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# User:A. Lamek

## My interests

I come from Germany, NRW.

I'm interested in numbers, especially Mersenne Primes and Wieferich Primes. However my level of understanding of number theory is nowhere near that of a mathematicians. Please have little bit of patience with me. Email: blackpuma65@googlemail.com

## My favored sequences

1. Mersenne Primes A000043, A247473
2. Wieferich Primes A001220, A001576

## My assumptions

Mersenne primes: Factors

Mersenne prime numbers ${\displaystyle {2^{p}-1}}$, which are not prime, divided by ${\displaystyle {2kp+1}}$. In this known formula can ${\displaystyle {|k|}}$ the values 1, 2, 3, 4, etc.
So that the divider ${\displaystyle {2p+1}}$, ${\displaystyle {4p+1}}$, ${\displaystyle {6p+1}}$, ${\displaystyle {8p+1}}$ etc could be.
According to my observation, there are, however, gaps in ${\displaystyle {|k|}}$, where no divider occur. These gaps have a mathematically calculable uniformity:

${\displaystyle {k\equiv 2{\pmod {4}}}\longrightarrow 2kp+1\nmid 2^{p}-1}$

ie.

${\displaystyle {|k|\neq 2,6,10,14,18\dots 4n-2\quad or\quad |2k|\neq 4,12,20,28,36\dots 8n-4\quad |n\geq 1,n\in \mathbb {N} _{1}}}$

Mersenne numbers (nonprimes): binomially or triangular !?

${\displaystyle {2^{p}-1=q_{1}*q_{2}\quad {\color {Red}\longrightarrow }\quad q_{1}=c+b,\quad q_{2}=c-b},\quad c={\frac {(c+b)+(c-b)}{2}}={\frac {q_{1}+q_{2}}{2}},\quad b={\frac {(c+b)-(c-b)}{2}}={\frac {q_{1}-q_{2}}{2}}}$

${\displaystyle {2^{p}-1=(c+b)*(c-b)\quad {\color {Red}\longrightarrow }\quad 2^{p}-1=c^{2}-b^{2}}}$

${\displaystyle {\color {Red}{\mbox{for example: }}}}$

${\displaystyle {\color {Red}p=11\longrightarrow 2^{11}-1=c^{2}-b^{2}=({\frac {q_{1}+q_{2}}{2}})^{2}-({\frac {q_{1}-q_{2}}{2}})^{2}=({\frac {89+23}{2}})^{2}-({\frac {89-23}{2}})^{2}=56^{2}-33^{2}\longrightarrow 2047=3136-1089}}$

${\displaystyle {\color {Red}p=23\longrightarrow 2^{23}-1=89264^{2}-89217^{2}\longrightarrow 8388607=7968061696-7959673089}}$

${\displaystyle {c=mp+1,\quad b=np,\quad 2^{p}-1=xp+1\quad {\color {Red}\longrightarrow }\quad x={\frac {2^{p}-2}{p}}}}$

${\displaystyle {2^{p}-1=(mp+1)^{2}-n^{2}p^{2}\quad {\color {Red}\longrightarrow }\quad xp+1=m^{2}p^{2}+2mp+1-n^{2}p^{2}\quad {\color {Red}\longrightarrow }}}$

${\displaystyle {n^{2}p^{2}=m^{2}p^{2}+2mp+1-(xp+1)\quad {\color {Red}\longrightarrow }\quad n^{2}p^{2}=m^{2}p^{2}+2mp-xp\quad {\color {Red}\longrightarrow }}}$

${\displaystyle {n^{2}p=m^{2}p+2m-x\quad {\color {Red}\longrightarrow }\quad n^{2}p=m^{2}p-(x-2m)\quad {\color {Red}\longrightarrow }}}$

${\displaystyle {n^{2}=m^{2}-{\frac {x-2m}{p}}\quad {\color {Red}\longleftarrow }\quad p\mid (x-2m),\quad 2\nmid m,n}}$

${\displaystyle {\color {Red}{\mbox{for example: }}}}$

${\displaystyle {\color {Red}p=11,\quad n=3,\quad m=5,\quad x=2046/11=186}}$

${\displaystyle {\color {Red}3^{2}=5^{2}-{\frac {186-2*5}{11}}\longrightarrow 9=25-{\frac {176}{11}}\longrightarrow 9=25-16\quad {\color {Red}\longleftarrow }\quad {\mbox{ (probably the only Pythagorean form of }}a^{2}+b^{2}=c^{2})}}$

