where $n$ and $k$ are coprime.
 +
:This result is known as the prime number theorem for arithmetic progressions. [Wikipedia](https://en.wikipedia.org/wiki/Prime_number_theorem#Prime_number_theorem_for_arithmetic_progressions) discusses it with a number of references and one can find a proof of it by Ivan Soprounov [here](http://academic.csuohio.edu/soprunov_i/pdf/primes.pdf), which makes use of the Dirichlet theorem on arithmetic progressions (which just says that $\pi_1$ and $\pi_2$ are unbounded) to prove this stronger result.
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Question $3$: For each neutral infinite subset $A$ of $\Bbb N$, does exist a cyclic group like $(\Bbb N,\star)$ such that $A$ is a maximal subgroup of $\Bbb N$?
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Question $4$: If $(\Bbb N,\star_1)$ is a cyclic group and $n\in\Bbb N$ and $A=\{a_i\mid i\in\Bbb N\}$ is a non-trivial subgroup of $\Bbb N$ then does exist another cyclic group $(\Bbb N,\star_2)$ such that $\prod _{i=1}^{\infty}a_i=a_1\star_2a_2\star_2a_3\star_2...=n$?
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Question $5$: If $(\Bbb N,\star)$ is a cyclic group and $n\in\Bbb N$ then does exist a non-trivial subset $A=\{a_i\mid i\in\Bbb N\}$ of $\Bbb P$ with $\#(\Bbb P\setminus A)=\aleph_0$ and $\prod _{i=1}^{\infty}a_i=a_1\star a_2\star a_3\star...=n$?
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Question $6$: If $(\Bbb N,\star_1)$ and $(\Bbb N,\star_2)$ are cyclic groups and $A=\{a_i\mid i\in\Bbb N\}$ is a non-trivial subgroup of $(\Bbb N,\star_1)$ and $B=A\cap\Bbb P$ then does $\prod_{i=1}^{\infty}a_i=a_1\star_2a_2\star_2a_3\star_2...\in\Bbb N$?
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Alireza Badali 12:34, 28 April 2018 (CEST)
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== Collatz conjecture ==
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The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined as follows: start with any positive integer $n$. Then each term is obtained from the previous term as follows: if the previous term is even, the next term is one half the previous term. Otherwise, the next term is $3$ times the previous term plus $1$. The conjecture is that no matter what value of $n$, the sequence will always reach $1$. The conjecture is named after German mathematician [https://en.wikipedia.org/wiki/Lothar_Collatz Lothar Collatz], who introduced the idea in $1937$, two years after receiving his doctorate. It is also known as the $3n + 1$ conjecture.
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'''Theorem''' $1$: If $(\Bbb N,\star_{\Bbb N})$ is a cyclic group with $e_{\Bbb N}=1$ & $\langle m_1\rangle=\langle m_2\rangle=(\Bbb N,\star_{\Bbb N})$ and $f:\Bbb N\to\Bbb N$ is a bijection such that $f(1)=1$ then $(\Bbb N,\star _f)$ is a cyclic group with: $e_f=1$ & $\langle f(m_1)\rangle=\langle f(m_2)\rangle=(\Bbb N,\star_f)$ & $\forall m,n\in\Bbb N,$ $f(m)\star _ff(n)=f(m\star_{\Bbb N}n)$ & $(f(n))^{-1}=f(n^{-1})$ that $n\star_{\Bbb N}n^{-1}=1$.
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I want make a group in accordance with [https://www.jasondavies.com/collatz-graph/ Collatz graph] but [https://math.stackexchange.com/users/334732/robert-frost $@$RobertFrost] from stackexchange site advised me in addition, it needs to be a torsion group because then it can be used to show convergence, meantime I like apply lines in the Euclidean plane $\Bbb R^2$ too.
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Question $1$: What is function of this sequence on to natural numbers? $1,2,4,3,6,5,10,7,14,8,16,9,18,11,22,12,24,13,26,15,30,17,34,19,38,20,40,21,42,23,46,25,50,...$ such that we begin from $1$ and then write $2$ then $2\times2$ then $3$ then $2\times3$ then ... but if $n$ is even and previously we have written $0.5n$ and then $n$ then ignore $n$ and continue and write $n+1$ and then $2n+2$ and so on for example we have $1,2,4,3,6,5,10$ so after $10$ we should write $7,14,...$ because previously we have written $3,6$.
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:[https://math.stackexchange.com/questions/2779491/what-is-function-of-this-sequence-on-to-natural-numbers/2779815#2779815 Answer] given by [https://math.stackexchange.com/users/16397/r-e-s $@$r.e.s] from stackexchange site: Following is a definition of your sequence without using recursion.
 +
:Let $S=(S_0,S_1,S_2,\ldots)$ be the increasing sequence of positive integers that are expressible as either $2^e$ or as $o_1\cdot 2^{o_2}$, where $e$ is an even nonnegative integer, $o_1>1$ is an odd positive integer and $o_2$ is an odd positive integer. Thus $$S=(1, 4, 6, 10, 14, 16, 18, 22, 24, 26, 30, 34, 38, 40,42,\ldots).$$ Let $\bar{S}$ be the complement of $S$ with respect to the positive integers; i.e., $$\bar{S}=(2, 3, 5, 7, 8, 9, 11, 12, 13, 15, 17, 19, 20, 21, 23, 25,\ldots).$$ Your sequence is then $T=(T_0,T_1,T_2,\ldots)$, where
 +
$$T_n:=\begin{cases}S_{n\over 2}&\text{ if $n$ is even}\\
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\bar{S}_{n-1\over 2}&\text{ if $n$ is odd.}
 +
\end{cases}
 +
$$
 +
:Thus $T=(1, 2, 4, 3, 6, 5, 10, 7, 14, 8, 16, 9, 18, 11, 22, 12, 24, 13, 26, 15, 30, 17, 34, 19, 38, 20, \ldots).$
 +
 +
----------------------------
 +
 +
:References:
 +
:Sequences $S,\bar{S},T$ are OEIS [A171945](http://oeis.org/A171945), [A053661](http://oeis.org/A053661), [A034701](http://oeis.org/A034701) respectively. These are all discussed in ["The vile, dopey, evil and odious game players"](https://www.sciencedirect.com/science/article/pii/S0012365X11001427).
 +
 +
--------------------------
 +
 +
:Sage code:
 +
    def is_in_S(n): return ( (n.valuation(2) % 2 == 0) and (n.is_power_of(2))  ) or ( (n.valuation(2) % 2 == 1) and not(n.is_power_of(2))  )
 +
    S = [n for n in [1..50] if is_in_S(n)]
 +
    S_ = [n for n in [1..50] if not is_in_S(n)]
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    T = []
 +
    for i in range(max(len(S),len(S_))):
 +
        if i % 2 == 0: T += [S[i/2]]
 +
        else: T += [S_[(i-1)/2]]
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    print S
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    print S_
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    print T
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 +
    [1, 4, 6, 10, 14, 16, 18, 22, 24, 26, 30, 34, 38, 40, 42, 46, 50]
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    [2, 3, 5, 7, 8, 9, 11, 12, 13, 15, 17, 19, 20, 21, 23, 25, 27, 28, 29, 31, 32, 33, 35, 36, 37, 39, 41, 43, 44, 45, 47, 48, 49]
 +
    [1, 2, 4, 3, 6, 5, 10, 7, 14, 8, 16, 9, 18, 11, 22, 12, 24, 13, 26, 15, 30, 17, 34, 19, 38, 20, 40, 21, 42, 23, 46, 25, 50]
 +
 +
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'''Theorem''' $2$: If $(\Bbb N,\star_1)$ & $(\Bbb N,\star_2)$ are cyclic groups with generators respectively $u_1$ & $v_1$ and $u_2$ & $v_2$ then $C_1=\{(m,2m)\mid m\in\Bbb N\}$ is a cyclic group with: $\begin{cases} e_{C_1}=(1,2)\\ \\\forall m,n\in\Bbb N,\,(m,2m)\star_{C_1}(n,2n)=(m\star_1n,2(m\star_1n))\\ (m,2m)^{-1}=(m^{-1},2\times m^{-1})\qquad\text{that}\quad m\star_1m^{-1}=1\\ \\C_1=\langle(u_1,2u_1)\rangle=\langle(v_1,2v_1)\rangle\end{cases}$ and $C_2=\{(3m-1,2m-1)\mid m\in\Bbb N\}$ is a cyclic group with: $\begin{cases} e_{C_2}=(2,1)\\ \\\forall m,n\in\Bbb N,\,(3m-1,2m-1)\star_{C_2}(3n-1,2n-1)=(3(m\star_2n)-1,2(m\star_2n)-1)\\ (3m-1,2m-1)^{-1}=(3\times m^{-1}-1,2\times m^{-1}-1)\qquad\text{that}\quad m\star_2 m^{-1}=1\\ \\C_2=\langle(3u_2-1,2u_2-1)\rangle=\langle(3v_2-1,2v_2-1)\rangle\end{cases}$•
 +
:And let $C:=C_1\oplus C_2$ be external direct sum of the groups $C_1$ & $C_2$. '''Question''' $2$: What are maximal subgroups of $C$?
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'''Theorem''' $3$: If $(\Bbb N,\star)$ is a cyclic group with generators $u,v$ and identity element $e=1$ and $f:\Bbb N\to\Bbb R$ is an injection then $(f(\Bbb N),\star_f)$ is a cyclic group with generators $f(u),f(v)$ and identity element $e_f=f(1)$ and operation law: $\forall m,n\in\Bbb N,$ $f(m)\star_ff(n)=f(m\star n)$ and inverse law: $\forall n\in\Bbb N,$ $(f(n))^{-1}=f(n^{-1})$ that $n\star n^{-1}=1$.
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'''Suppose''' $\forall m,n\in\Bbb N,\qquad$ $\begin{cases} m\star 1=m\\ (4m)\star (4m-2)=1=(4m+1)\star (4m-1)\\ (4m-2)\star (4n-2)=4m+4n-5\\ (4m-2)\star (4n-1)=4m+4n-2\\ (4m-2)\star (4n)=\begin{cases} 4m-4n-1 & 4m-2\gt 4n\\ 4n-4m+1 & 4n\gt 4m-2\\ 3 & m=n+1\end{cases}\\ (4m-2)\star (4n+1)=\begin{cases} 4m-4n-2 & 4m-2\gt 4n+1\\ 4n-4m+4 & 4n+1\gt 4m-2\end{cases}\\ (4m-1)\star (4n-1)=4m+4n-1\\ (4m-1)\star (4n)=\begin{cases} 4m-4n+2 & 4m-1\gt 4n\\ 4n-4m & 4n\gt 4m-1\\ 2 & m=n\end{cases}\\ (4m-1)\star (4n+1)=\begin{cases} 4m-4n-1 & 4m-1\gt 4n+1\\ 4n-4m+1 & 4n+1\gt 4m-1\\ 3 & m=n+1\end{cases}\\ (4m)\star (4n)=4m+4n-3\\ (4m)\star (4n+1)=4m+4n\\ (4m+1)\star  (4n+1)=4m+4n+1\\ \Bbb N=\langle 2\rangle=\langle 4\rangle\end{cases}$
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and let $C_1=\{(m,2m)\mid m\in\Bbb N\}$ is a cyclic group with: $\begin{cases} e_{C_1}=(1,2)\\ \\\forall m,n\in\Bbb N,\,(m,2m)\star_{C_1}(n,2n)=(m\star n,2(m\star n))\\ (m,2m)^{-1}=(m^{-1},2\times m^{-1})\qquad\text{that}\quad m\star m^{-1}=1\\ \\C_1=\langle(2,4)\rangle=\langle(4,8)\rangle\end{cases}$
