Numeris primis ludere
An odd natural number \(n>1\) is a prime iff \(\displaystyle\prod_{k=1}^{n-1} k \equiv n-1 \pmod {\displaystyle\sum_{k=1}^{n-1} k}.\)
Let \(p \equiv 5 \pmod 6\) be prime then , \( 2p+1\) is prime iff \( 2p+1 \mid 3^p-1\) .
Let \( p\equiv 1 \pmod 6\) be prime and let \( 5 \nmid 4p+1\) , then \( 4p+1\) is prime iff \( 4p+1 \mid 2^{2p}+1\) .
Let \(p_n\) be the nth prime , then
\( p_n=1+\displaystyle\sum_{k=1}^{2 \cdot (\lfloor n\ln(n) \rfloor+1)}\left(1-\left\lfloor \frac{1}{n} \cdot \displaystyle\sum_{j=2}^k \left\lceil \frac{3-\displaystyle\sum_{i=1}^j \left\lfloor \frac{\left\lfloor \frac{j}{i} \right\rfloor}{\left\lceil \frac{j}{i} \right\rceil} \right\rfloor}{j} \right\rceil \right\rfloor\right)\)
Let \( P_j(x)=2^{-j}\cdot \left(\left(x-\sqrt{x^2-4}\right)^{j}+\left(x+\sqrt{x^2-4}\right)^{j}\right)\)
Let $N=k\cdot 2^m-1$ such that $m>2$ , $3 \mid k$ , $0< k <2^m$ and
$\begin{cases}k \equiv 1 \pmod{10} \text{ with } m \equiv 2,3 \pmod{4} \\k \equiv 3 \pmod{10} \text{ with } m \equiv 0,3 \pmod{4} \\k \equiv 7 \pmod{10} \text{ with } m \equiv 1,2 \pmod{4} \\k \equiv 9 \pmod{10} \text{ with } m \equiv 0,1 \pmod{4}\end{cases}$
Let \( S_i=S_{i-1}^2-2\) with \( S_0=P_k(3)\) , then \( N\) is prime iff \( S_{m-2} \equiv 0 \pmod N\) .
Let \( P_j(x)=2^{-j}\cdot \left(\left(x-\sqrt{x^2-4}\right)^{j}+\left(x+\sqrt{x^2-4}\right)^{j}\right)\)
Let $N=k\cdot 2^m-1$ such that $m>2$ , $3 \mid k$ , $ 0< k <2^m $ and
$\begin{cases}k \equiv 3 \pmod{42} \text{ with } m \equiv 0,2 \pmod{3} \\k \equiv 9 \pmod{42} \text{ with } m \equiv 0 \pmod{3} \\k \equiv 15 \pmod{42} \text{ with } m \equiv 1 \pmod{3} \\k \equiv 27 \pmod{42} \text{ with } m \equiv 1,2 \pmod{3} \\k \equiv 33 \pmod{42} \text{ with } m \equiv 0,1 \pmod{3} \\k \equiv 39 \pmod{42} \text{ with } m \equiv 2 \pmod{3} \end{cases}$
Let \( S_i=S_{i-1}^2-2\) with \( S_0=P_k(5)\) , then \( N\) is prime iff \( S_{m-2} \equiv 0 \pmod N\) .