${\displaystyle {\color {Red}p=23,\quad n=3879,\quad m=3881,\quad x=8388606/23=364722}}$

${\displaystyle {\color {Red}3879^{2}=3881^{2}-{\frac {364722-2*3881}{23}}\longrightarrow 15046641=15062161-{\frac {356960}{23}}\longrightarrow 15046641=15062161-15520}}$

Wieferich primes

${\displaystyle {WP_{1}-1=1092_{10}=0100.0100.0100_{2}=444_{16}{^{[1]}}=444_{\color {Red}2^{4}}}}$
${\displaystyle {WP_{2}-1=3510_{10}=110.110.110.110_{2}=6666_{8}{^{[1]}}=6666_{\color {Red}2^{3}}}}$

A new property of Wieferich primes:

${\displaystyle {{\mbox{base of exponent }}E\longrightarrow {\color {Red}2^{4}}\quad and\quad {\color {Red}2^{3}}\longrightarrow {\color {Red}2^{(4*3)}-1}={\color {Red}2^{12}-1}}}$

${\displaystyle {{\mbox{base of divider }}D_{1}={\color {Red}2^{4}-1}\quad and\quad D_{2}={\color {Red}2^{3}-1}}}$

${\displaystyle {{\mbox{base of factor }}F_{1}={\color {Red}4}\quad and\quad F_{2}={\color {Red}3}*2\quad {\big |}{\mbox{ odd factor multiplied by 2}}}}$

${\displaystyle {WP_{1;2}={\frac {E}{D_{1;2}}}*F_{1;2}+1}\quad \longrightarrow {\frac {E}{D_{1;2}}}={\frac {2^{m}}{2^{n_{1;2}}}}=(2^{k-1})^{n_{1;2}}+(2^{k-2})^{n_{1;2}}+...+(2^{0})^{n_{1;2}}\quad {\big |}k={\frac {m}{n_{1;2}}},\quad m=n_{1}*n_{2}}$

${\displaystyle {WP_{1}={\frac {\color {Red}2^{12}-1}{\color {Red}2^{4}-1}}*{\color {Red}4}+1={\frac {4095}{15}}*4+1=273*4+1=1093}\quad \longrightarrow {\frac {2^{12}}{2^{4}}}=(2^{3-1})^{4}+(2^{3-2})^{4}+...+(2^{0})^{4}=273\quad {\big |}k={\frac {12}{4}}=3}$

${\displaystyle {WP_{2}={\frac {\color {Red}2^{12}-1}{\color {Red}2^{3}-1}}*{\color {Red}3}*2+1={\frac {4095}{7}}*6+1=585*6+1=3511}\quad \longrightarrow {\frac {2^{12}}{2^{3}}}=(2^{4-1})^{3}+(2^{4-2})^{3}+...+(2^{0})^{3}=585\quad {\big |}k={\frac {12}{3}}=4}$

This formula shows that from one base of exponent |E|

${\displaystyle {E=2^{b}-1\longrightarrow 2^{12}-1}}$

2 Wieferich primes (1093, 3511) can be calculated.

My last hunch is that the exponent |${\displaystyle {b}}$| is calculated as follows:

${\displaystyle {D_{1}=2^{d_{1}}-1\longrightarrow 2^{4}-1}\quad {\big |}d_{1}=2^{n}}$

${\displaystyle {D_{2}=2^{d_{2}}-1\longrightarrow 2^{3}-1}\quad {\big |}d_{2}=3^{n-1}}$

${\displaystyle {{\big |}b{\big |}\longrightarrow b+1=p\longrightarrow M_{p}=M_{b+1}\quad and\quad b=d_{1}*d_{2}=2^{n}*3^{n-1}\quad {\big |}n\geq 1,n\in \mathbb {N} _{1}}}$

Only for ${\displaystyle {n=2}}$ there an appropriate solution. If ${\displaystyle {2^{b+1}-1}}$ the next Mersenne prime, than gives the next 2 Wieferich primes with

${\displaystyle {WP_{1}={\frac {2^{b}-1}{2^{d_{1}}-1}}*d_{1}+1\quad and\quad WP_{2}={\frac {2^{b}-1}{2^{d_{2}}-1}}*d_{2}*2+1}}$

footnote:
[1] by Johnson (1977) (see A note on the two known Wieferich Primes)

• his formula for the Wieferich primes is
• "${\displaystyle {1092=4*(16^{3}-1)/(16-1)}}$ (base 10)"
• "${\displaystyle {3510=6*(8^{4}-1)/(8-1)}}$ (base 10)"