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 +
and $C_2=\{(3m-1,2m-1)\mid m\in\Bbb N\}$ is a cyclic group with: $\begin{cases} e_{C_2}=(2,1)\\ \\\forall m,n\in\Bbb N,\, (3m-1,2m-1)\star_{C_2}(3n-1,2n-1)=(3(m\star n)-1,2(m\star n)-1)\\ (3m-1,2m-1)^{-1}=(3\times m^{-1}-1,2\times m^{-1}-1)\qquad\text{that}\quad m\star m^{-1}=1\\ \\C_2=\langle(5,3)\rangle=\langle(11,7)\rangle\end{cases}$.
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and let $C:=C_1\oplus C_2$ be external direct sum of the groups $C_1$ & $C_2$, '''Question''' $3$: What are maximal subgroups of $C$?
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Alireza Badali 10:02, 12 May 2018 (CEST)
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== Erdős–Straus conjecture ==
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'''Theorem''': If $(\Bbb N,\star)$ is a cyclic group with identity element $e=1$ and generators $a,b$ then $E=\{({1\over x},{1\over y},{1\over z},{-4\over n+1},n)\mid x,y,z,n\in\Bbb N\}$ is an Abelian group with: $\forall x,y,z,n,x_1,y_1,z_1,n_1\in\Bbb N$ $\begin{cases} e_E=(1,1,1,-2,1)=({1\over 1},{1\over 1},{1\over 1},{-4\over 1+1},1)\\ \\({1\over x},{1\over y},{1\over z},{-4\over n+1},n)^{-1}=({1\over x^{-1}},{1\over y^{-1}},{1\over z^{-1}},\frac{-4}{n^{-1}+1},n^{-1})\quad\text{that}\\ x\star x^{-1}=1=y\star y^{-1}=z\star z^{-1}=n\star n^{-1}\\ \\({1\over x},{1\over y},{1\over z},\frac{-4}{n+1},n)\star_E({1\over x_1},{1\over y_1},{1\over z_1},\frac{-4}{n_1+1},n_1)=(\frac{1}{x\star x_1},\frac{1}{y\star y_1},\frac{1}{z\star z_1},\frac{-4}{n\star {n_1}+1},n\star n_1)\\ \\E=\langle({1\over a},1,1,-2,1),(1,{1\over a},1,-2,1),(1,1,{1\over a},-2,1),(1,1,1,\frac{-4}{a+1},1),(1,1,1,-2,a)\rangle=\\ \langle({1\over b},1,1,-2,1),(1,{1\over b},1,-2,1),(1,1,{1\over b},-2,1),(1,1,1,\frac{-4}{b+1},1),(1,1,1,-2,b)\rangle\end{cases}$•
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 +
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Let $(\Bbb N,\star)$ is a cyclic group with: $\begin{cases} n\star 1=n\\ (2n)\star (2n+1)=1\\ (2n)\star (2m)=2n+2m\\ (2n+1)\star (2m+1)=2n+2m+1\\ (2n)\star (2m+1)=\begin{cases} 2m-2n+1 & 2m+1\gt 2n\\ 2n-2m & 2n\gt 2m+1\end{cases}\\\Bbb N=\langle 2\rangle =\langle 3\rangle \end{cases}$
 +
:Question: Is $E_0=\{({1\over x},{1\over y},{1\over z},\frac{-4}{n+1},n)\mid x,y,z,n\in\Bbb N,\, {1\over x}+{1\over y}+{1\over z}-{4\over n+1}=0\}$ a subgroup of $E$?
 +
 +
Alireza Badali 17:34, 25 May 2018 (CEST)
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== Landaus forth problem ==
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Friedlander–Iwaniec theorem: there are infinitely many prime numbers of the form $a^2+b^4$.
 +
:I want use this theorem for [https://en.wikipedia.org/wiki/Landau%27s_problems Landaus forth problem] but prime numbers properties have been applied for Friedlander–Iwaniec theorem hence no need to prime number theorem or its other forms or extensions.
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'''Theorem''': If $(\Bbb N,\star)$ is a cyclic group with identity element $e=1$ and generators $u,v$ then $F=\{(a^2,b^4)\mid a,b\in\Bbb N\}$ is a group with: $\forall a,b,c,d\in\Bbb N\,$ $\begin{cases} e_F=(1,1)\\ (a^2,b^4)\star_F(c^2,d^4)=((a\star c)^2,(b\star d)^4)\\ (a^2,b^4)^{-1}=((a^{-1})^2,(b^{-1})^4)\qquad\text{that}\quad a\star a^{-1}=1=b\star b^{-1}\\ F=\langle (1,u^4),(u^2,1)\rangle=\langle (1,v^4),(v^2,1)\rangle\end{cases}$
 +
 +
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'''now''' let $H=\langle\{(a^2,b^4)\mid a,b\in\Bbb N,\,b\neq 1\}\rangle$ and $G=F/H$ is quotient group of $F$ by $H$. ($G$ is a group including prime numbers properties only of the form $1+n^2$.)
 +
 +
and also $L=\{1+n^2\mid n\in\Bbb N\}$ is a cyclic group with: $\forall m,n\in\Bbb N$ $\begin{cases} e_L=2=1+1^2\\ (1+n^2)\star_L(1+m^2)=1+(n\star m)^2\\ (1+n^2)^{-1}=1+(n^{-1})^2\quad\text{that}\;n\star n^{-1}=1\\ L=\langle 1+u^2\rangle=\langle 1+v^2\rangle\end{cases}$
 +
 +
but on the other hand we have: $L\simeq G$ hence we can apply $L$ instead $G$ of course since we are working on natural numbers generally we could consider from the beginning the group $L$ without involvement with the group $G$ anyhow.
 +
:Question $1$: For each neutral cyclic group on $\Bbb N$ then what are maximal subgroups of $L$?
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 +
 +
'''Guess''' $1$: For each cyclic group structure on $\Bbb N$ like $(\Bbb N,\star)$ then for each non-trivial subgroup of $\Bbb N$ like $T$ we have $T\cap\Bbb P\neq\emptyset$.
 +
:I think this guess must be proved via prime number theorem.
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 +
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'''For''' each neutral cyclic group on $\Bbb N$ if $L\cap\Bbb P=\{1+n_1^2,1+n_2^2,...,1+n_k^2\},\,k\in\Bbb N$ and if $A=\bigcap _{i=1}^k\langle 1+n_i^2\rangle$ so $\exists m\in\Bbb N$ that $A=\langle 1+m^2\rangle$ & $m\neq n_i$ for $i=1,2,3,...,k$ (intelligibly $k\gt1$) so we have: $A\cap\Bbb P=\emptyset$.
 +
:Question $2$: Is $A$ only unique greatest subgroup of $L$ such that $A\cap\Bbb P=\emptyset$?
 +
 +
Alireza Badali 16:49, 28 May 2018 (CEST)
 +
 +
== Lemoine's conjecture ==
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 +
'''Theorem''': If $(\Bbb N,\star)$ is a cyclic group with identity element $e=1$ & generators $u,v$ then $L=\{(p_{n_1},p_{n_2},p_{n_3},-2n-5)\mid n,n_1,n_2,n_3\in\Bbb N,\,p_{n_i}$ is $n_i$_th prime for $i=1,2,3\}$ is an Abelian group with: $\forall n_1,n_2,n_3,n,m_1,m_2,m_3,m\in\Bbb N$ $\begin{cases} e_L=(2,2,2,-7)=(2,2,2,-2\times 1-5)\\ \\(p_{n_1},p_{n_2},p_{n_3},-2n-5)\star_L(p_{m_1},p_{m_2},p_{m_3},-2m-5)=(p_{n_1\star m_1},p_{n_2\star m_2},p_{n_3\star m_3},-2\times(n\star m)-5)\\ \\(p_{n_1},p_{n_2},p_{n_3},-2n-5)^{-1}=(p_{n_1^{-1}},p_{n^{-1}_2},p_{n_3^{-1}},-2\times n^{-1}-5)\quad\text{that}\\ n_1\star n_1^{-1}=1=n_2\star n_2^{-1}=n_3\star n_3^{-1}=n\star n^{-1}\\ \\L=\langle(p_u,2,2,-7),(2,p_u,2,-7),(2,2,p_u,-7),(2,2,2,-2u-5)\rangle=\\\langle(p_v,2,2,-7),(2,p_v,2,-7),(2,2,p_v,-7),(2,2,2,-2v-5)\rangle\end{cases}$•
 +
 +
 +
'''Theorem''': $\forall n\in\Bbb N,\,\exists (p_{m_1},p_{m_2},p_{m_3},-2n-5)\in(L,\star_L)$ such that $p_{m_1}+p_{m_2}+p_{m_3}-2n-5=0$.
 +
:Proof using Goldbach's weak conjecture.
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'''Question''': Is $L_0=\{(p_{m_1},p_{m_2},p_{m_2},-2n-5)\mid\forall m_1,m_2\in\Bbb N,\,\exists n\in\Bbb N,$ such that $p_{m_1}+2p_{m_2}-2n-5=0\}$ a subgroup of $L$?
 +
 +
Alireza Badali 19:30, 3 June 2018 (CEST)
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 +
== Primes in beatty sequences ==
 +
 +
How can we understand $\infty$? we humans only can think on natural numbers and other issues are only theorizing, algebraic theories can be some features for this aim.
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 +
 +
[http://oeis.org/A184774 Conjecture]: If $r$ is an irrational number and $1\lt r\lt 2$, then there are infinitely many primes in the set $L=\{\text{floor}(n\cdot r)\mid n\in\Bbb N\}$.
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'''Theorem''' $1$: If $(\Bbb N,\star)$ is a cyclic group with identity element $e=1$ & generators $u,v$ and $r\in[1,2]\setminus\Bbb Q$ then $L=\{\lfloor n\cdot r\rfloor\mid n\in\Bbb N\}$ is another cyclic group with: $\forall m,n\in\Bbb N$ $\begin{cases} e_L=1\\ \lfloor n\cdot r\rfloor\star_L\lfloor m\cdot r\rfloor=\lfloor (n\star m)\cdot r\rfloor\\ (\lfloor n\cdot r\rfloor)^{-1}=\lfloor n^{-1}\cdot r\rfloor\qquad\text{that}\quad n\star n^{-1}=1\\ L=\langle\lfloor u\cdot r\rfloor\rangle=\langle\lfloor v\cdot r\rfloor\rangle\end{cases}$.
 +
:Guess $1$: $\prod_{n=1}^{\infty}\lfloor n\cdot r\rfloor=\lfloor 1\cdot r\rfloor\star\lfloor 2\cdot r\rfloor\star\lfloor 3\cdot r\rfloor\star...\in\Bbb N$.
 +
 +
 +
The conjecture generalized: if $r$ is a positive irrational number and $h$ is a real number, then each of the sets $\{\text{floor}(n\cdot r+h)\mid n\in\Bbb N\}$, $\{\text{round}(n\cdot r+h)\mid n\in\Bbb N\}$, and $\{\text{ceiling}(n\cdot r+h)\mid n\in\Bbb N\}$ contains infinitely many primes.
 +
 +
 +
'''Theorem''' $2$: If $(\Bbb N,\star)$ is a cyclic group with identity element $e=1$ & generators $u,v$ & $r$ is a positive irrational number & $h\in\Bbb R$ then $G=\{n\cdot r+h\mid n\in\Bbb N\}$ is another cyclic group with: $\forall m,n\in\Bbb N$ $\begin{cases} e_G=\lfloor r+h\rfloor\\ \lfloor n\cdot r+h\rfloor\star_G\lfloor m\cdot r+h\rfloor=\lfloor (n\star m)\cdot r+h\rfloor\\ (\lfloor n\cdot r+h\rfloor)^{-1}=\lfloor n^{-1}\cdot r+h\rfloor\qquad\text{that}\quad n\star n^{-1}=1\\ L=\langle\lfloor u\cdot r+h\rfloor\rangle=\langle\lfloor v\cdot r+h\rfloor\rangle\end{cases}$.
 +
:Guess $2$: $\prod_{n=k}^{\infty}\lfloor n\cdot r+h\rfloor=\lfloor k\cdot r+h\rfloor\star\lfloor (k+1)\cdot r+h\rfloor\star\lfloor (k+2)\cdot r+h\rfloor\star...\in\Bbb N$ in which $\lfloor k\cdot r+h\rfloor\in\Bbb N$ & $\lfloor (k-1)\cdot r+h\rfloor\lt1$.
 +
 +
Alireza Badali 19:09, 7 June 2018 (CEST)
 +
 +
== Comments ==
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 +
Please just insert your comment here! Alireza Badali 20:47, 15 April 2018 (CEST)

