A246065 a(n) = Sum_{k=0..n}C(n,k)^2*C(2k,k)/(2k-1), where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!).

-1, 1, 9, 39, 177, 927, 5463, 34857, 234657, 1641471, 11820135, 87080265, 653499135, 4979882385, 38441107305, 300027646647, 2364113123073, 18784242756927, 150351698420247, 1211310469545081, 9816017765368671, 79963826730913809, 654504197331971961, 5380270242617370951

Offset: 0

Author

Zhi-Wei Sun, Aug 24 2014

Keywords

• sign

Comments

a(n) is always an integer since (2k-1)|C(2k,k) for any nonnegative integer k.

Conjecture: (i) The sequence a(n+1)/a(n) (n = 2,3,...) is strictly increasing to the limit 9, and the sequence a(n+1)^(1/(n+1))/a(n)^(1/n) (n = 1,2,3,...) is strictly decreasing to the limit 1.

(ii) sum_{k=0}^{n-1}a(k) == 0 (mod n^2) for all n > 0. Moreover, for any prime p we have sum_{k=0}^{p-1}a(k) == -p^2*(1+9*(p/3))/2 (mod p^3), where (p/3) is the Legendre symbol.

We are able to prove n | sum_{k=0}^{n-1}a(k). Note also that sum_{k=0}^{n-1}a(k)*9^(n-1-k) = -n^2*A086618(n-1) for all n > 0 since both sides satisfy the same recurrence via the Zeilberger algorithm.

The congruence (0 mod n^2) in (ii) is true, see the formula for A246138 in terms of A005802. - Mark van Hoeij, Nov 07 2023

Examples

			a(2) = 9 since Sum_{k=0,1,2}C(2,k)^2*C(2k,k)/(2k-1) = -1 + 8 + 6/3 = 9.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 0..150
• Zhi-Wei Sun, A new kind of numbers and their arithmetic properties, preprint, arXiv:1408.5381 [math.NT], 2014-2018.

Crossrefs

Cf. A086618, A245769.

Programs

Maple

a := n -> -hypergeom([-1/2, -n, -n], [1, 1], 4):
seq(simplify(a(n)), n=0..23); # Peter Luschny, Nov 07 2023
ogf := -(1-9*x)^(1/4)*hypergeom([-1/4, 3/4],[1],64*x^3/((1-9*x)*(x-1)^3))/(1-x)^(5/4);
series(ogf, x=0, 25); # Mark van Hoeij, Nov 12 2023

Mathematica

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

Formula

Recurrence (obtained via the Zeilberger algorithm):

9*(n+1)^2*a(n) -(19n^2+58n+63)*a(n+1) + (11n^2+46n+47)*a(n+2)-(n+3)^2*a(n+3) = 0.

a(n) ~ A086618(n)/2 ~ 3^(2*n + 5/2)/(16*Pi*n^2) as n tends to the infinity.

a(n) = (9*(2*n+1)^2*A002893(n) - 4*(n+1)^2*A002893(n+1))/3. - Mark van Hoeij, Nov 07 2023

a(n) = -hypergeom([-1/2, -n, -n], [1, 1], 4). - Peter Luschny, Nov 07 2023