A188685 Partial alternating sums of binomial(3n,n)^2/(2n+1)^2.
1, 0, 9, 135, 2890, 71639, 1967545, 58125959, 1813561210, 59034994415, 1987910416810, 68818255912790, 2437897047570874, 88061136002276310, 3234416650430634090, 120525771933269446806, 4548292982313797644875
Offset: 0
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..608
Crossrefs
Programs
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Magma
[ &+[(-1)^(n-k)*Binomial(3*k, k)^2/(2*k+1)^2: k in [0..n]]: n in [0..16]]; // Bruno Berselli, Apr 11 2011
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Maple
A001764 := proc(n) binomial(3*n,n)/(2*n+1) ; end proc: A188685 := proc(n) add( (-1)^(n-k)*A001764(k)^2,k=0..n) ; end proc: # R. J. Mathar, Apr 11 2011
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Mathematica
Table[Sum[Binomial[3k,k]^2(-1)^(n-k)/(2k+1)^2,{k,0,n}],{n,0,20}]
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Maxima
makelist(sum(binomial(3*k,k)^2*(-1)^(n-k)/(2*k+1)^2,k,0,n),n,0,20);
Formula
a(n) = Sum_{k=0..n} (-1)^(n-k)*A001764(k)^2.
4*(2*n^2 + 9*n + 10)^2*a(n+2) - (713*n^4 + 4230*n^3 + 9317*n^2 + 9000*n + 3200)*a(n+1) - 9*(9*n^2 + 27*n + 20)^2*a(n) = 0.
a(n) ~ 3^(6*n+7)/(745*Pi*n^3*2^(4*n+4)). - Vaclav Kotesovec, Aug 06 2013