Revision as of 15:19, 9 June 2018

Whole my previous notes is visible in the revision as of 18:42, 13 April 2018 Alireza Badali 21:52, 13 April 2018 (CEST)

$\mathscr B$ $theory$ (algebraic topological analytical number theory)

Logarithm function as an inverse of the function $f:\Bbb N\to\Bbb R,\,f(n)=a^n,\,a\in\Bbb R$ has prime numbers properties because in usual definition of prime numbers multiplication operation is a point meantime we have $a^n=a\times a\times ...a,$ $n$ times, hence prime number theorem or its extensions or some other forms is applied in $B$ theory for solving problems on prime numbers exclusively and not all natural numbers.

Goldbach's conjecture

Lemma: For each subinterval $(a,b)$ of $[0.1,1),\,\exists m\in \Bbb N$ that $\forall k\in \Bbb N$ with $k\ge m$ then $\exists t\in (a,b)$ that $t\cdot 10^k\in \Bbb P$.

Proof given by @Adayah from stackexchange site: Without loss of generality (by passing to a smaller subinterval) we can assume that $(a, b) = \left( \frac{s}{10^r}, \frac{t}{10^r} \right)$, where $s, t, r$ are positive integers and $s < t$. Let $\alpha = \frac{t}{s}$.
The statement is now equivalent to saying that there is $m \in \mathbb{N}$ such that for every $k \geqslant m$ there is a prime $p$ with $10^{k-r} \cdot s < p < 10^{k-r} \cdot t$.
We will prove a stronger statement: there is $m \in \mathbb{N}$ such that for every $n \geqslant m$ there is a prime $p$ such that $n < p < \alpha \cdot n$. By taking a little smaller $\alpha$ we can relax the restriction to $n < p \leqslant \alpha \cdot n$.
Now comes the prime number theorem: $$\lim_{n \to \infty} \frac{\pi(n)}{\frac{n}{\log n}} = 1$$
where $\pi(n) = \# \{ p \leqslant n : p$ is prime$\}.$ By the above we have $$\frac{\pi(\alpha n)}{\pi(n)} \sim \frac{\frac{\alpha n}{\log(\alpha n)}}{\frac{n}{\log(n)}} = \alpha \cdot \frac{\log n}{\log(\alpha n)} \xrightarrow{n \to \infty} \alpha$$
hence $\displaystyle \lim_{n \to \infty} \frac{\pi(\alpha n)}{\pi(n)} = \alpha$. So there is $m \in \mathbb{N}$ such that $\pi(\alpha n) > \pi(n)$ whenever $n \geqslant m$, which means there is a prime $p$ such that $n < p \leqslant \alpha \cdot n$, and that is what we wanted♦


Now we can define function $f:\{(c,d)\mid (c,d)\subseteq [0.01,0.1)\}\to\Bbb N$ that $f((c,d))$ is the least $n\in\Bbb N$ that $\exists t\in(c,d),\,\exists k\in\Bbb N$ that $p_n=t\cdot 10^{k+1}$ that $p_n$ is $n$_th prime and $\forall m\ge f((c,d))\,\,\exists u\in (c,d)$ that $u\cdot 10^{m+1}\in\Bbb P$

and $g:(0,0.09)\cap (\bigcup _{k\in\Bbb N} r_k(\Bbb N))\to\Bbb N,$ is a function by $\forall\epsilon\in (0,0.09)\cap (\bigcup _{k\in\Bbb N} r_k(\Bbb N))$ $g(\epsilon)=max(\{f((c,d))\mid d-c=\epsilon,$ $(c,d)\subseteq [0.01,0.1)\})$.

Guess $1$: $g$ isn't an injective function.

Question $1$: Assuming guess $1$, let $[a,a]:=\{a\}$ and $\forall n\in\Bbb N,\, h_n$ is the least subinterval of $[0.01,0.1)$ like $[a,b]$ in terms of size of $b-a$ such that $\{\epsilon\in (0,0.09)\cap (\bigcup _{k\in\Bbb N} r_k(\Bbb N))\mid g(\epsilon)=n\}\subsetneq h_n$ and obviously $g(a)=n=g(b)$ now the question is $\forall n,m\in\Bbb N$ that $m\neq n$ is $h_n\cap h_m=\emptyset$?

Guidance given by @reuns from stackexchange site:
  • For $n \in \mathbb{N}$ then $r(n) = 10^{-\lceil \log_{10}(n) \rceil} n$, ie. $r(19) = 0.19$. We look at the image by $r$ of the primes $\mathbb{P}$.
  • Let $F((c,d)) = \min \{ p \in \mathbb{P}, r(p) \in (c,d)\}$ and $f((c,d)) = \pi(F(c,d))= \min \{ n, r(p_n) \in (c,d)\}$ ($\pi$ is the prime counting function)
  • If you set $g(\epsilon) = \max_a \{ f((a,a+\epsilon))\}$ then try seing how $g(\epsilon)$ is constant on some intervals defined in term of the prime gap $g(p) = -p+\min \{ q \in \mathbb{P}, q > p\}$ and things like $ \max \{ g(p), p > 10^i, p+g(p) < 10^{i+1}\}$
Another guidance: The affirmative answer is given by Liouville's theorem on approximation of algebraic numbers.


Suppose $r:\Bbb N\to (0,1)$ is a function given by $r(n)$ is obtained by putting a point at the beginning of $n$ instance $r(34880)=0.34880$ and similarly consider $\forall k\in\Bbb N,\, w_k:\Bbb N\to (0,1)$ is a function given by $\forall n\in\Bbb N,$ $w_k(n)=10^{1-k}\cdot r(n)$ and let $S=\bigcup _{k\in\Bbb N}w_k(\Bbb P)$.

Theorem $1$: $r(\Bbb P)$ is dense in the interval $[0.1,1]$. (proof using lemma above)

Regarding to expression form of Goldbach's conjecture, by using this theorem, I wanted enmesh prime numbers properties (prime number theorem should be used for proving this theorem and there is no way except using prime number theorem to prove this density because there is no deference between a prime $p$ and its image $r(p)$ other than a sign or a mark as a point for instance $59$ & $0.59$.) towards Goldbach hence I planned this method.
A corollary: For each natural number like $a=a_1a_2a_3...a_k$ that $a_j$ is $j$_th digit for $j=1,2,3,...,k$, there is a natural number like $b=b_1b_2b_3...b_r$ such that the number $c=a_1a_2a_3...a_kb_1b_2b_3...b_r$ is a prime number.

Theorem $2$: $S$ is dense in the interval $[0,1]$ and $S\times S$ is dense in the $[0,1]\times [0,1]$.


