25336Re: The primeness of even n compared to odd n in k*2^n+/-1
- Aug 21 2:37 PM--- In email@example.com, "robert44444uk" <robert_smith44@...> wrote:
>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.
> 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.
> 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.
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.
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