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 A219196 A subsequence of the denominators of the Bernoulli numbers: a(n) = A027642(A131577(n)). 0
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 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 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 *) PROG (PARI) a(n) = denominator(bernfrac(1<

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