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Numerator of 8^(2n-1) |B(2n)| / (2n)!, where B() are the Bernoulli numbers.
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%I #17 May 01 2013 21:06:47

%S 8,128,4096,32768,1048576,11593056256,536870912,7767448354816,

%T 3014517285978112,191986824837595136,2733227576976736256,

%U 66530577009460375453696,5926115612870995607552,488951148984934932554973184,7946710949908368748447692488704,71105936114697022329949662478336

%N Numerator of 8^(2n-1) |B(2n)| / (2n)!, where B() are the Bernoulli numbers.

%C Start with zeta(2n) = (2Pi)^(2n) |B(2n)| /(2 (2n)!) and replace Pi by 4 arctan(1) and take the numerator of the rational part. The denominator is given by A036278.

%e 8/3,128/45,4096/945,32768/4725,...

%p f:=n->8^(2*n)*abs(B(2*n))/(2*(2*n)!); [seq(numer(f(n)),n=1..60)];

%Y Cf. A002432, A046988, A036278.

%K nonn

%O 1,1

%A _Eric Desbiaux_, Apr 15 2011

%E Entry revised by _N. J. A. Sloane_, Apr 17 2011