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A177735
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a(0)=1, a(n)=A002445(n)/6 for n>=1.
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3
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1, 1, 5, 7, 5, 11, 455, 1, 85, 133, 55, 23, 455, 1, 145, 2387, 85, 1, 319865, 1, 2255, 301, 115, 47, 7735, 11, 265, 133, 145, 59, 9464455, 1, 85, 10787, 5, 781, 23350145, 1, 5, 553, 38335, 83, 567385, 1, 10235, 45353, 235, 1, 750295, 1, 5555, 721, 265, 107
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OFFSET
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0,3
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COMMENTS
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For n>=1: denominators of the Bernoulli numbers (A002445) divided by 6.
All entries are odd.
5 divides a(2*n) for n>=1.
These numbers also equal to the lengths of the repeating patterns for the excluded integer values of c/6, when both p^n + c and p^n - c are prime, for an infinite number of primes p>2, and a given integer n>0, arising from the union of one or more prime-based modulo cycles, determined by the divisors of n. See A005097 for details and connection to the von Staudt-Clausen Theorem below. - Richard R. Forberg, Jul 19 2016
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LINKS
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FORMULA
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A simple direct calculation of the denominators, for n>=1, is based on the von Staudt-Clausen Theorem: Product{d|n}(2d+1), for d>1 and 2d+1 prime. See in the Mathematica section below. - Richard R. Forberg, Jul 19 2016
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MAPLE
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A002445 := proc(n) bernoulli(2*n) ; denom(%) ; end proc:
A177735 := proc(n) if n = 0 then 1; else A002445(n)/6 ; end if; end proc:
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MATHEMATICA
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Join[{1}, Denominator[BernoulliB[Range[2, 120, 2]]]/6] (* Harvey P. Dale, Oct 19 2012 *)
result = {}; Do[prod = 1; Do[If[PrimeQ[2*Divisors[n][[i]] + 1], prod *= (2*Divisors[n][[i]] + 1)], {i, 2, Length[Divisors[n]]}];
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PROG
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(PARI)
a(n)=
{
my(bd=1);
forprime (p=5, 2*n+1, if( (2*n)%(p-1)==0, bd*=p ) );
bd;
}
(PARI) a(n)=if(n<2, return(1)); my(s=1); fordiv(n, d, if(isprime(2*d+1) && d>1, s *= 2*d+1)); s \\ Charles R Greathouse IV, Jul 20 2016
(Sage)
if n == 0: return 1
M = map(lambda i: i+1, divisors(2*n))
return mul(filter(lambda s: is_prime(s), M))//6
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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