An algorithm that makes new cyclic groups on $\Bbb N$:

Let $\Bbb N$ be that group and at first write integers as a sequence with starting from $0$ and let identity element $e=1$ be corresponding with $0$ and two generating elements $m$ & $n$ be corresponding with $1$ & $-1$ so we have $\Bbb N=\langle m\rangle=\langle n\rangle$ for instance: $$0,1,2,-1,-2,3,4,-3,-4,5,6,-5,-6,7,8,-7,-8,9,10,-9,-10,11,12,-11,-12,...$$ $$1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,...$$ then regarding to the sequence find a rotation number that for this sequence is $4$ and hence equations should be written with module $4$, then consider $4m-2,4m-1,4m,4m+1$ that the last should be $km+1$ and initial be $km+(2-k)$ otherwise equations won't match with definitions of members inverse, and make a table of products of those $k$ elements but during writing equations pay attention if an equation is right for given numbers it will be right generally for other numbers and of course if integers corresponding with two member don't have same signs then product will be a piecewise-defined function for example $12\star _u 15=6$ or $(4\times 3)\star _u (4\times 4-1)=6$ because $(-5)+8=3$ & $-5\to 12,\,\, 8\to 15,\,\, 3\to 6,$ that implies $(4n)\star _u (4m-1)=4m-4n+2$ if $4m-1\gt 4n$ of course it's better at first members inverse be defined for example since $(-9)+9=0$ & $0\to 1,\,\, -9\to 20,\,\, 9\to 18$ so $20\star _u 18=1$, that implies $(4m)\star _u (4m-2)=1$, and with a little bit addition and multiplication all equations will be obtained simply that for this example is:

$\begin{cases} m\star _u 1=m\\ (4m)\star _u (4m-2)=1=(4m+1)\star _u (4m-1)\\ (4m-2)\star _u (4n-2)=4m+4n-5\\ (4m-2)\star _u (4n-1)=4m+4n-2\\ (4m-2)\star _u (4n)=\begin{cases} 4m-4n-1 & 4m-2\gt 4n\\ 4n-4m+1 & 4n\gt 4m-2\\ 3 & m=n+1\end{cases}\\ (4m-2)\star _u (4n+1)=\begin{cases} 4m-4n-2 & 4m-2\gt 4n+1\\ 4n-4m+4 & 4n+1\gt 4m-2\end{cases}\\ (4m-1)\star _u (4n-1)=4m+4n-1\\ (4m-1)\star _u (4n)=\begin{cases} 4m-4n+2 & 4m-1\gt 4n\\ 4n-4m & 4n\gt 4m-1\\ 2 & m=n\end{cases}\\ (4m-1)\star _u (4n+1)=\begin{cases} 4m-4n-1 & 4m-1\gt 4n+1\\ 4n-4m+1 & 4n+1\gt 4m-1\\ 3 & m=n+1\end{cases}\\ (4m)\star _u (4n)=4m+4n-3\\ (4m)\star _u (4n+1)=4m+4n\\ (4m+1)\star _u (4n+1)=4m+4n+1\\ \Bbb N=\langle 2\rangle=\langle 4\rangle\end{cases}$


Problem $1$: By using matrices rewrite operation of every group on $\Bbb N$.


Assume $\forall m,n\in\Bbb N$: $\begin{cases} n\star 1=n\\ (2n)\star (2n+1)=1\\ (2n)\star (2m)=2n+2m\\ (2n+1)\star (2m+1)=2n+2m+1\\ (2n)\star (2m+1)=\begin{cases} 2m-2n+1 & 2m+1\gt 2n\\ 2n-2m & 2n\gt 2m+1\end{cases}\end{cases}$

and $p_n\star _1p_m=p_{n\star m}$ that $p_n$ is $n$_th prime with $e=p_1=2$, obviously $(\Bbb N,\star)$ & $(\Bbb P,\star _1)$ are groups and $\langle 2\rangle =\langle 3\rangle =(\Bbb N,\star)\simeq (\Bbb Z,+)\simeq (\Bbb P,\star _1)=\langle 3\rangle=\langle 5\rangle$.


Theorem $3$: $(S,\star _S)$ is a group as: $\forall p,q\in\Bbb P,\,\forall m,n\in\Bbb N,\,\forall w_m(p),w_n(q)\in S,$

$\begin{cases} e=0.2\\ \\(w_m(p))^{-1}=w_{m^{-1}}(p^{-1}) & m\star m^{-1}=1,\, p\star _1 p^{-1}=2\\ \\w_m(p)\star _S w_n(q)=w_{m\star n} (p\star _1 q)\end{cases}$

hence $\langle 0.02,0.3\rangle=(S,\star _S)\simeq\Bbb Z\oplus\Bbb Z$.

of course using algorithm above to generate cyclic groups on $\Bbb N$, we can impose another group structure on $\Bbb N$ and consequently on $\Bbb P$ but eventually $S$ with an operation analogous above operation $\star _S$ will be an Abelian group.


Theorem $4$: $(S\times S,\star _{S\times S})$ is a group as: $\forall m_1,n_1,m_2,n_2\in\Bbb N,\,\forall p_1,p_2,q_1,q_2\in\Bbb P,$ $\forall (w_{m_1}(p_1),w_{m_2}(p_2)),(w_{n_1}(q_1),w_{n_2}(q_2))\in S\times S,$

$\begin{cases} e=(0.2,0.2)\\ \\(w_{m_1}(p_1),w_{m_2}(p_2))^{-1}=(w_{m_1^{-1}}(p_1^{-1}),w_{m_2^{-1}}(p_2^{-1}))\\ \text{such that}\quad m_1\star m_1^{-1}=1=m_2\star m_2^{-1},\, p_1\star _1p_1^{-1}=2=p_2\star _1p_2^{-1}\\ \\(w_{m_1}(p_1),w_{m_2}(p_2))\star _{S\times S} (w_{n_1}(q_1),w_{n_2}(q_2))=(w_{m_1\star n_1} (p_1\star _1 q_1),w_{m_2\star n_2}(p_2\star _1 q_2))\end{cases}$

hence $\langle (0.02,0.2),(0.2,0.02),(0.3,0.2),(0.2,0.3)\rangle=(S\times S,\star _{S\times S})\simeq\Bbb Z\oplus\Bbb Z\oplus\Bbb Z\oplus\Bbb Z$.

of course using algorithm above to generate cyclic groups on $\Bbb N$, we can impose another group structure on $\Bbb N$ and consequently on $\Bbb P$ but eventually $S\times S$ with an operation analogous above operation $\star _{S\times S}$ will be an Abelian group.


I want make some topologies having prime numbers properties presentable in the collection of open sets, in principle when we image a prime $p$ to real numbers as $w_k(p)$ indeed we accompany prime numbers properties among real numbers which regarding to the expression form of prime number theorem for this aim we should use an important mathematical technique as logarithm function into some planned topologies: question $2$: Let $M$ be a topological space and $A,B$ are subsets of $M$ with $A\subset B$ and $A$ is dense in $B,$ since $A$ is dense in $B,$ is there some way in which a topology on $B$ may be induced other than the subspace topology? I am also interested in specialisations, for example if $M$ is Hausdorff or Euclidean. ($M=\Bbb R,\,B=[0,1],\,A=S$ or $M=\Bbb R^2,$ $B=[0,1]\times[0,1],$ $A=S\times S$)

