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Numerator of a(n) = (integral_{x=0..1/3} (1-x^2)^n dx).
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%I #7 Jul 14 2015 01:46:37

%S 1,26,1128,68592,5368704,514149120,58253091840,7623288207360,

%T 1131761338122240,187970402507489280,34537682442564403200,

%U 6956566802152095744000,1524349874113331960217600

%N Numerator of a(n) = (integral_{x=0..1/3} (1-x^2)^n dx).

%C The denominator is b(n)= (2*n+2)!*3^(2*n+1)/((n+1)!*2^(n+1)).

%F c(n)=[(2n+2)!*3^(2n+1)/[(n+1)!*2^(n+1)]]int((1-x^2)^n, x=0..1/3). - _Emeric Deutsch_, Mar 15 2004

%p c := n->((2*n+2)!*3^(2*n+1)/((n+1)!*2^(n+1)))*int((1-x^2)^n,x=0..1/3): seq(c(n),n=0..18);

%t A091429[n_] := Integrate[(1 - x^2)^n, {x, 0, 1/3}](2n + 2)!*3^(2n + 1)/((n + 1)!*2^(n + 1)); Table[ A091429[n], {n, 0, 13}] (* _Robert G. Wilson v_, Mar 15 2004 *)

%K nonn

%O 0,2

%A Al Hakanson (hawkuu(AT)excite.com), Mar 02 2004

%E More terms from _Robert G. Wilson v_ and _Emeric Deutsch_, Mar 15 2004