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1, 11, 136, 1787, 24376, 341048, 4859968, 70223483, 1025790616, 15116164136, 224365547968, 3350371999928, 50287277411008, 758124098549696, 11473331826459136, 174221578556572283, 2653437885092286808, 40520013896165905928
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OFFSET
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1,2
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COMMENTS
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It appears that a(n) == 16^n/Pi^3 * Integrate[x=0..1, x^n*F(x)*F(1-x)], where F(x) = Pi/2 * hypergeometric([1/2, 1/2], [1], x) (== elliptic K(sqrt(x))). - Vladimir Reshetnikov, Jan 20 2011
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LINKS
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FORMULA
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G.f.: 7/8 + (1/2)*(K(16x)/pi)^2, where K(x) is the elliptic integral of the first kind (as defined in Mathematica). - Emanuele Munarini, Mar 12 2011
a(n) = (1/8)*sum(binomial(2k,k)^2*binomial(2n-2k,n-k)^2, k=0..n) for n >= 1. - Emanuele Munarini, Mar 12 2011
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MAPLE
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seq(add(binomial(2*k, k)^2*binomial(2*(n-k), n-k)^2, k=0..n)/8, n=1..12); # Emanuele Munarini, Mar 12 2011
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MATHEMATICA
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Table[Sum[Binomial[2k, k]^2 Binomial[2n-2k, n-k]^2, {k, 0, n}]/8, {n, 1, 12}] (* Emanuele Munarini, Mar 12 2011 *)
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PROG
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(Maxima) makelist(sum(binomial(2*k, k)^2*binomial(2*(n-k), n-k)^2, k, 0, n)/8, n, 1, 12); /* Emanuele Munarini, Mar 12 2011 */
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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