A246138 a(n) = (Sum_{k=0..n-1} A246065(k)) / n^2.

-1, 0, 1, 3, 9, 32, 135, 648, 3409, 19176, 113535, 700125, 4463415, 29256120, 196334697, 1344542787, 9371335905, 66335058128, 476022873279, 3457886816997, 25394948961831, 188353304179920, 1409578821465129, 10635308054118792, 80845157085234975

Offset: 1

Author

Zhi-Wei Sun, Aug 25 2014

Keywords

• sign

Comments

Part (ii) of the conjecture in A246065 implies that all the terms in the current sequence are integers.

Conjecture: The sequence a(n+1)/a(n) (n = 4,5,...) is strictly increasing to the limit 9, and the sequence a(n+1)^(1/(n+1))/a(n)^(1/n) (n = 3,4,...) is strictly decreasing to the limit 1.

Examples

			a(5) = 9 since sum_{k=0}^{5-1}A246065(k) = -1 + 1 + 9 + 39 + 177 = 225 = 5^2*9.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..170

Crossrefs

Cf. A005802, A246065.

Programs

Maple

ogf := (1-((9*x-1)/(x-1))^(3/4)*hypergeom([-1/4, 3/4],[1],-64*x/(9*x-1)^3/(x-1)))/6;
series(ogf, x=0, 25); # Mark van Hoeij, Nov 12 2023

Mathematica

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

Formula

Recurrence: n^2*a(n) = 2*(n-2)*(5*n-8)*a(n-1) - 9*(n-2)^2*a(n-2). - Vaclav Kotesovec, Aug 27 2014

a(n) ~ 3^(2*n+5/2) / (128*Pi*n^4). - Vaclav Kotesovec, Aug 27 2014

a(n) = ((3*n+2)*(3*n-2)*A005802(n-1) - (n+2)^2*A005802(n))/4. - Mark van Hoeij, Nov 06 2023