A243101 a(n) = (sum_{k=0}^{n-1}(4*k^3-1)*C(n-1,k)*C(n+k,k))/n^2, where C(m,k) denotes the binomial coefficient m!/(k!*(m-k)!).

-1, 2, 37, 324, 2403, 16582, 109961, 711176, 4521607, 28394442, 176648877, 1090974156, 6698429931, 40930511502, 249105886737, 1510954029072, 9138180665871, 55128765109906, 331851698655797, 1993747216968788

Offset: 1

Author

Zhi-Wei Sun, Aug 29 2014

Keywords

• sign

Comments

Conjecture: Let n be any positive integer.

(i) For any m = 0,1,2,..., we have sum_{k=0}^{n-1}(-1)^k*(4*k^3-1)*C(n-1,k)^m*C(-n-1,k)^m == 0 (mod n^2). Also, we may replace 4*k^3-1 by 2*k^2+2*k+1 or 2*k^3+k^2+k.

(ii) For any s,t = 0,1,2,..., both sum_{k=0}^{n-1}(4*k^3-1)*C(n-1,k)^s*C(-n-1,k)^t and sum_{k=0}^{n-1}(-1)^k*(4*k^3-1)*C(n-1,k)^s*C(-n-1,k)^t are multiples of n.

The author proved the conjecture in the latest version of arXiv:1408.5381, thus a(n) is always an integer. - Zhi-Wei Sun, Sep 01 2014

Examples

			a(2) = 2 since (sum_{k=0,1}(4*k^3-1)*C(1,k)*C(2+k,k))/2^2 = (-1 + 3*3)/4 = 2.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..200
• Zhi-Wei Sun, Two new kinds of numbers and related divisibility results, arXiv:1408.5381, 2014.

Crossrefs

Cf. A246542, A246543.

Programs

Maple

A243101:=n->add((4*k^3-1)*binomial(n-1,k)*binomial(n+k,k), k=0..n-1)/n^2: seq(A243101(n), n=1..20); # Wesley Ivan Hurt, Sep 01 2014

Mathematica

a[n_]:=Sum[(4k^3-1)*Binomial[n-1,k]Binomial[n+k,k],{k,0,n-1}]/n^2
Table[a[n],{n,1,20}]

Formula

Recurrence (obtained via the Zeilberger algorithm):

n*(2*n+3)*(n^2+3*n+1)*a(n) - 2*(2*n^2+4*n+1)*(3*n^2+6*n+2)*a(n+1) + (n+2)*(2*n+1)*(n^2+n-1)*a(n+2) = 0.

a(n) ~ (3+2*sqrt(2))^n * sqrt(n) / (2^(5/4) * sqrt(Pi)). - Vaclav Kotesovec, Sep 02 2014