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)!).

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

Zhi-Wei Sun, Table of n, a(n) for n = 0..200

Zhi-Wei Sun, Two kinds of numbers and their arithmetic properties, arXiv:1408.5381, 2014.

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

Cf. A002002, A246460, A246542, A246543.

Sequence in context: A155670 A085770 A002501 * A125407 A289423 A278556

Adjacent sequences: A180361 A180362 A180363 * A180365 A180366 A180367

Keyword

nonn

Author

Zhi-Wei Sun, Aug 29 2014

Status

approved