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A180364
a(n) = sum_{k=0..n} C(n,k)*C(n+k,k)*(2*k+1)^2, where C(m,k) denotes the binomial coefficient m!/(k!*(m-k)!).
2
1, 19, 205, 1839, 14961, 114483, 839917, 5975455, 41524897, 283272723, 1903686093, 12636115407, 83007985425, 540484102707, 3492471392493, 22418010385983, 143062290575937, 908253002030355, 5739641232682957, 36121371405797743, 226475167518421681
OFFSET
0,2
COMMENTS
For any n > 0, we have a(0)+ ... + a(n-1) = n*sum_{k=0}^{n-1} (2*k+1)*C(n-1,k)*C(n+k,k) = n^2*A002002(n). The first equality can be easily deduced, and the second equality holds since both sides satisfy the same recurrence by the Zeilberger algorithm.
Conjecture: The sequence a(n+1)/a(n) (n = 0,1,...) is strictly decreasing to the limit 3+2*sqrt(2), and the sequence a(n+1)^(1/(n+1))/a(n)^(1/n) (n = 1,2,3,..) is strictly increasing to the limit 1.
LINKS
FORMULA
Recurrence (obtained via the Zeilberger algorithm):
-(n+1)*(2*n^2+10*n+11)*a(n) + (2*n+3)*(6*n^2+18*n-7)*a(n+1) - (n+2)*(2*n^2+2*n-1)*a(n+2) = 0.
a(n) ~ sqrt(8+6*sqrt(2)) * (3+2*sqrt(2))^n * n^(3/2) / (2*sqrt(Pi)). - Vaclav Kotesovec, Sep 02 2014
EXAMPLE
a(1) = 19 since sum_{k=0,1} C(1,k)*C(1+k,k)*(2k+1)^2 = 1 + 2*3^2 = 19.
MATHEMATICA
a[n_] := Sum[Binomial[n, k] Binomial[n + k, k] (2 k + 1)^2, {k, 0, n}]
Table[a[n], {n, 0, 20}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Aug 29 2014
STATUS
approved