A165949 -id:A165949 - cp's OEIS frontend

A177735 a(0)=1, a(n)=A002445(n)/6 for n>=1.

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

Offset: 0

Paul Curtz, May 12 2010

nonn

For n>=1: denominators of the Bernoulli numbers (A002445) divided by 6.
All entries are odd.
a(n)= A002445(n) / A020793(n).
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

Harvey P. Dale, Table of n, a(n) for n = 0..1000
C. M. Bender and K. A. Milton, Continued fraction as a discrete nonlinear transform, arXiv:hep-th/9304062, 1993.
Eric Weisstein's World of Mathematics, von Staudt-Clausen Theorem

Cf. A002445, A165949.

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:
seq(A177735(n),n=0..60) ; # R. J. Mathar, Aug 15 2010

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]]}];
AppendTo[result, prod] , {n, 1, 100}]  ; result (* Richard R. Forberg, Jul 19 2016 *)

a(n)=
{
    my(bd=1);
    forprime (p=5, 2*n+1, if( (2*n)%(p-1)==0, bd*=p ) );
    bd;
}
/* Joerg Arndt, May 06 2012 */

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

def A177735(n):
    if n == 0: return 1
    M = map(lambda i: i+1, divisors(2*n))
    return mul(filter(lambda s: is_prime(s), M))//6
print([A177735(n) for n in (0..53)]) # Peter Luschny, Feb 20 2016

a(n) = denominator(BernoulliB(2*n, 1/2))/(3*2^(2*n)). - Jean-François Alcover, Apr 16 2013

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

A216639 A027642(6*n+6)/(sequence of period 2:repeat 42,210).

Original entry on oeis.org

1, 13, 19, 13, 341, 9139, 43, 221, 19, 270413, 1541, 667147, 79, 16211, 6479, 21437, 103, 996151, 1, 11086933, 103759, 20033, 6533, 11341499, 51491, 8545667, 3097, 16211, 59, 34408161359, 1, 4137341, 5826521, 1339, 219666403, 72719023, 223, 2977, 1501, 45423164501, 83

Offset: 0

Paul Curtz, Sep 12 2012

nonn

Is a(n) always an integer?  Is there an a(n) ending with 5?
It appears (tested for n <= 800) that a(n) mod 9 is always one of {1, 2, 4, 5, 7, 8}.
There is a similar sequence of ratios A027642(10n+1)/(66*A010686(n)) which starts 1, 1, 217, 41, 1, 172081, 71, 697, 4123, 101, 23, 7055321, 131, 2059, 32767, 697, 1, 21896102683,...
a(n) is always an integer: 42 = 2*3*7 and 1, 2, and 6 divide 12n+6; 210 = 2*3*5*7 and 1, 2, 4, and 6 divide 12n+12. a(n) never ends in 5 (or 0) since 12n+6 is not divisible by 4 hence the (12n+6)-th Bernoulli denominator is not divisible by 5, and Bernoulli denominators are squarefree and hence the (12n+12)-th Bernoulli denominator, divided by 210, cannot be divisible by 5. - Charles R Greathouse IV, Sep 12 2012
The previous comments argue that 3 or 5 are never prime divisors of a(n). In addition (tested up to n <=900), 7 apparently is also a non-divisor of a(n). In summary, the prime divisors appear all to be in A140461. - Jean-François Alcover, Sep 17 2012

Charles R Greathouse IV, Table of n, a(n) for n = 0..10000
A. Joyal, Les nombres de Bernoulli (in French), 2003.

Cf. A002445, A027762, A165734, A165949.

A010686 := proc(n)
    op((n mod 2)+1,[1,5]) ;
end proc:
A216639 := proc(n)
    A027642(6*n+6)/42/A010686(n) ;
end proc: # R. J. Mathar, Sep 21 2012

a(n)=denominator(bernfrac(6*n+6))/if(n%2,210,42) \\ Charles R Greathouse IV, Sep 12 2012

a(n)=my(t=1);fordiv(3*n+3,d,if(isprime(2*d+1),t*=2*d+1));t/if(n%2,105,21) \\ Charles R Greathouse IV, Sep 12 2012

a(n) = A027642(6*n+6)/(42*A010686(n)).

a(20)-a(40) from Charles R Greathouse IV, Sep 12 2012

A212655 Denominator of Bernoulli(2*n,1/2) / Period of length 2: repeat 12, 60.

Original entry on oeis.org

1, 4, 112, 64, 2816, 93184, 4096, 278528, 8716288, 2883584

Offset: 1

Paul Curtz, Apr 14 2013

See A165949(n) = (A027642(n+1)=A027762(n))/A165734(n).

a(n) is divisible by 4^(n-1).

			a(1) = (B(2,1/2)=12)/12=1, a(2)=240/60=4, a(3)=1344/12=112, a(4)=3840/60=64.

Cf. A000302.

a(n) = A033469(n)/A040874(n).

a(n) = 4^(n-1) * A165949(n).

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A177735 a(0)=1, a(n)=A002445(n)/6 for n>=1.

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A216639 A027642(6*n+6)/(sequence of period 2:repeat 42,210).

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A212655 Denominator of Bernoulli(2*n,1/2) / Period of length 2: repeat 12, 60.

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