A328808 Constant term in the expansion of (3 + x + y + z + 1/x + 1/y + 1/z + x*y + y*z + z*x + 1/(x*y) + 1/(y*z) + 1/(z*x) + x*y*z + 1/(x*y*z))^n.

1, 3, 23, 225, 2583, 32133, 422069, 5757699, 80790775, 1158593589, 16905540753, 250185539079, 3746205581589, 56652844671855, 864032059578879, 13274539401672345, 205252378269637815, 3191578469685269925, 49876569284504593505, 782943268394316187815

Offset

0,2

Links

Table of n, a(n) for n=0..19.

Formula

a(n) = Sum_{i=0..n} binomial(n,i)*Sum_{j=0..i} binomial(i,j)^4.

From Vaclav Kotesovec, Oct 28 2019: (Start)

Recurrence: n^3*a(n) = (2*n - 1)*(8*n^2 - 8*n + 3)*a(n-1) + (n-1)*(22*n^2 - 44*n + 13)*a(n-2) - 44*(n-2)*(n-1)*(2*n - 3)*a(n-3) + 51*(n-3)*(n-2)*(n-1)*a(n-4).

a(n) ~ sqrt(2) * 17^(n + 3/2) / (64 * Pi^(3/2) * n^(3/2)). (End)

Mathematica

Table[Sum[Binomial[n, i]*Sum[Binomial[i, j]^4, {j, 0, i}], {i, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Oct 28 2019 *)

Prog

(PARI) {a(n) = polcoef(polcoef(polcoef((1+(1+x)*(1+y)*(1+z)+(1+1/x)*(1+1/y)*(1+1/z))^n, 0), 0), 0)}
(PARI) {a(n) = sum(i=0, n, binomial(n, i)*sum(j=0, i, binomial(i, j)^4))}

Crossrefs

Column k=4 of A328807.

Cf. A005260, A328725, A328735.

Sequence in context: A305754 A202997 A093162 * A206763 A306154 A201205

Adjacent sequences: A328805 A328806 A328807 * A328809 A328810 A328811

Keyword

nonn

Author

Seiichi Manyama, Oct 28 2019

Status

approved