Perhaps this technique is useful: an extension of prime number theorem: $\forall n\in\Bbb N,$ and for each subinterval $(a,b)$ of $[0.1,1),$ that $a\neq b,$
$\begin{cases} U_{(a,b)}:=\{n\in\Bbb N\mid a\le r(n)\le b\},\\ \\V_{(a,b)}:=\{p\in\Bbb P\mid a\le r(p)\le b\},\\ \\U_{(a,b),n}:=\{m\in U_{(a,b)}\mid m\le n\},\\ \\V_{(a,b),n}:=\{m\in V_{(a,b)}\mid m\le n\},\\ \\w_{(a,b),n}:=(\#U_{(a,b),n})^{-1}\cdot\#V_{(a,b),n}\cdot\log n,\\ \\w_{(a,b)}:=\lim _{n\to\infty} w_{(a,b),n}\end{cases}$ ::Guess $2$: $\forall (a,b)\subset [0.1,1),\,w_{(a,b)}=0.9^{-1}\cdot (b-a)$. :::[https://math.stackexchange.com/questions/2683513/an-extension-of-prime-number-theorem/2683561#2683561 Answer] given by [https://math.stackexchange.com/users/82961/peter $@$Peter] from stackexchange site: Imagine a very large number $N$ and consider the range $[10^N,10^{N+1}]$. The natural logarithms of $10^N$ and $10^{N+1}$ only differ by $\ln(10)\approx 2.3$ Hence the reciprocals of the logarithms of all primes in this range virtually coincicde. Because of the approximation '"`UNIQ-MathJax5-QINU`"' for the number of primes in the range $[a,b]$ the number of primes is approximately the length of the interval divided by $\frac{1}{\ln(10^N)}$, so is approximately equally distributed. Hence your conjecture is true. :::Benfords law seems to contradict this result , but this only applies to sequences producing primes as the Mersenne primes and not if the primes are chosen randomly in the range above. '''Theorem''' $5$: Let $t_n:\Bbb N\to\Bbb N\setminus\{n\in\Bbb N: 10\mid n\}$ is a surjective strictly monotonically increasing sequence now $\{t_n\}_{n\in\Bbb N}$ is a cyclic group with: $\begin{cases} e=1\\ t_n^{-1}=t_{n^{-1}}\quad\text{that}\quad n\star n^{-1}=1\\ t_n\star _tt_m=t_{n\star m}\end{cases}$ that $(\{t_n\}_{n\in\Bbb N},\star _t)=\langle 2\rangle=\langle 3\rangle$ and let $E:=\bigcup _{k\in\Bbb N} w_k(\Bbb N\setminus\{n\in\Bbb N: 10\mid n\})$ so $(E,\star _E)$ is an Abelian group with $\forall m,n\in\Bbb N,$ $\forall a,b\in\Bbb N\setminus\{n\in\Bbb N: 10\mid n\}$: $\,\,\begin{cases} e=0.1\\ w_n(a)^{-1}=w_{n^{-1}}(a^{-1})\quad\text{that}\quad n\star n^{-1}=1,\, a\star _tb=1\\ w_n(a)\star _Ew_m(b)=w_{n\star m}(a\star _tb)\end{cases}$ that $\langle 0.01,0.2\rangle=E\simeq\Bbb Z\oplus\Bbb Z$ ♦ '''now''' assume $(S\times S)\oplus E$ is external direct sum of the groups $S\times S$ and $E$ with $e=(0.2,0.2,0.1)$ and $\langle (0.02,0.2,0.1),(0.2,0.02,0.1),(0.3,0.2,0.1),(0.2,0.3,0.1),(0.2,0.2,0.01),(0.2,0.2,0.2)\rangle=$ $(S\times S)\oplus E\simeq\Bbb Z\oplus\Bbb Z\oplus\Bbb Z\oplus\Bbb Z\oplus\Bbb Z\oplus\Bbb Z$. '''Theorem''' $6$: $(S,\lt _1)$ is a well-ordering set with order relation $\lt _1$ as: $\forall i,n,k\in\Bbb N$ if $p_n$ be $n$-th prime number, relation $\lt _1$ is defined with: $w_i(p_n)\lt _1w_i(p_{n+k})\lt _1w_{i+1}(p_n)$ or '"`UNIQ-MathJax6-QINU`"' '"`UNIQ-MathJax7-QINU`"' and $(E,\lt _2)$ is another well ordering set with order relation $\lt _2$ as: $\forall i,n,k\in\Bbb N$ that $10\nmid n,\, 10\nmid n+k,$ $w_i(n)\lt _2w_i(n+k)\lt _2w_{i+1}(n)$ or '"`UNIQ-MathJax8-QINU`"' '"`UNIQ-MathJax9-QINU`"' now $M:=S\times S\times E$ is a well-ordering set with order relation $\lt _3$ as: $\forall (a,b,t),(c,d,u)\in S\times S\times E,$ $(a,b,t)\lt _3(c,d,u)$ if $\,\,\begin{cases} t\lt _2u & or\\ t=u,\,\, a+b\lt _2c+d & or\\ t=u,\,\, a+b=c+d,\,\, b\lt _1 d\end{cases}$ ♦ '''now''' assume $M$ is a topological space ('''Hausdorff space''') induced by order relation $\lt _3$. '''Question''' $3$: Is $S$ a topological group under topology induced by order relation $\lt_1$ and is $(S\times S)\oplus E$ a topological group under topology of $M$? '''A new version of Goldbach's conjecture''': For each even natural number $t$ greater than $4$ and $\forall c,m\in\Bbb N\cup\{0\}$ that $10^c\mid t,\, 10^{1+c}\nmid t$, $A_m=\{(a,b)\mid a,b\in S,\, 10^{-1-m}\le a+b\lt 10^{-m}\}$ and if $u$ is the number of digits in $t$ then $\exists (a,b)\in A_c$ such that $t=10^{c+u}\cdot (a+b),\, 10^{c+u}\cdot a,10^{c+u}\cdot b\in\Bbb P\setminus\{2\},\, (a,b,10^{-c-u}\cdot t)\in M$. :Using homotopy groups Goldbach's conjecture will be proved. Alireza Badali 08:27, 31 March 2018 (CEST) ==='"`UNIQ--h-2--QINU`"' Polignac's conjecture === In previous chapter above I used an important technique by theorem $1$ for presentment of prime numbers properties as density in discussion that using prime number theorem it became applicable, anyway, but now I want perform another method for Twin prime conjecture (Polignac) in principle prime numbers properties are ubiquitous in own natural numbers. '''Theorem''' $1$: $(\Bbb N,\star _T)$ is a group with: $\forall m,n\in\Bbb N,$ $\begin{cases} (12m-10)\star_T(12m-9)=1=(12m-8) \star_T(12m-5)=(12m-7) \star_T(12m-4)=\\ (12m-6) \star_T(12m-1)=(12m-3) \star_T(12m)=(12m-2) \star_T(12m+1)\\ (12m-10) \star_T(12n-10)=12m+12n-19\\ (12m-10) \star_T(12n-9)=\begin{cases} 12m-12n+1 & 12m-10\gt 12n-9\\ 12n-12m-2 & 12n-9\gt 12m-10\end{cases}\\ (12m-10) \star_T(12n-8)=12m+12n-15\\ (12m-10) \star_T(12n-7)=12m+12n-20\\ (12m-10) \star_T(12n-6)=12m+12n-11\\ (12m-10) \star_T(12n-5)=\begin{cases} 12m-12n-3 & 12m-10\gt 12n-5\\ 12n-12m+8 & 12n-5\gt 12m-10\end{cases}\\ (12m-10) \star_T(12n-4)=\begin{cases} 12m-12n-6 & 12m-10\gt 12n-4\\ 12n-12m+3 & 12n-4\gt 12m-10\end{cases}\\ (12m-10) \star_T(12n-3)=12m+12n-18\\ (12m-10) \star_T(12n-2)=\begin{cases} 12m-12n-10 & 12m-10\gt 12n-2\\ 12n-12m+11 & 12n-2\gt 12m-10\end{cases}\\ (12m-10) \star_T(12n-1)=\begin{cases} 12m-12n-7 & 12m-10\gt 12n-1\\ 12n-12m+12 & 12n-1\gt 12m-10\end{cases}\\ (12m-10) \star_T(12n)=\begin{cases} 12m-12n-8 & 12m-10\gt 12n\\ 12n-12m+7 & 12n\gt 12m-10\end{cases}\\ (12m-10) \star_T(12n+1)=12m+12n-10\\ (12m-9) \star_T(12n-9)=12m+12n-16\\ (12m-9) \star_T(12n-8)=\begin{cases} 12m-12n & 12m-9\gt 12n-8\\ 12n-12m+5 & 12n-8\gt 12m-9\end{cases}\\ (12m-9) \star_T(12n-7)=\begin{cases} 12m-12n-1 & 12m-9\gt 12n-7\\ 12n-12m+2 & 12n-7\gt 12m-9\end{cases}\\ (12m-9) \star_T(12n-6)=\begin{cases} 12m-12n-4 & 12m-9\gt 12n-6\\ 12n-12m+9 & 12n-6\gt 12m-9\end{cases}\\ (12m-9) \star_T(12n-5)=12m+12n-12\\ (12m-9) \star_T(12n-4)=12m+12n-17\\ (12m-9) \star_T(12n-3)=\begin{cases} 12m-12n-5 & 12m-9\gt 12n-3\\ 12n-12m+4 & 12n-3\gt 12m-9\end{cases}\\ (12m-9) \star_T(12n-2)=12m+12n-9\\ (12m-9) \star_T(12n-1)=12m+12n-14\\ (12m-9) \star_T(12n)=12m+12n-13\\ (12m-9)\star_T(12n+1)=\begin{cases} 12m-12n-9 & 12m-9\gt 12n+1\\ 12n-12m+6 & 12n+1\gt 12m-9\end{cases}\\ (12m-8) \star_T(12n-8)=12m+12n-11\\ (12m-8) \star_T(12n-7)=12m+12n-18\\ (12m-8) \star_T(12n-6)=12m+12n-7\\ (12m-8) \star_T(12n-5)=\begin{cases} 12m-12n+1 & 12m-8\gt 12n-5\\ 12n-12m-2 & 12n-5\gt 12m-8\end{cases}\\ (12m-8) \star_T(12n-4)=\begin{cases} 12m-12n+2 & 12m-8\gt 12n-4\\ 12n-12m-1 & 12n-4\gt 12m-8\\ 2 & m=n\end{cases}\\ (12m-8) \star_T(12n-3)=12m+12n-10\\ (12m-8) \star_T(12n-2)=\begin{cases} 12m-12n-8 & 12m-8\gt 12n-2\\ 12n-12m+7 & 12n-2\gt 12m-8\end{cases}\\ (12m-8) \star_T(12n-1)=\begin{cases} 12m-12n-3 & 12m-8\gt 12n-1\\ 12n-12m+8 & 12n-1\gt 12m-8\end{cases}\\ (12m-8) \star_T(12n)=\begin{cases} 12m-12n-6 & 12m-8\gt 12n\\ 12n-12m+3 & 12n\gt 12m-8\end{cases}\\ (12m-8) \star_T(12n+1)=12m+12n-8\\ (12m-7) \star_T(12n-7)=12m+12n-15\\ (12m-7) \star_T(12n-6)=12m+12n-10\\ (12m-7) \star_T(12n-5)=\begin{cases} 12m-12n-6 & 12m-7\gt 12n-5\\ 12n-12m+3 & 12n-5\gt 12m-7\end{cases}\\ (12m-7) \star_T(12n-4)=\begin{cases} 12m-12n+1 & 12m-7\gt 12n-4\\ 12n-12m-2 & 12n-4\gt 12m-7\end{cases}\\ (12m-7) \star_T(12n-3)=12m+12n-11\\ (12m-7) \star_T(12n-2)=\begin{cases} 12m-12n-7 & 12m-7\gt 12n-2\\ 12n-12m+12 & 12n-2\gt 12m-7\end{cases}\\ (12m-7) \star_T(12n-1)=\begin{cases} 12m-12n-8 & 12m-7\gt 12n-1\\ 12n-12m+7 & 12n-1\gt 12m-7\end{cases}\\ (12m-7) \star_T(12n)=\begin{cases} 12m-12n-3 & 12m-7\gt 12n\\ 12n-12m+8 & 12n\gt 12m-7\end{cases}\\ (12m-7) \star_T(12n+1)=12m+12n-7\\ (12m-6) \star_T(12n-6)=12m+12n-3\\ (12m-6) \star_T(12n-5)=\begin{cases} 12m-12n+5 & 12m-6\gt 12n-5\\ 12n-12m & 12n-5\gt 12m-6\\ 5 & m=n\end{cases}\\ (12m-6) \star_T(12n-4)=\begin{cases} 12m-12n+4 & 12m-6\gt 12n-4\\ 12n-12m-5 & 12n-4\gt 12m-6\\ 4 & m=n\end{cases}\\ (12m-6) \star_T(12n-3)=12m+12n-8\\ (12m-6) \star_T(12n-2)=\begin{cases} 12m-12n-6 & 12m-6\gt 12n-2\\ 12n-12m+3 & 12n-2\gt 12m-6\end{cases}\\ (12m-6) \star_T(12n-1)=\begin{cases} 12m-12n+1 & 12m-6\gt 12n-1\\ 12n-12m-2 & 12n-1\gt 12m-6\end{cases}\\ (12m-6) \star_T(12n)=\begin{cases} 12m-12n+2 & 12m-6\gt 12n\\ 12n-12m-1 & 12n\gt 12m-6\\ 2 & m=n\end{cases}\\ (12m-6) \star_T(12n+1)=12m+12n-6\\ (12m-5) \star_T(12n-5)=12m+12n-14\\ (12m-5) \star_T(12n-4)=12m+12n-13\\ (12m-5) \star_T(12n-3)=\begin{cases} 12m-12n-1 & 12m-5\gt 12n-3\\ 12n-12m+2 & 12n-3\gt 12m-5\end{cases}\\ (12m-5) \star_T(12n-2)=12m+12n-5\\ (12m-5) \star_T(12n-1)=12m+12n-4\\ (12m-5) \star_T(12n)=12m+12n-9\\ (12m-5) \star_T(12n+1)=\begin{cases} 12m-12n-5 & 12m-5\gt 12n+1\\ 12n-12m+4 & 12n+1\gt 12m-5\end{cases}\\ (12m-4) \star_T(12n-4)=12m+12n-12\\ (12m-4) \star_T(12n-3)=\begin{cases} 12m-12n & 12m-4\gt 12n-3\\ 12n-12m+5 & 12n-3\gt 12m-4\end{cases}\\ (12m-4) \star_T(12n-2)=12m+12n-4\\ (12m-4) \star_T(12n-1)=12m+12n-9\\ (12m-4) \star_T(12n)=12m+12n-14\\ (12m-4) \star_T(12n+1)=\begin{cases} 12m-12n-4 & 12m-4\gt 12n+1\\ 12n-12m+9 & 12n+1\gt 12m-4\end{cases}\\ (12m-3) \star_T(12n-3)=12m+12n-7\\ (12m-3) \star_T(12n-2)=\begin{cases} 12m-12n-3 & 12m-3\gt 12n-2\\ 12n-12m+8 & 12n-2\gt 12m-3\end{cases}\\ (12m-3) \star_T(12n-1)=\begin{cases} 12m-12n-6 & 12m-3\gt 12n-1\\ 12n-12m+3 & 12n-1\gt 12m-3\end{cases}\\ (12m-3) \star_T(12n)=\begin{cases} 12m-12n+1 & 12m-3\gt 12n\\ 12n-12m-2 & 12n\gt 12m-3\end{cases}\\ (12m-3) \star_T(12n+1)=12m+12n-3\\ (12m-2) \star_T(12n-2)=12m+12n-2\\ (12m-2) \star_T(12n-1)=12m+12n-1\\ (12m-2) \star_T(12n)=12m+12n\\ (12m-2) \star_T(12n+1)=\begin{cases} 12m-12n-2 & 12m-2\gt 12n+1\\ 12n-12m+1 & 12n+1\gt 12m-2\end{cases}\\ (12m-1) \star_T(12n-1)=12m+12n\\ (12m-1) \star_T(12n)=12m+12n-5\\ (12m-1) \star_T(12n+1)=\begin{cases} 12m-12n-1 & 12m-1\gt 12n+1\\ 12n-12m+2 & 12n+1\gt 12m-1\end{cases}\\ (12m) \star_T(12n)=12m+12n-4\\ (12m) \star_T(12n+1)=\begin{cases} 12m-12n & 12m\gt 12n+1\\ 12n-12m+5 & 12n+1\gt 12m\end{cases}\\ (12m+1) \star_T(12n+1)=12m+12n+1\end{cases}$ that $\forall k\in\Bbb N,\,\langle 2\rangle =\langle 3\rangle =\langle (2k+1)\star _T (2k+3)\rangle=(\Bbb N,\star _T)\simeq (\Bbb Z,+)$ and $\langle (2k)\star _T(2k+2)\rangle\neq\Bbb N$ and each prime in $\langle 5\rangle$ is to form of $5+12k$ or $13+12k$, $k\in\Bbb N\cup\{0\}$ and each prime in $\langle 7\rangle$ is to form of $7+12k$ or $13+12k$, $k\in\Bbb N\cup\{0\}$ and $\langle 5\rangle\cap\langle 7\rangle=\langle 13\rangle$ and $\Bbb N=\langle 5\rangle\oplus\langle 7\rangle$ but there isn't any proper subgroup including all primes of the form $11+12k,$ $k\in\Bbb N\cup\{0\}$ (probably I have to make another better). :Proof: '"`UNIQ-MathJax10-QINU`"' '"`UNIQ-MathJax11-QINU`"' '"`UNIQ-MathJax12-QINU`"' '"`UNIQ-MathJax13-QINU`"' '''Guess''' $1$: For each group on $\Bbb N$ like $(\Bbb N,\star)$ generated from algorithm above, if $p_i$ be $i$_th prime number and $x_i$ be $i$_th composite number then $\exists m\in\Bbb N,\,\forall n\in\Bbb N$ that $n\ge m$ we have: $2\star3\star5\star7...\star p_n=\prod_{i=1}^{n}p_i\gt\prod _{i=1}^{n}x_i=4\star6\star8\star9...\star x_n$ '''Guess''' $2$: For each group on $\Bbb N$ like $(\Bbb N,\star)$ generated from algorithm above, we have: $\lim_{n\to\infty}\prod _{n=1}^{\infty}p_n,\lim_{n\to\infty}\prod _{n=1}^{\infty}x_n\in\Bbb N,\,\,(\lim_{n\to\infty}\prod _{n=1}^{\infty}p_n)\star(\lim_{n\to\infty}\prod _{n=1}^{\infty}x_n)=1$. now let the group $G$ be external direct sum of three copies of the group $(\Bbb N,\star _T)$, hence $G=\Bbb N\oplus\Bbb N\oplus\Bbb N$. '''Theorem''' $2$: $(\Bbb N\times\Bbb N\times\Bbb N,\lt _T)$ is a well ordering set with order relation $\lt _T$ as: $\forall (m_1,n_1,t_1),(m_2,n_2,t_2)\in\Bbb N\times\Bbb N\times\Bbb N,\quad (m_1,n_1,t_1)\lt _T(m_2,n_2,t_2)$ if $\begin{cases} t_1\lt t_2 & or\\ t_1=t_2,\, m_1-n_1\lt m_2-n_2 & or\\ t_1=t_2,\, m_1-n_1=m_2-n_2,\, n_1\lt n_2\end{cases}$ and suppose $M=\Bbb N\times\Bbb N\times\Bbb N$ is a topological space ('''Hausdorff space''') induced by order relation $\lt _T$. '''Question''' $1$: Is $G$ a topological group with topology of $M$? '''Now''' regarding to the group $(\Bbb N,\star_T)$, I am planning an algebraic form of prime number theorem towards twin prime conjecture: Recall the statement of the prime number theorem: Let $x$ be a positive real number, and let $\pi(x)$ denote the number of primes that are less than or equal to $x$. Then the ratio $\pi(x)\cdot{\log x\over x}$ can be made arbitrarily close to $1$ by taking $x$ sufficiently large. Question $2$: Suppose $\pi_1(x)$ is all prime numbers of the form $4k+1$ and less than $x$ and $\pi_2(x)$ is all prime numbers of the form $4k+3$ and less than $x$. Do $\lim_{x\to\infty}\pi_1(x)\cdot{\log x\over x}=0.5=\lim_{x\to\infty}\pi_2(x)\cdot{\log x\over x}\ ?$ :[https://math.stackexchange.com/questions/2769471/another-extension-of-prime-number-theorem/2769494#2769494 Answer] given by [https://math.stackexchange.com/users/174927/milo-brandt $@$Milo Brandt] from stackexchange site: Basically, for any $k$, the primes are equally distributed across the congruence classes $\langle n\rangle$ mod $k$ where $n$ and $k$ are coprime. :This result is known as the prime number theorem for arithmetic progressions. [Wikipedia](https://en.wikipedia.org/wiki/Prime_number_theorem#Prime_number_theorem_for_arithmetic_progressions) discusses it with a number of references and one can find a proof of it by Ivan Soprounov [here](http://academic.csuohio.edu/soprunov_i/pdf/primes.pdf), which makes use of the Dirichlet theorem on arithmetic progressions (which just says that $\pi_1$ and $\pi_2$ are unbounded) to prove this stronger result. Question $3$: For each neutral infinite subset $A$ of $\Bbb N$, does exist a cyclic group like $(\Bbb N,\star)$ such that $A$ is a maximal subgroup of $\Bbb N$? Question $4$: If $(\Bbb N,\star_1)$ is a cyclic group and $n\in\Bbb N$ and $A=\{a_i\mid i\in\Bbb N\}$ is a non-trivial subgroup of $\Bbb N$ then does exist another cyclic group $(\Bbb N,\star_2)$ such that $\prod _{i=1}^{\infty}a_i=a_1\star_2a_2\star_2a_3\star_2...=n$? Question $5$: If $(\Bbb N,\star)$ is a cyclic group and $n\in\Bbb N$ then does exist a non-trivial subset $A=\{a_i\mid i\in\Bbb N\}$ of $\Bbb P$ with $\#(\Bbb P\setminus A)=\aleph_0$ and $\prod _{i=1}^{\infty}a_i=a_1\star a_2\star a_3\star...=n$? Question $6$: If $(\Bbb N,\star_1)$ and $(\Bbb N,\star_2)$ are cyclic groups and $A=\{a_i\mid i\in\Bbb N\}$ is a non-trivial subgroup of $(\Bbb N,\star_1)$ and $B=A\cap\Bbb P$ then does $\prod_{i=1}^{\infty}a_i=a_1\star_2a_2\star_2a_3\star_2...