Let \( P_j(x)=2^{-j}\cdot \left(\left(x-\sqrt{x^2-4}\right)^{j}+\left(x+\sqrt{x^2-4}\right)^{j}\right)\)
Let \( N=k\cdot 2^m+1\) such that \(m>2\) , \( 0< k <2^m\) and
\( \begin{cases} k \equiv 1 \pmod{42} \text{ with } m \equiv 2,4 \pmod{6} \\ k \equiv 5 \pmod{42} \text{ with } m \equiv 3 \pmod{6} \\ k \equiv 11 \pmod{42} \text{ with } m \equiv 3,5 \pmod{6} \\ k \equiv 13 \pmod{42} \text{ with } m \equiv 4 \pmod{6} \\ k \equiv 17 \pmod{42} \text{ with } m \equiv 5 \pmod{6} \\ k \equiv 19 \pmod{42} \text{ with } m \equiv 0 \pmod{6} \\ k \equiv 23 \pmod{42} \text{ with } m \equiv 1,3 \pmod{6} \\ k \equiv 25 \pmod{42} \text{ with } m \equiv 0,2 \pmod{6} \\ k \equiv 29 \pmod{42} \text{ with } m \equiv 1,5 \pmod{6} \\ k \equiv 31 \pmod{42} \text{ with } m \equiv 2 \pmod{6} \\ k \equiv 37 \pmod{42} \text{ with } m \equiv 0,4 \pmod{6} \\ k \equiv 41 \pmod{42} \text{ with } m \equiv 1 \pmod{6} \end{cases}\)
Let \( S_i=S_{i-1}^2-2\) with \( S_0=P_k(5)\) , then \( N\) is prime iff \( S_{m-2} \equiv 0 \pmod N\) .
Let \( P_j(x)=2^{-j}\cdot \left(\left(x-\sqrt{x^2-4}\right)^{j}+\left(x+\sqrt{x^2-4}\right)^{j}\right)\)
Let \( N=k\cdot 2^m+1\) such that \(m>2\) , \( 0< k <2^m\) and
\( \begin{cases} k \equiv 1,7 \pmod{30} \text{ with } m \equiv 0 \pmod{4} \\ k \equiv 11,23 \pmod{30} \text{ with } m \equiv 1 \pmod{4} \\ k \equiv 13,19 \pmod{30} \text{ with } m \equiv 2 \pmod{4} \\ k \equiv 17,29 \pmod{30} \text{ with } m \equiv 3 \pmod{4} \end{cases}\)
Let \( S_i=S_{i-1}^2-2\) with \( S_0=P_k(8)\) , then \( N\) is prime iff \( S_{m-2} \equiv 0 \pmod N\) .
Let \( P_m(x)=2^{-m}\cdot \left(\left(x-\sqrt{x^2-4}\right)^{m}+\left(x+\sqrt{x^2-4}\right)^{m}\right)\)
Let \( F_n(b)=b^{2^n}+1\) such that \( n \ge 2\) and \( b\) is even number .
Let \( S_i=P_b(S_{i-1})\) with \( S_0=P_{b}(6)\), then If \( F_n(b)\) is prime, then \( S_{2^n-1} \equiv 2\pmod{F_n(b)}\).
Let \( P_m(x)=2^{-m}\cdot \left(\left(x-\sqrt{x^2-4}\right)^{m}+\left(x+\sqrt{x^2-4}\right)^{m}\right)\)
Let \( E_n(b)=\Large\frac{b^{2^n}+1}{2}\) such that \( n>1\), \( b\) is odd number greater than one .
Let \( S_i=P_b(S_{i-1})\) with \( S_0=P_{b}(6)\), then If \( E_n(b)\) is prime, then \( S_{2^n-1} \equiv 6\pmod{E_n(b)}\).
Let \( P_m(x)=2^{-m}\cdot \left(\left(x-\sqrt{x^2-4}\right)^{m}+\left(x+\sqrt{x^2-4}\right)^{m}\right)\)
Let \( N_p(b)=\frac{b^p+1}{b+1}\) , where \( p\) is an odd prime and \( b\) is an odd natural number greater than one .
CASE(1). \( b \equiv 1,9 \pmod{12}\) , or \( b \equiv 3,7 \pmod{12}\) and \( p \equiv 1 \pmod 4\) , or \( b\equiv 5 \pmod{12}\) and
\( p \equiv 1,7 \pmod{12}\) , or \( b \equiv 11 \pmod{12}\) and \( p \equiv 1,11 \pmod{12}\) .
