A246513 a(n) = (4/n^2)*( Sum_{k=0..n-1} k*A246459(k) ).

0, 7, 52, 378, 2832, 21785, 171036, 1364391, 11023264, 89985681, 740894700, 6144227430, 51267563280, 430045297695, 3623966778180, 30662599042530, 260367332354496, 2217928838577641, 18947382204700044, 162281586037920126

Offset: 1

Author

Zhi-Wei Sun, Aug 28 2014

Keywords

• nonn

Comments

Conjecture: a(n) is always an integer.

Note: the formula for a(n) in terms of A005802 proves that a(n) is an integer, divisible by n-1. - Mark van Hoeij, Nov 06 2023

Examples

			a(2) = 7 since (4/2^2)*( Sum_{k=0..1} k*A246459(k) ) = A246459(1) = 7.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..150
• Zhi-Wei Sun, Two new kinds of numbers and their arithmetic properties, arXiv:1408.5381 [math.NT], 2014-2018.

Crossrefs

Cf. A246459, A246460.

Programs

Maple

h := n -> hypergeom([1/2, 1 - n, -n], [2, 2], 4):
a := n -> (n - 1) * ((n + 1)^2 * h(n) / n - n * h(n - 1)):
seq(simplify(a(n)), n = 1..20);  # Peter Luschny, Nov 06 2023
ogf := (((-54*x^4+18*x^3+30*x^2+6*x)*hypergeom([4/3, 4/3],[2],-27*x*(x-1)^2/(9*x-1)^2)+(-1701*x^3+783*x^2-111*x+5)*hypergeom([1/3, 1/3],[1],-27*x*(x-1)^2/(9*x-1)^2))/(1-9*x)^(8/3) - 5)/6;
series(ogf, x=0, 25); # Mark van Hoeij, Nov 12 2023

Mathematica

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

Formula

Recurrence: (n-2)*n^2*(2*n-7)*(4*n-5)*a(n) = (n-1)*(80*n^4 - 532*n^3 + 1126*n^2 - 893*n + 195)*a(n-1) - 9*(n-2)^2*(n-1)*(2*n-5)*(4*n-1)*a(n-2). - Vaclav Kotesovec, Aug 28 2014

a(n) ~ 3^(2*n+1/2) / (2*Pi*n). - Vaclav Kotesovec, Aug 28 2014

a(n) = (n-1) * ((n+1)^2 * A005802(n-1) - (n-1)*n * A005802(n-2)). - Mark van Hoeij, Nov 06 2023