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%I #11 Oct 05 2013 12:07:58
%S 1,4,38,582,12354,335730,11127150,435300390,19633815810,1003121039970,
%T 57259773499950,3611583223860150,249441581246630850,
%U 18723487284033181650,1517668796159163197550,132117536404977132759750
%N Numerator of integral_{x=1..2} (x^2-1)^n dx.
%C Denominator is (2n+1)!/(n! 2^n).
%C Note that these fractions are not reduced. The reduced fractions are 1, 4/3, 38/15, 194/35, 4118/315, 22382/693, 247270/3003, 1381906/6435, etc. and lead to a different sequence of numerators. [From _R. J. Mathar_, Nov 24 2008]
%F (-1)^n*(2*n+1)!!*(2*hypergeom([1/2, -n], [3/2], 4)-hypergeom([1/2, -n], [3/2], 1)). - _Vladeta Jovovic_, Dec 05 2002
%F E.g.f.: (2/sqrt(1-6*x)-1)/(1+2*x). - _Vladeta Jovovic_, Dec 14 2003
%F a(n) ~ 3*(6*n)^n/(sqrt(2)*exp(n)). - _Vaclav Kotesovec_, Oct 05 2013
%e If n=3 the integral is 194/35, so a(3) = 7!/(3! 2^3) * 194/35 = 582.
%t a[n_] := (2n+1)!/n!/2^n*Integrate[(x^2-1)^n, {x, 1, 2}]
%Y Cf. A076729.
%K frac,nonn
%O 0,2
%A Al Hakanson (hawkuu(AT)excite.com), Dec 02 2002
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