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244Mersenne Primes

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  • Paul Mills
    Jan 22, 2001
      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

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


      Paul Mills.
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