CASE(2). \( b \equiv 3,7 \pmod{12}\) and \( p \equiv 3 \pmod 4\) , or \( b\equiv 5 \pmod{12}\) and \( p \equiv 5,11 \pmod{12}\) , or
\( b \equiv 11 \pmod{12}\) and \( p \equiv 5,7 \pmod{12}\) .
Let \( S_i=P_b(S_{i-1})\) with \( S_0=P_b(4)\) . Suppose \( N_p(b)\) is prime , then :
\( \bullet\) \( S_{p-1} \equiv P_{b}(4) \pmod {N_p(b)}\) if Case(1) holds ;
\( \bullet\) \( S_{p-1} \equiv P_{b+2}(4) \pmod {N_p(b)}\) if Case(2) holds ;
Let \( P_m(x)=2^{-m}\cdot \left(\left(x-\sqrt{x^2-4}\right)^{m}+\left(x+\sqrt{x^2-4}\right)^{m}\right)\)
Let \( M_p(a)=\frac{a^p-1}{a-1}\) , where \( p\) is an odd prime and \( a\) is an odd natural number greater than one .
CASE(1). \( a \equiv 3,11 \pmod{12}\) , or \( a \equiv 5,9 \pmod{12}\) and \( p \equiv 1 \pmod 4\) , or \( a\equiv 7 \pmod{12}\) and
\( p \equiv 1,7 \pmod{12}\) , or \( a \equiv 1 \pmod{12}\) and \( p \equiv 1,11 \pmod{12}\) .
CASE(2). \( a \equiv 5,9 \pmod{12}\) and \( p \equiv 3 \pmod 4\) , or \( a\equiv 7 \pmod{12}\) and \( p \equiv 5,11 \pmod{12}\) , or
\( a \equiv 1 \pmod{12}\) and \( p \equiv 5,7 \pmod{12}\) .
Let \( S_i=P_a(S_{i-1})\) with \( S_0=P_a(4)\) . Suppose \( M_p(a)\) is prime , then :
\( \bullet\) \( S_{p-1} \equiv P_{a}(4) \pmod {M_p(a)}\) if Case(1) holds ;
\( \bullet\) \( S_{p-1} \equiv P_{a-2}(4) \pmod {M_p(a)}\) if Case(2) holds ;
Given integer $P$ , where $P>1$ let $S_k=P \cdot S_{k-1}-(2P-1) \cdot S_{k-2}+P \cdot S_{k-3}$ with $S_0=0$ , $S_1=1$ , $S_2=P-1$ .
Let $n$ be an odd natural number greater than $2$ such that $\operatorname{gcd}(P,n)=1$. Let $\left(\frac{D}{n}\right)$ be the Jacobi symbol where $D$ represents the discriminant of the characteristic polynomial: $x^3-Px^2+(2P-1)x-P$, and let $\delta(n)=n-\left(\frac{D}{n}\right)$ , then if $n$ is a prime then $S_{\delta(n)} \equiv 0 \pmod{n}$ .
Let \( b\) and \( n\) be a natural numbers , \( b\geq 2\) , \( n>2\) and \( n \neq 9\) . Then \( n\) is prime if and only if
\( \displaystyle\sum_{k=1}^{n-1}\left(b^k-1\right)^{n-1} \equiv n \pmod{\frac{b^n-1}{b-1}}\)
Let \( b\) and \( n\) be a natural numbers , \( b\geq 2\) , \( n>1\) and \( n \not\in \{4,8,9\}\) . Then \( n\) is prime if and only if
\( \displaystyle\sum_{k=1}^{n}\left(b^k+1\right)^{n-1} \equiv n \pmod{\frac{b^n-1}{b-1}}\)
If \( q\) is the smallest prime greater than \( \displaystyle\prod_{i=1}^n C_i+1\) , where \( \displaystyle\prod_{i=1}^n C_i\) is the product of the first \( n\) composite numbers,
then \( q-\displaystyle\prod_{i=1}^n C_i\) is prime.
