Loading ...
Sorry, an error occurred while loading the content.

25336Re: The primeness of even n compared to odd n in k*2^n+/-1

Expand Messages
  • thomas_ritschel
    Aug 21 2:37 PM
    • 0 Attachment
      --- In primenumbers@yahoogroups.com, "robert44444uk" <robert_smith44@...> wrote:
      > Recently I announced the discovery of a super-prime power series of the form k*2^n-1, n variable, k fixed, with over 200 primes to the Yahoo Primeform group.
      > http://tech.groups.yahoo.com/group/primeform/message/11407
      > An analysis of the primes discovered to date shows that there are significantly more even n in the list of primes (119) than odd (94). The complete list of the 213 primes to date is at:
      > http://www.mersenneforum.org/showthread.php?t=18407 message 5.

      The occurence of more even exponents in the list of primes is a direct result of the distrubution of even and odd n in the input file(s). After sieving to p=6.34B the input file for the range n=500,001-800,000 contains 30872 even and 26912 odd n candidates.
      That's a ratio of 53.4% : 46.6%.
      The ratio for the 213 primes is quite similar: 55.9% : 44.1%.

      The occurence of a specific n in a sieve file for a given k (and therefore even and/or odd n) is determined by the residue classes (k mod p) and (b^n mod p) and the relations between them (e.g. for eliminating a specific n in k*b^n-1 we need to have (k mod p) == 1/(b^n mod p) ).

      For simpler (smaller) k there might be cases with no even or odd exponents at all, for example: 5*2^n-1 (no odd exponents) or 7*2^n-1 (no even ones).

      I would therefore expect also some Payam sequences with an opposite distribution, e.g. more odd than even exponents.

      -- Thomas
    • Show all 2 messages in this topic