A357407 a(n) = coefficient of x^n, n >= 0, in A(x) = exp( Sum_{n>=1} A183204(n)*x^n/n ), where A183204 equals the central terms of triangle A181544.

1, 4, 32, 360, 4964, 78064, 1344020, 24708928, 477282794, 9580852360, 198322047840, 4209371498256, 91221481924426, 2011834246746792, 45039165331725264, 1021419638492387856, 23426910170090512779, 542666070296546760492, 12681393784980089971368

Offset: 0

Author

Paul D. Hanna, Oct 19 2022

Keywords

• nonn

Comments

A183204(n) = Sum_{k=floor(n/2)..n} C(n,k)^2 * C(2*k,n) * C(n+k,k).

A183204(n) equals the coefficient of x^n in (1-x)^(3*n+1) * Sum_{k>=0} binomial(n+k-1,k)^3 * x^k, which is the central term of row n of triangle A181544.

Examples

			G.f.: A(x) = 1 + 4*x + 32*x^2 + 360*x^3 + 4964*x^4 + 78064*x^5 + 1344020*x^6 + 24708928*x^7 + 477282794*x^8 + 9580852360*x^9 + 198322047840*x^10 + ...
where
log(A(x)) = 4*x + 48*x^2/2 + 760*x^3/3 + 13840*x^4/4 + 273504*x^5/5 + 5703096*x^6/6 + 123519792*x^7/7 + 2751843600*x^8/8 + 62659854400*x^9/9 + ... + A183204(n)*x^n/n + ...
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 0..500

Crossrefs

Cf. A183204, A181544.

Programs

PARI

{A183204(n) = sum(k=n\2,n, binomial(n,k)^2 * binomial(2*k,n) * binomial(n+k,k) )}
{a(n) = polcoeff( exp( sum(m=1,n, A183204(m)*x^m/m ) + x*O(x^n) ),n)}
for(n=0,20,print1(a(n),", "))

Formula

a(n) ~ c * 3^(3*n) / n^(5/2), where c = 0.289447274610263555814082139782101227837126089347468995035938970190651243... - Vaclav Kotesovec, Mar 14 2023