If \( q\) is the greatest prime less than \( \displaystyle\prod_{i=1}^n C_i-1\) , where \( \displaystyle\prod_{i=1}^n C_i\) is the product of the first \( n\) composite numbers,
then \( \displaystyle\prod_{i=1}^n C_i-q\) is prime .
Let \( n\) be a natural number greater than two . Let \( r\) be the smallest odd prime number such that \( r \nmid n\) and \( n^2 \not\equiv 1 \pmod r\) .
Let \( T_n(x)\) be Chebyshev polynomial of the first kind , then \( n\) is a prime number if and only if \( T_n(x) \equiv x^n \pmod {x^r-1,n}\) .
Let \( n\) be a natural number greater than one and let \( F_{n}(x)\) be Fibonacci polynomial , then \( n\) is prime if and only if :
\( \displaystyle\sum_{k=0}^{n-1}F_{n}(k) \equiv -1 \pmod n\) .
Let \( P_m(x)=2^{-m}\cdot \left(\left(x-\sqrt{x^2-4}\right)^{m}+\left(x+\sqrt{x^2-4}\right)^{m}\right)\)
Let \( F_n(b)= b^{2^n}+1 \) where \( b\) is an even natural number and \( n\ge2\) . Let \( a\) be a natural number greater
than two such that \( \left(\frac{a-2}{F_n(b)}\right)=-1\) and \( \left(\frac{a+2}{F_n(b)}\right)=-1\) where \( \left(\frac{}{}\right)\) denotes Jacobi symbol. Let \( S_i=P_b(S_{i-1})\)
with \( S_0\) equal to the modular \( P_{b/2}(P_{b/2}(a))\phantom{5} \text{mod} \phantom{5} F_n(b)\). Then \( F_n(b)\) is prime if and only if \( S_{2^n-2} \equiv 0 \pmod{F_n(b)}\) .
Let \( P_m(x)=2^{-m}\cdot \left(\left(x-\sqrt{x^2-4}\right)^{m}+\left(x+\sqrt{x^2-4}\right)^{m}\right)\)
Let $N= 4kp^{n} - 1 $ where $k$ is a positive natural number , $ 4k<2^n$ , $p$ is a prime number and $n\ge3$ .
Let $a$ be a natural number greater than two such that $\left(\frac{a-2}{N}\right)=1$ and $\left(\frac{a+2}{N}\right)=-1$ where $\left(\frac{}{}\right)$ denotes Jacobi symbol.
Let $S_i=P_p(S_{i-1})$ with $S_0$ equal to the modular $P_{kp^2}(a)\phantom{5} \text{mod} \phantom{5} N$. Then $N$ is prime if and only if $S_{n-2} \equiv 0 \pmod{N}$ .
Let \( P_m(x)=2^{-m}\cdot \left(\left(x-\sqrt{x^2-4}\right)^{m}+\left(x+\sqrt{x^2-4}\right)^{m}\right)\)
Let $N= 4kp^{n} + 1 $ where $k$ is a positive natural number , $ 4k<2^n$ , $p$ is a prime number and $n\ge3$ .
Let $a$ be a natural number greater than two such that $\left(\frac{a-2}{N}\right)=-1$ and $\left(\frac{a+2}{N}\right)=-1$ where $\left(\frac{}{}\right)$ denotes Jacobi symbol.
Let $S_i=P_p(S_{i-1})$ with $S_0$ equal to the modular $P_{kp^2}(a)\phantom{5} \text{mod} \phantom{5} N$. Then $N$ is prime if and only if $S_{n-2} \equiv 0 \pmod{N}$ .
