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%I #38 Sep 27 2015 10:32:31
%S 1,2,6,30,30,510,510,510,510,131070,131070,131070,131070,131070,
%T 131070,131070,131070,8589934590,8589934590,8589934590,8589934590,
%U 8589934590,8589934590,8589934590,8589934590,8589934590,8589934590,8589934590,8589934590,8589934590,8589934590
%N A subsequence of the denominators of the Bernoulli numbers: a(n) = A027642(A131577(n)).
%C Conjecture: a(15) = a(16) = 131070, a(17) through a(32) = 8589934590.
%C Number of different terms: 1, 1, 1, 2, 4, ... = abs(A141531)?
%C Factorization of terms from 2:
%C 2 = 2
%C 6 = 2*3
%C 30 = 2*3*5
%C 510 = 2*3*5*17
%C 131070 = 2*3*5*17*257
%C 8589934590 = 2*3*5*17*257*65537.
%C Note that all factors shown are 2 or Fermat numbers (see A092506, A019434, A000215).
%C Empirical: using the von Staudt-Clausen theorem, terms a(17) through a(4215) are all 8589934590. - _Simon Plouffe_, Sep 20 2015
%C 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
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/vonStaudt-ClausenTheorem.html">von Staudt-Clausen Theorem</a>
%t 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 *)
%o (PARI) a(n) = denominator(bernfrac(1<<n--)); \\ _Michel Marcus_, Aug 14 2013
%Y Cf. A092506, A019434, A000215.
%K nonn
%O 0,2
%A _Paul Curtz_, Nov 14 2012
%E Extended up to a(20) by _Jean-François Alcover_, Dec 07 2012
%E More terms from _Michel Marcus_, Sep 27 2015