A246459 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, 7, 55, 465, 4047, 35673, 316521, 2819295, 25173855, 225157881, 2016242265, 18070920255, 162071863425, 1454320387575, 13055422263255, 117237213829953, 1053070838993151, 9461217421304505, 85019389336077225, 764113545253570191, 6868417199986308129

Offset: 0

Author

Zhi-Wei Sun, Aug 26 2014

Keywords

• nonn

Comments

Zhi-Wei Sun proved that for any n > 0 we have Sum_{k=0..n-1} a(k) = n^2*A086618(n-1), and (Sum_{k=0..n-1}a(k,x))/n is a polynomial with integer coefficients, where a(k,x) = sum_{j=0..k}C(k,j)^2*C(2j,j)*(2j+1)*x^j.

Examples

			a(2) = 55 since Sum_{k=0,1,2} C(2,k)^2*C(2k,k)(2k+1) = 1 + 8*3 + 6*5 = 55.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 0..160
• Zhi-Wei Sun, Two new kinds of numbers and their arithmetic properties, arXiv:1408.5381 [math.NT], 2014-2018.

Crossrefs

Cf. A086618, A245769, A246065, A246138.

Programs

Maple

A246459:=n->add(binomial(n,k)^2*binomial(2*k,k)*(2*k+1), k=0..n): seq(A246459(n), n=0..20); # Wesley Ivan Hurt, Aug 26 2014

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) - (19*n^2+74*n+87)*a(n+1) + (n+3)*(11*n+29)*a(n+2) - (n+3)^2*a(n+3) = 0.

a(n) ~ 3^(2*n+1/2) / Pi. - Vaclav Kotesovec, Aug 27 2014

a(n) = (4*n+3)*A002893(n)/3. - Mark van Hoeij, Nov 12 2023