A246542 a(n) = 2/n^2*( sum_{k=0..n-1} (2*k+1)*C(n-1,k)^2*C(n+k,k) ), where C(m,k) denotes the binomial coefficient m!/(k!*(m-k)!).

2, 5, 22, 132, 918, 6981, 56390, 475796, 4149286, 37133043, 339307098, 3154030050, 29741815998, 283896719073, 2738445478758, 26656533873204, 261561469613190, 2584718580416919, 25703179602581234, 257046296680889600, 2583719988283365322, 26090463844931102715, 264570229302222957162, 2693187696469413499902, 27511970457139362253598

Offset

1,1

Comments

The following general conjecture implies that a(n) is always an integer.

Conjecture: Let a and b be nonnegative integers, and let n be a positive integer. If a == b (mod 2), then sum_{k=0}^{n-1}(2k+1)*C(n-1,k)^a*C(-n-1,k)^b == 0 (mod n^2); if a - b is odd, then 2*sum_{k=0}^{n-1}(-1)^k*(2k+1)C(n-1,k)^a*C(-n-1,k)^b == 0 (mod n^2).

An extension of the conjecture was proved in the latest version of arXiv:1408.5381. - Zhi-Wei Sun, Sep 14 2014

Links

Zhi-Wei Sun, Table of n, a(n) for n = 1..150

Zhi-Wei Sun, Two new kinds of numbers and related divisibility results, arXiv:1408.5381, 2014.

Formula

Recurrence (obtained via the Zeilberger algorithm):

-n*(n-2)*(20*n^2+50*n+29)*a(n) - (220*n^4+550*n^3+369*n^2+59*n-10)*a(n+1)

+ (n+2)^2*(20*n^2+10*n-1)*a(n+2) = 0.

Example

a(2) = 5 since 2/2^2*( sum_{k=0,1} (2k+1)*C(1,k)^2*C(2+k,k) ) = 1/2*(1+3*3) = 5.

Mathematica

a[n_] := Sum[(2 k + 1)Binomial[n - 1, k]^2 Binomial[n + k, k], {k, 0, n - 1}] 2/n^2
Table[a[n], {n, 1, 25}]

Crossrefs

Cf. A180364, A246460, A246511, A246543.

Sequence in context: A329260 A111542 A342432 * A001437 A067549 A262842

Adjacent sequences: A246539 A246540 A246541 * A246543 A246544 A246545

Keyword

nonn

Author

Zhi-Wei Sun, Aug 29 2014

Status

approved