A328725 - OEIS

%I #37 May 16 2020 14:26:15

%S 1,1,15,115,1255,13671,160461,1936425,24071895,305313415,3939158905,

%T 51521082405,681635916325,9105864515125,122657982366375,

%U 1664151758259915,22720725637684215,311933068664333175,4303704125389134825,59640225721889127525,829774531966386480705

%N Constant term in the expansion of (1 + 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.

%H Seiichi Manyama, <a href="/A328725/b328725.txt">Table of n, a(n) for n = 0..854</a>

%F a(n) = Sum_{i=0..n} (-1)^(n-i)*binomial(n,i)*Sum_{j=0..i} binomial(i,j)^4.

%F From _Vaclav Kotesovec_, Oct 28 2019: (Start)

%F Recurrence: n^3*a(n) = (2*n - 1)^3*a(n-1) + (n-1)*(94*n^2 - 188*n + 93)*a(n-2) + 80*(n-2)*(n-1)*(2*n - 3)*a(n-3) + 75*(n-3)*(n-2)*(n-1)*a(n-4).

%F a(n) ~ 15^(n + 3/2) / (2^(11/2) * Pi^(3/2) * n^(3/2)). (End)

%o (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)}

%o (PARI) {a(n) = sum(i=0, n, (-1)^(n-i)*binomial(n,i)*sum(j=0, i, binomial(i, j)^4))}

%Y Sum_{i=0..n} (-1)^(n-i)*binomial(n,i)*Sum_{j=0..i} binomial(i,j)^m: A002426 (m=2), A172634 (m=3), this sequence (m=4), A328750 (m=5).

%Y Cf. A002899, A005260, A328713.

%K nonn

%O 0,3

%A _Seiichi Manyama_, Oct 26 2019