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.