Let \( P_m(x)=2^{-m}\cdot \left(\left(x-\sqrt{x^2-4}\right)^{m}+\left(x+\sqrt{x^2-4}\right)^{m}\right)\)
Let \( M_p(a)= \frac{a^p-1}{a-1} \) where \( a\) is a natural number greater than one and \( p\) is an odd prime number . Let \( c\) be
a natural number greater than two such that \( \left(\frac{c-2}{M_p(a)}\right)=-1\) and \( \left(\frac{c+2}{M_p(a)}\right)=1\) where \( \left(\frac{}{}\right)\) denotes Jacobi
symbol. Let \( S_i=P_a(S_{i-1})\) with \( S_0 =P_{a}(c)\). Then , if \( M_p(a)\) is prime then \( S_{p-1} \equiv P_{a-2}(c) \pmod{M_p(a)}\) .
Let \( P_m(x)=2^{-m}\cdot \left(\left(x-\sqrt{x^2-4}\right)^{m}+\left(x+\sqrt{x^2-4}\right)^{m}\right)\)
Let \( N_p(b)= \frac{b^p+1}{b+1} \) where \( b\) is a natural number greater than one and \( p\) is an odd prime number . Let \( c\) be
a natural number greater than two such that \( \left(\frac{c-2}{N_p(b)}\right)=-1\) and \( \left(\frac{c+2}{N_p(b)}\right)=1\) where \( \left(\frac{}{}\right)\) denotes Jacobi
symbol. Let \( S_i=P_b(S_{i-1})\) with \( S_0 =P_{b}(c)\). Then , if \( N_p(b)\) is prime then \( S_{p-1} \equiv P_{b+2}(c) \pmod{N_p(b)}\) .
Let \( P_m(x)=2^{-m}\cdot \left(\left(x-\sqrt{x^2-4}\right)^{m}+\left(x+\sqrt{x^2-4}\right)^{m}\right)\)
Let \( M= k \cdot b^{n}-c \) where \( k,b,n,c\) are natural numbers such that \( k>0\) , \( b>1\) , \( n>1\) and \( c>0\). Let \( a\) be a
natural number greater than two such that \( \left(\frac{a-2}{M}\right)=-1\) and \( \left(\frac{a+2}{M}\right)=1\) where \( \left(\frac{}{}\right)\) denotes Jacobi symbol.
Let \( S_i=P_b(S_{i-1})\) with \( S_0\) equal to the modular \( P_{kb}(a)\phantom{5} \text{mod} \phantom{5} M\). Then, if \( M\) is prime then
\( S_{n-1} \equiv P_{c-1}(a) \pmod{M}\) .
Let \( P_m(x)=2^{-m}\cdot \left(\left(x-\sqrt{x^2-4}\right)^{m}+\left(x+\sqrt{x^2-4}\right)^{m}\right)\)
Let \( N= k \cdot b^{n}+c \) where \( k,b,n,c\) are natural numbers such that \( k>0\) , \( b>1\) , \( n>1\) and \( c>0\) . Let \( a\) be a
natural number greater than two such that \( \left(\frac{a-2}{N}\right)=1\) and \( \left(\frac{a+2}{N}\right)=1\) where \( \left(\frac{}{}\right)\) denotes Jacobi symbol.
Let \( S_i=P_b(S_{i-1})\) with \( S_0\) equal to the modular \( P_{kb}(a)\phantom{5} \text{mod} \phantom{5} N\). Then, if \( N\) is prime then
\( S_{n-1} \equiv P_{c-1}(a) \pmod{N}\) .
Let $b_n=b_{n-1}+\operatorname{lcm}(\lfloor \sqrt{2} \cdot n \rfloor , b_{n-1})$ with $b_1=2$ and $n>1$ . Let $a_n=b_{n+1}/b_n-1$ .
1. Every term of this sequence $a_i$ is either prime or $1$ .
2. Every odd prime of the form $\left\lfloor \sqrt{2}\cdot n \right\rfloor$ is a term of this sequence.
3. At the first appearance of each prime of the form $\left\lfloor \sqrt{2}\cdot n \right\rfloor$ greater than $5$, it is the next prime of the given form after the largest prime that has already appeared.
