A219196 A subsequence of the denominators of the Bernoulli numbers: a(n) = A027642(A131577(n)).

1, 2, 6, 30, 30, 510, 510, 510, 510, 131070, 131070, 131070, 131070, 131070, 131070, 131070, 131070, 8589934590, 8589934590, 8589934590, 8589934590, 8589934590, 8589934590, 8589934590, 8589934590, 8589934590, 8589934590, 8589934590, 8589934590, 8589934590, 8589934590

Offset: 0

Author

Paul Curtz, Nov 14 2012

Keywords

• nonn

Comments

Conjecture: a(15) = a(16) = 131070, a(17) through a(32) = 8589934590.

Number of different terms: 1, 1, 1, 2, 4, ... = abs(A141531)?

Factorization of terms from 2:

2 = 2

6 = 2*3

30 = 2*3*5

510 = 2*3*5*17

131070 = 2*3*5*17*257

8589934590 = 2*3*5*17*257*65537.

Note that all factors shown are 2 or Fermat numbers (see A092506, A019434, A000215).

Empirical: using the von Staudt-Clausen theorem, terms a(17) through a(4215) are all 8589934590. - Simon Plouffe, Sep 20 2015

Using the von Staudt-Clausen theorem, a(n) is the product of 2 and all Fermat primes <= 2^(n-1)+1: see A019434. The only known Fermat primes are 3,5,17,257,65537; it is known that there are no others < 2^(2^33)+1, so that a(n) = 8589934590 for n <= 2^33 = 8589934592. - Robert Israel, Sep 21 2015

Links

• Eric Weisstein's World of Mathematics, von Staudt-Clausen Theorem

Crossrefs

Cf. A092506, A019434, A000215.

Programs

Mathematica

a[n_] := a[n] = Times @@ Select[ Divisors[2^(n-1)] + 1, PrimeQ]; a[0] = 1; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Dec 07 2012 *)

PARI

a(n) = denominator(bernfrac(1<Michel Marcus, Aug 14 2013

Extensions

Extended up to a(20) by Jean-François Alcover, Dec 07 2012

More terms from Michel Marcus, Sep 27 2015