OFFSET
0,2
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.
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.
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
EXAMPLE
a(2) = 55 since Sum_{k=0,1,2} C(2,k)^2*C(2k,k)(2k+1) = 1 + 8*3 + 6*5 = 55.
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}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Aug 26 2014
STATUS
approved