\in\Bbb N$? Alireza Badali 12:34, 28 April 2018 (CEST) =='"`UNIQ--h-3--QINU`"' Collatz conjecture == The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined as follows: start with any positive integer $n$. Then each term is obtained from the previous term as follows: if the previous term is even, the next term is one half the previous term. Otherwise, the next term is $3$ times the previous term plus $1$. The conjecture is that no matter what value of $n$, the sequence will always reach $1$. The conjecture is named after German mathematician [https://en.wikipedia.org/wiki/Lothar_Collatz Lothar Collatz], who introduced the idea in $1937$, two years after receiving his doctorate. It is also known as the $3n + 1$ conjecture. '''Theorem''' $1$: If $(\Bbb N,\star_{\Bbb N})$ is a cyclic group with $e_{\Bbb N}=1$ & $\langle m_1\rangle=\langle m_2\rangle=(\Bbb N,\star_{\Bbb N})$ and $f:\Bbb N\to\Bbb N$ is a bijection such that $f(1)=1$ then $(\Bbb N,\star _f)$ is a cyclic group with: $e_f=1$ & $\langle f(m_1)\rangle=\langle f(m_2)\rangle=(\Bbb N,\star_f)$ & $\forall m,n\in\Bbb N,$ $f(m)\star _ff(n)=f(m\star_{\Bbb N}n)$ & $(f(n))^{-1}=f(n^{-1})$ that $n\star_{\Bbb N}n^{-1}=1$. I want make a group in accordance with [https://www.jasondavies.com/collatz-graph/ Collatz graph] but [https://math.stackexchange.com/users/334732/robert-frost $@$RobertFrost] from stackexchange site advised me in addition, it needs to be a torsion group because then it can be used to show convergence, meantime I like apply lines in the Euclidean plane $\Bbb R^2$ too. Question $1$: What is function of this sequence on to natural numbers? $1,2,4,3,6,5,10,7,14,8,16,9,18,11,22,12,24,13,26,15,30,17,34,19,38,20,40,21,42,23,46,25,50,...$ such that we begin from $1$ and then write $2$ then $2\times2$ then $3$ then $2\times3$ then ... but if $n$ is even and previously we have written $0.5n$ and then $n$ then ignore $n$ and continue and write $n+1$ and then $2n+2$ and so on for example we have $1,2,4,3,6,5,10$ so after $10$ we should write $7,14,...$ because previously we have written $3,6$. :[https://math.stackexchange.com/questions/2779491/what-is-function-of-this-sequence-on-to-natural-numbers/2779815#2779815 Answer] given by [https://math.stackexchange.com/users/16397/r-e-s $@$r.e.s] from stackexchange site: Following is a definition of your sequence without using recursion. :Let $S=(S_0,S_1,S_2,\ldots)$ be the increasing sequence of positive integers that are expressible as either $2^e$ or as $o_1\cdot 2^{o_2}$, where $e$ is an even nonnegative integer, $o_1>1$ is an odd positive integer and $o_2$ is an odd positive integer. Thus '"`UNIQ-MathJax14-QINU`"' Let $\bar{S}$ be the complement of $S$ with respect to the positive integers; i.e., '"`UNIQ-MathJax15-QINU`"' Your sequence is then $T=(T_0,T_1,T_2,\ldots)$, where '"`UNIQ-MathJax16-QINU`"' :Thus $T=(1, 2, 4, 3, 6, 5, 10, 7, 14, 8, 16, 9, 18, 11, 22, 12, 24, 13, 26, 15, 30, 17, 34, 19, 38, 20, \ldots).$ ---------------------------- :References: :Sequences $S,\bar{S},T$ are OEIS [A171945](http://oeis.org/A171945), [A053661](http://oeis.org/A053661), [A034701](http://oeis.org/A034701) respectively. These are all discussed in ["The vile, dopey, evil and odious game players"](https://www.sciencedirect.com/science/article/pii/S0012365X11001427). -------------------------- :Sage code: def is_in_S(n): return ( (n.valuation(2) % 2 == 0) and (n.is_power_of(2)) ) or ( (n.valuation(2) % 2 == 1) and not(n.is_power_of(2)) ) S = [n for n in [1..50] if is_in_S(n)] S_ = [n for n in [1..50] if not is_in_S(n)] T = [] for i in range(max(len(S),len(S_))): if i % 2 == 0: T += [S[i/2]] else: T += [S_[(i-1)/2]] print S print S_ print T [1, 4, 6, 10, 14, 16, 18, 22, 24, 26, 30, 34, 38, 40, 42, 46, 50] [2, 3, 5, 7, 8, 9, 11, 12, 13, 15, 17, 19, 20, 21, 23, 25, 27, 28, 29, 31, 32, 33, 35, 36, 37, 39, 41, 43, 44, 45, 47, 48, 49] [1, 2, 4, 3, 6, 5, 10, 7, 14, 8, 16, 9, 18, 11, 22, 12, 24, 13, 26, 15, 30, 17, 34, 19, 38, 20, 40, 21, 42, 23, 46, 25, 50] '''Theorem''' $2$: If $(\Bbb N,\star_1)$ & $(\Bbb N,\star_2)$ are cyclic groups with generators respectively $u_1$ & $v_1$ and $u_2$ & $v_2$ then $C_1=\{(m,2m)\mid m\in\Bbb N\}$ is a cyclic group with: $\begin{cases} e_{C_1}=(1,2)\\ \\\forall m,n\in\Bbb N,\,(m,2m)\star_{C_1}(n,2n)=(m\star_1n,2(m\star_1n))\\ (m,2m)^{-1}=(m^{-1},2\times m^{-1})\qquad\text{that}\quad m\star_1m^{-1}=1\\ \\C_1=\langle(u_1,2u_1)\rangle=\langle(v_1,2v_1)\rangle\end{cases}$ and $C_2=\{(3m-1,2m-1)\mid m\in\Bbb N\}$ is a cyclic group with: $\begin{cases} e_{C_2}=(2,1)\\ \\\forall m,n\in\Bbb N,\,(3m-1,2m-1)\star_{C_2}(3n-1,2n-1)=(3(m\star_2n)-1,2(m\star_2n)-1)\\ (3m-1,2m-1)^{-1}=(3\times m^{-1}-1,2\times m^{-1}-1)\qquad\text{that}\quad m\star_2 m^{-1}=1\\ \\C_2=\langle(3u_2-1,2u_2-1)\rangle=\langle(3v_2-1,2v_2-1)\rangle\end{cases}$• :And let $C:=C_1\oplus C_2$ be external direct sum of the groups $C_1$ & $C_2$. '''Question''' $2$: What are maximal subgroups of $C$? '''Theorem''' $3$: If $(\Bbb N,\star)$ is a cyclic group with generators $u,v$ and identity element $e=1$ and $f:\Bbb N\to\Bbb R$ is an injection then $(f(\Bbb N),\star_f)$ is a cyclic group with generators $f(u),f(v)$ and identity element $e_f=f(1)$ and operation law: $\forall m,n\in\Bbb N,$ $f(m)\star_ff(n)=f(m\star n)$ and inverse law: $\forall n\in\Bbb N,$ $(f(n))^{-1}=f(n^{-1})$ that $n\star n^{-1}=1$. '''Suppose''' $\forall m,n\in\Bbb N,\qquad$ $\begin{cases} m\star 1=m\\ (4m)\star (4m-2)=1=(4m+1)\star (4m-1)\\ (4m-2)\star (4n-2)=4m+4n-5\\ (4m-2)\star (4n-1)=4m+4n-2\\ (4m-2)\star (4n)=\begin{cases} 4m-4n-1 & 4m-2\gt 4n\\ 4n-4m+1 & 4n\gt 4m-2\\ 3 & m=n+1\end{cases}\\ (4m-2)\star (4n+1)=\begin{cases} 4m-4n-2 & 4m-2\gt 4n+1\\ 4n-4m+4 & 4n+1\gt 4m-2\end{cases}\\ (4m-1)\star (4n-1)=4m+4n-1\\ (4m-1)\star (4n)=\begin{cases} 4m-4n+2 & 4m-1\gt 4n\\ 4n-4m & 4n\gt 4m-1\\ 2 & m=n\end{cases}\\ (4m-1)\star (4n+1)=\begin{cases} 4m-4n-1 & 4m-1\gt 4n+1\\ 4n-4m+1 & 4n+1\gt 4m-1\\ 3 & m=n+1\end{cases}\\ (4m)\star (4n)=4m+4n-3\\ (4m)\star (4n+1)=4m+4n\\ (4m+1)\star (4n+1)=4m+4n+1\\ \Bbb N=\langle 2\rangle=\langle 4\rangle\end{cases}$ and let $C_1=\{(m,2m)\mid m\in\Bbb N\}$ is a cyclic group with: $\begin{cases} e_{C_1}=(1,2)\\ \\\forall m,n\in\Bbb N,\,(m,2m)\star_{C_1}(n,2n)=(m\star n,2(m\star n))\\ (m,2m)^{-1}=(m^{-1},2\times m^{-1})\qquad\text{that}\quad m\star m^{-1}=1\\ \\C_1=\langle(2,4)\rangle=\langle(4,8)\rangle\end{cases}$ and $C_2=\{(3m-1,2m-1)\mid m\in\Bbb N\}$ is a cyclic group with: $\begin{cases} e_{C_2}=(2,1)\\ \\\forall m,n\in\Bbb N,\, (3m-1,2m-1)\star_{C_2}(3n-1,2n-1)=(3(m\star n)-1,2(m\star n)-1)\\ (3m-1,2m-1)^{-1}=(3\times m^{-1}-1,2\times m^{-1}-1)\qquad\text{that}\quad m\star m^{-1}=1\\ \\C_2=\langle(5,3)\rangle=\langle(11,7)\rangle\end{cases}$. and let $C:=C_1\oplus C_2$ be external direct sum of the groups $C_1$ & $C_2$, '''Question''' $3$: What are maximal subgroups of $C$? Alireza Badali 10:02, 12 May 2018 (CEST) =='"`UNIQ--h-4--QINU`"' Erdős–Straus conjecture == '''Theorem''': If $(\Bbb N,\star)$ is a cyclic group with identity element $e=1$ and generators $a,b$ then $E=\{({1\over x},{1\over y},{1\over z},{-4\over n+1},n)\mid x,y,z,n\in\Bbb N\}$ is an Abelian group with: $\forall x,y,z,n,x_1,y_1,z_1,n_1\in\Bbb N$ $\begin{cases} e_E=(1,1,1,-2,1)=({1\over 1},{1\over 