## 244Mersenne Primes

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• Jan 22, 2001
Hello,
Just to bring the Nathan Russell observation about a relationship
between Mersenne primes and Pascal's triangle to a neat
conclusion ( or is it the beginning..)  here is the
Russell-Mills formula for Mersenne primes.

Let [n,m] be the binomial coefficient = n!/(n-m)!m!

A = B * C
A = 2* `the sum from k= 0 to k = n/2 of' { [n,n-2k] *2^n-2k * 3^k }
B = `the sum from k = 0 to k = n-1 of' {2^k }
C = `the sum from k = 0 to k = m of' { a_k*2^k }

A is the Lucas term S_n-1 = (2 + sqrt(3))^2^n-2 +
(2  sqrt(3))^2^n-2 expanded using the Binomial Theorem.

B is the Mn Mersenne prime, you know it as (2^n) 1, here it is
the sum of
1 + 2 + 4 + ..2^(n-1) (neat, eh!)

C is the binary number ,what you get if Mn divides S_n-1 .

So, The Russell-Mills formula states that if A = B* C has a solution
for some values of the coefficients a_k and n is prime, then 2^n  1
is prime (a Mersenne prime).

Note that B and C are natural binary numbers and A would like to be a
binary number but we have to `unfold' the binomial
coefficients and the `powers of 3' onto a binary number which
factorises into B and C. Also the summation term m in C is ` a
large enough value' for the `power of 2' terms to balance on either
side of the
equation.

Would anyone like to investigate the relationships between the
coefficients a_k so as to ensure a solution?

Regards,

Paul Mills.
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