This site is supported by donations to The OEIS Foundation.

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

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

ie.

${|k| \ne 2, 6, 10, 14, 18 \dots 4n-2 \quad or \quad |2k| \ne 4, 12, 20, 28, 36 \dots 8n-4 \quad | n \ge 1, n \in \mathbb{N}_1}$

Mersenne numbers (nonprimes): binomially or triangular !?

${2^p-1 = q_1 * q_2 \quad {\color{Red}\longrightarrow} \quad q_1 = c + b, \quad q_2 = c - b}, \quad c=\frac{(c+b)+(c-b)}{2}=\frac{q_1+q_2}{2}, \quad b=\frac{(c+b)-(c-b)}{2}=\frac{q_1-q_2}{2}$

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

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

${{\color{Red}p = 11 \longrightarrow 2^{11}-1 = c^2-b^2=(\frac{q_1+q_2}{2})^2-(\frac{q_1-q_2}{2})^2=(\frac{89+23}{2})^2-(\frac{89-23}{2})^2=56^2 - 33^2 \longrightarrow 2047 = 3136 - 1089}}$

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

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

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

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

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

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

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

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

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

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

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

Wieferich primes

${WP_1-1 = 1092_{10} = 0100.0100.0100_2 = 444_{16}{^{[1]}} = 444_{{\color{Red}2^4}}}$
${WP_2-1 = 3510_{10} = 110.110.110.110_2 = 6666_8{^{[1]}} = 6666_{{\color{Red}2^3}}}$

A new property of Wieferich primes:

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

${\mbox{base of divider } D_1={\color{Red}2^4-1} \quad and \quad D_2={\color{Red}2^3-1}}$

${\mbox{base of factor } F_1={\color{Red}4} \quad and \quad F_2={\color{Red}3}*2 \quad \big| \mbox{ odd factor multiplied by 2}}$

${WP_{1;2} = \frac{E}{D_{1;2}}*F_{1;2}+1} \quad \longrightarrow \frac{E}{D_{1;2}}=\frac{2^m}{2^{n_{1;2}}}=(2^{k-1})^{n_{1;2}}+(2^{k-2})^{n_{1;2}}+...+(2^0)^{n_{1;2}} \quad \big| k=\frac{m}{n_{1;2}}, \quad m=n_1*n_2$

${WP_1= \frac{{\color{Red}2^{12}-1}}{{\color{Red}2^4-1}}*{\color{Red}4}+1= \frac{4095}{15}*4+1=273*4+1=1093} \quad \longrightarrow \frac{2^{12}}{2^4}=(2^{3-1})^4+(2^{3-2})^4+...+(2^0)^4=273 \quad \big| k=\frac{12}{4}=3$

${WP_2= \frac{{\color{Red}2^{12}-1}}{{\color{Red}2^3-1}}*{\color{Red}3}*2+1= \frac{4095}{7}*6+1=585*6+1=3511} \quad \longrightarrow \frac{2^{12}}{2^3}=(2^{4-1})^3+(2^{4-2})^3+...+(2^0)^3=585 \quad \big| k=\frac{12}{3}=4$

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

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

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

My last hunch is that the exponent |b| is calculated as follows:

${D_1=2^{d_1}-1 \longrightarrow 2^4-1} \quad \big| d_1=2^n$

${D_2=2^{d_2}-1 \longrightarrow 2^3-1} \quad \big| d_2=3^{n-1}$

${\big| b \big| \longrightarrow b+1=p \longrightarrow M_p=M_{b+1} \quad and \quad b=d_1*d_2=2^n*3^{n-1} \quad \big| n \ge 1, n \in \mathbb{N}_1}$

Only for n = 2 there an appropriate solution. If 2b + 1 − 1 the next Mersenne prime, than gives the next 2 Wieferich primes with

${WP_1= \frac{2^b-1}{2^{d_1}-1}*d_1+1 \quad and \quad WP_2= \frac{2^b-1}{2^{d_2}-1}*d_2*2+1}$

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

• his formula for the Wieferich primes is
• "1092 = 4 * (163 − 1) / (16 − 1) (base 10)"
• "3510 = 6 * (84 − 1) / (8 − 1) (base 10)"