1},{1\over 1},{-4\over 1+1},1)\\ \\({1\over x},{1\over y},{1\over z},{-4\over n+1},n)^{-1}=({1\over x^{-1}},{1\over y^{-1}},{1\over z^{-1}},\frac{-4}{n^{-1}+1},n^{-1})\quad\text{that}\\ x\star x^{-1}=1=y\star y^{-1}=z\star z^{-1}=n\star n^{-1}\\ \\({1\over x},{1\over y},{1\over z},\frac{-4}{n+1},n)\star_E({1\over x_1},{1\over y_1},{1\over z_1},\frac{-4}{n_1+1},n_1)=(\frac{1}{x\star x_1},\frac{1}{y\star y_1},\frac{1}{z\star z_1},\frac{-4}{n\star {n_1}+1},n\star n_1)\\ \\E=\langle({1\over a},1,1,-2,1),(1,{1\over a},1,-2,1),(1,1,{1\over a},-2,1),(1,1,1,\frac{-4}{a+1},1),(1,1,1,-2,a)\rangle=\\ \langle({1\over b},1,1,-2,1),(1,{1\over b},1,-2,1),(1,1,{1\over b},-2,1),(1,1,1,\frac{-4}{b+1},1),(1,1,1,-2,b)\rangle\end{cases}$• Let $(\Bbb N,\star)$ is a cyclic group with: $\begin{cases} n\star 1=n\\ (2n)\star (2n+1)=1\\ (2n)\star (2m)=2n+2m\\ (2n+1)\star (2m+1)=2n+2m+1\\ (2n)\star (2m+1)=\begin{cases} 2m-2n+1 & 2m+1\gt 2n\\ 2n-2m & 2n\gt 2m+1\end{cases}\\\Bbb N=\langle 2\rangle =\langle 3\rangle \end{cases}$ :Question: Is $E_0=\{({1\over x},{1\over y},{1\over z},\frac{-4}{n+1},n)\mid x,y,z,n\in\Bbb N,\, {1\over x}+{1\over y}+{1\over z}-{4\over n+1}=0\}$ a subgroup of $E$? Alireza Badali 17:34, 25 May 2018 (CEST) =='"`UNIQ--h-5--QINU`"' Landaus forth problem == Friedlander–Iwaniec theorem: there are infinitely many prime numbers of the form $a^2+b^4$. :I want use this theorem for [https://en.wikipedia.org/wiki/Landau%27s_problems Landaus forth problem] but prime numbers properties have been applied for Friedlander–Iwaniec theorem hence no need to prime number theorem or its other forms or extensions. '''Theorem''': If $(\Bbb N,\star)$ is a cyclic group with identity element $e=1$ and generators $u,v$ then $F=\{(a^2,b^4)\mid a,b\in\Bbb N\}$ is a group with: $\forall a,b,c,d\in\Bbb N\,$ $\begin{cases} e_F=(1,1)\\ (a^2,b^4)\star_F(c^2,d^4)=((a\star c)^2,(b\star d)^4)\\ (a^2,b^4)^{-1}=((a^{-1})^2,(b^{-1})^4)\qquad\text{that}\quad a\star a^{-1}=1=b\star b^{-1}\\ F=\langle (1,u^4),(u^2,1)\rangle=\langle (1,v^4),(v^2,1)\rangle\end{cases}$ '''now''' let $H=\langle\{(a^2,b^4)\mid a,b\in\Bbb N,\,b\neq 1\}\rangle$ and $G=F/H$ is quotient group of $F$ by $H$. ($G$ is a group including prime numbers properties only of the form $1+n^2$.) and also $L=\{1+n^2\mid n\in\Bbb N\}$ is a cyclic group with: $\forall m,n\in\Bbb N$ $\begin{cases} e_L=2=1+1^2\\ (1+n^2)\star_L(1+m^2)=1+(n\star m)^2\\ (1+n^2)^{-1}=1+(n^{-1})^2\quad\text{that}\;n\star n^{-1}=1\\ L=\langle 1+u^2\rangle=\langle 1+v^2\rangle\end{cases}$ but on the other hand we have: $L\simeq G$ hence we can apply $L$ instead $G$ of course since we are working on natural numbers generally we could consider from the beginning the group $L$ without involvement with the group $G$ anyhow. :Question $1$: For each neutral cyclic group on $\Bbb N$ then what are maximal subgroups of $L$? '''Guess''' $1$: For each cyclic group structure on $\Bbb N$ like $(\Bbb N,\star)$ then for each non-trivial subgroup of $\Bbb N$ like $T$ we have $T\cap\Bbb P\neq\emptyset$. :I think this guess must be proved via prime number theorem. '''For''' each neutral cyclic group on $\Bbb N$ if $L\cap\Bbb P=\{1+n_1^2,1+n_2^2,...,1+n_k^2\},\,k\in\Bbb N$ and if $A=\bigcap _{i=1}^k\langle 1+n_i^2\rangle$ so $\exists m\in\Bbb N$ that $A=\langle 1+m^2\rangle$ & $m\neq n_i$ for $i=1,2,3,...,k$ (intelligibly $k\gt1$) so we have: $A\cap\Bbb P=\emptyset$. :Question $2$: Is $A$ only unique greatest subgroup of $L$ such that $A\cap\Bbb P=\emptyset$? Alireza Badali 16:49, 28 May 2018 (CEST) =='"`UNIQ--h-6--QINU`"' Lemoine's conjecture == '''Theorem''': If $(\Bbb N,\star)$ is a cyclic group with identity element $e=1$ & generators $u,v$ then $L=\{(p_{n_1},p_{n_2},p_{n_3},-2n-5)\mid n,n_1,n_2,n_3\in\Bbb N,\,p_{n_i}$ is $n_i$_th prime for $i=1,2,3\}$ is an Abelian group with: $\forall n_1,n_2,n_3,n,m_1,m_2,m_3,m\in\Bbb N$ $\begin{cases} e_L=(2,2,2,-7)=(2,2,2,-2\times 1-5)\\ \\(p_{n_1},p_{n_2},p_{n_3},-2n-5)\star_L(p_{m_1},p_{m_2},p_{m_3},-2m-5)=(p_{n_1\star m_1},p_{n_2\star m_2},p_{n_3\star m_3},-2\times(n\star m)-5)\\ \\(p_{n_1},p_{n_2},p_{n_3},-2n-5)^{-1}=(p_{n_1^{-1}},p_{n^{-1}_2},p_{n_3^{-1}},-2\times n^{-1}-5)\quad\text{that}\\ n_1\star n_1^{-1}=1=n_2\star n_2^{-1}=n_3\star n_3^{-1}=n\star n^{-1}\\ \\L=\langle(p_u,2,2,-7),(2,p_u,2,-7),(2,2,p_u,-7),(2,2,2,-2u-5)\rangle=\\\langle(p_v,2,2,-7),(2,p_v,2,-7),(2,2,p_v,-7),(2,2,2,-2v-5)\rangle\end{cases}$• '''Theorem''': $\forall n\in\Bbb N,\,\exists (p_{m_1},p_{m_2},p_{m_3},-2n-5)\in(L,\star_L)$ such that $p_{m_1}+p_{m_2}+p_{m_3}-2n-5=0$. :Proof using Goldbach's weak conjecture. '''Question''': Is $L_0=\{(p_{m_1},p_{m_2},p_{m_2},-2n-5)\mid\forall m_1,m_2\in\Bbb N,\,\exists n\in\Bbb N,$ such that $p_{m_1}+2p_{m_2}-2n-5=0\}$ a subgroup of $L$? Alireza Badali 19:30, 3 June 2018 (CEST) =='"`UNIQ--h-7--QINU`"' Primes in beatty sequences == How can we understand $\infty$? we humans only can think on natural numbers and other issues are only theorizing, algebraic theories can be some features for this aim. [http://oeis.org/A184774 Conjecture]: If $r$ is an irrational number and $1\lt r\lt 2$, then there are infinitely many primes in the set $L=\{\text{floor}(n\cdot r)\mid n\in\Bbb N\}$. '''Theorem''' $1$: If $(\Bbb N,\star)$ is a cyclic group with identity element $e=1$ & generators $u,v$ and $r\in[1,2]\setminus\Bbb Q$ then $L=\{\lfloor n\cdot r\rfloor\mid n\in\Bbb N\}$ is another cyclic group with: $\forall m,n\in\Bbb N$ $\begin{cases} e_L=1\\ \lfloor n\cdot r\rfloor\star_L\lfloor m\cdot r\rfloor=\lfloor (n\star m)\cdot r\rfloor\\ (\lfloor n\cdot r\rfloor)^{-1}=\lfloor n^{-1}\cdot r\rfloor\qquad\text{that}\quad n\star n^{-1}=1\\ L=\langle\lfloor u\cdot r\rfloor\rangle=\langle\lfloor v\cdot r\rfloor\rangle\end{cases}$. :Guess $1$: $\prod_{n=1}^{\infty}\lfloor n\cdot r\rfloor=\lfloor 1\cdot r\rfloor\star\lfloor 2\cdot r\rfloor\star\lfloor 3\cdot r\rfloor\star...\in\Bbb N$. The conjecture generalized: if $r$ is a positive irrational number and $h$ is a real number, then each of the sets $\{\text{floor}(n\cdot r+h)\mid n\in\Bbb N\}$, $\{\text{round}(n\cdot r+h)\mid n\in\Bbb N\}$, and $\{\text{ceiling}(n\cdot r+h)\mid n\in\Bbb N\}$ contains infinitely many primes. '''Theorem''' $2$: If $(\Bbb N,\star)$ is a cyclic group with identity element $e=1$ & generators $u,v$ & $r$ is a positive irrational number & $h\in\Bbb R$ then $G=\{n\cdot r+h\mid n\in\Bbb N\}$ is another cyclic group with: $\forall m,n\in\Bbb N$ $\begin{cases} e_G=\lfloor r+h\rfloor\\ \lfloor n\cdot r+h\rfloor\star_G\lfloor m\cdot r+h\rfloor=\lfloor (n\star m)\cdot r+h\rfloor\\ (\lfloor n\cdot r+h\rfloor)^{-1}=\lfloor n^{-1}\cdot r+h\rfloor\qquad\text{that}\quad n\star n^{-1}=1\\ L=\langle\lfloor u\cdot r+h\rfloor\rangle=\langle\lfloor v\cdot r+h\rfloor\rangle\end{cases}$. :Guess $2$: $\prod_{n=k}^{\infty}\lfloor n\cdot r+h\rfloor=\lfloor k\cdot r+h\rfloor\star\lfloor (k+1)\cdot r+h\rfloor\star\lfloor (k+2)\cdot r+h\rfloor\star...\in\Bbb N$ in which $\lfloor k\cdot r+h\rfloor\in\Bbb N$ & $\lfloor (k-1)\cdot r+h\rfloor\lt1$.
Alireza Badali 19:09, 7 June 2018 (CEST)

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Please just insert your comment here! Alireza Badali 20:47, 15 April 2018 (CEST)

How to Cite This Entry:
Musictheory2math. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Musictheory2math&oldid=43135