Let $b_n=b_{n-1}+\operatorname{lcm}(\lfloor \sqrt{3} \cdot n \rfloor , b_{n-1})$ with $b_1=3$ and $n>1$ . Let $a_n=b_{n+1}/b_n-1$ .
1. Every term of this sequence $a_i$ is either prime or $1$ .
2. Every odd prime of the form $\left\lfloor \sqrt{3}\cdot n \right\rfloor$ greater than $3$ is a term of this sequence.
3. At the first appearance of each prime of the form $\left\lfloor \sqrt{3}\cdot n \right\rfloor$ greater than $5$, it is the next prime of the given form after the largest prime that has already appeared.
Let $b_n=b_{n-1}+\operatorname{lcm}(\lfloor \sqrt{n^3} \rfloor , b_{n-1})$ with $b_1=2$ and $n>1$ . Let $a_n=b_{n+1}/b_n-1$ .
1. Every term of this sequence $a_i$ is either prime or $1$ .
2. Every odd prime of the form $\lfloor \sqrt{n^3} \rfloor $ is member of this sequence .
3. At the first appearance of each prime of the form $\lfloor \sqrt{n^3} \rfloor $ greater than $5$, it is the next prime of the given form after the largest prime that has already appeared.
Let $a$ , $b$ and $n$ be a natural numbers , $b>a>1$ , $n>2$ and $n \not\in \{4,9,25\}$ .
Then $n$ is prime iff $\displaystyle \prod_{k=1}^{n-1} \left(b^k-a\right) \equiv \frac{a^n-1}{a-1} \pmod{\frac{b^n-1}{b-1}}$
Let $a$ , $b$ and $n$ be a natural numbers , $b>a>0$ , $n>2$ and $n \not\in \{4,9,25\}$ .
Then $n$ is prime iff $\displaystyle \prod_{k=1}^{n-1} \left(b^k+a\right) \equiv \frac{a^n+1}{a+1} \pmod{\frac{b^n-1}{b-1}}$
Let $P_m(x)=2^{-m}\cdot \left(\left(x-\sqrt{x^2-4}\right)^{m}+\left(x+\sqrt{x^2-4}\right)^{m}\right)$ .
Let $N=8kp^n-1$ such that $k>0$ , $3 \not\mid k$ , $p$ is a prime number, $p \neq 3$ , $n > 2$ and $ 8k < p^n $ .
Let $S_i=P_p(S_{i-1})$ with $S_0=P_{2kp^2}(4)$ , then: $N$ is a prime iff $S_{n-2} \equiv 0\pmod{N}$
Let $P_m(x)=2^{-m}\cdot \left(\left(x-\sqrt{x^2-4}\right)^{m}+\left(x+\sqrt{x^2-4}\right)^{m}\right)$ .
Let $N=8k3^n-1$ such that $n>2$ , $k>0$ , $ 8k < 3^n $ and
$\begin{cases} k \equiv 1 \pmod{5} \text{ with } n \equiv 0,1 \pmod{4} \\ k \equiv 2 \pmod{5} \text{ with } n \equiv 1,2 \pmod{4} \\ k \equiv 3 \pmod{5} \text{ with } n \equiv 0,3 \pmod{4} \\ k \equiv 4 \pmod{5} \text{ with } n \equiv 2,3 \pmod{4} \end{cases}$
Let $S_i=S_{i-1}^3-3S_{i-1}$ with $S_0=P_{18k}(3)$ , then $N$ is prime iff $S_{n-2} \equiv 0 \pmod N$ .
Let $n$ be an odd natural number that is not a perfect square and let $c$ be the smallest odd prime number such that
$\left(\frac{c}{n}\right)=-1$, where $\left(\frac{}{}\right)$ denotes a Jacobi symbol . Let $T_m(x)$ be the mth Chebyshev polynomial of the first kind,
then $n$ is prime iff $T_n\left(1+\sqrt{c}\right) \equiv 1- \sqrt{c} \pmod{n}$ .
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