%I #42 Sep 21 2016 22:55:18

%S 1,12,240,1344,3840,33792,5591040,49152,16711680,104595456,173015040,

%T 289406976,22900899840,201326592,116769423360,7689065201664,

%U 1095216660480,51539607552,65942866278481920,824633720832,7438196161904640,3971435999526912

%N Denominator of Bernoulli(2n,1/2).

%C From the von Staudt-Clausen theorem it follows that a(n) can be computed without using Bernoulli polynomials or the 'denominator'-function (see the Sage implementation). - _Peter Luschny_, Mar 24 2014

%D J. R. Philip, The symmetrical Euler-Maclaurin summation formula, Math. Sci., 6, 1981, pp. 35-41.

%H Vincenzo Librandi, <a href="/A033469/b033469.txt">Table of n, a(n) for n = 0..250</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/vonStaudt-ClausenTheorem.html">von Staudt-Clausen Theorem</a>.

%H <a href="/index/Be#Bernoulli">Index entries for sequences related to Bernoulli numbers.</a>

%F a(n) = denominator(2*(2*Pi)^(-2*n)*(2*n)!*Li_{2*n}(-1)). - _Peter Luschny_, Jun 29 2012

%F a(n) = A081294(n) * A002445(n) for n > 0. - _Paul Curtz_, Apr 17 2013

%F Apparently, denominators of the fractions with e.g.f. (x/2) / sinh(x/2). - _Tom Copeland_, Sep 17 2016

%p with(numtheory); seq(denom(bernoulli(2*n, 1/2)), n=0..20);

%t Table[ BernoulliB[2*n, 1/2] // Denominator, {n, 0, 18}] (* _Jean-François Alcover_, Apr 15 2013 *)

%t a[ n_] := If[ n < 0, 0, (2 n)! SeriesCoefficient[ x/2 / Sinh[x/2], {x, 0, 2 n}] // Denominator]; (* _Michael Somos_, Sep 21 2016 *)

%o (PARI) a(n)=denominator(subst(bernpol(2*n,x),x,1/2)); \\ _Joerg Arndt_, Apr 17 2013

%o (Sage)

%o def A033469(n):

%o if n == 0: return 1

%o M = map(lambda i: i+1, divisors(2*n))

%o return 2^(2*n-1)*mul(filter(lambda s: is_prime(s), M))

%o [A033469(n) for n in (0..21)] # _Peter Luschny_, Mar 24 2014

%Y Cf. A001896.

%K nonn

%O 0,2

%A _N. J. A. Sloane_

%E More terms from _Joerg Arndt_, Apr 17 2013