A347276 Third column of A008296.

1, 6, 5, -15, 49, -196, 944, -5340, 34716, -254760, 2078856, -18620784, 180973584, -1887504768, 20887922304, -242111586816, 2889841121280, -34586897978880, 393722260047360, -3659128846433280, 5687630494110720, 1137542166526464000, -49644151627682304000

Offset: 3

Author

Luca Onnis, Aug 25 2021

Keywords

• sign

Links

• Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 139, b(n,3).

Crossrefs

Cf. A008296, A045406, A081048.

Cf. A001008, A002805, A007406, A007407.

Cf. A048994, A008275.

Programs

Maple

b:= proc(n, k) option remember; `if`(n=k, 1, `if`(k=0, 0,
      (n-1)*b(n-2, k-1)+b(n-1, k-1)+(k-n+1)*b(n-1, k)))
    end:
a:= n-> b(n, 3):
seq(a(n), n=3..30);  # Alois P. Heinz, Aug 25 2021

Mathematica

a[1, 1] = a[2, 1] = 1; a[n_, 1] = (-1)^n (n - 2)!;
a[n_, n_] = 1;
a[n_, k_] :=  a[n, k] = (n - 1) a[n - 2, k - 1] +
    a[n - 1, k - 1] + (k - n + 1) a[n - 1, k];
Flatten[Table[a[n + 3, 3], {n, 0, 30}]]

PARI

a(n) = sum(m=3, n, binomial(m, 3)*3^(m-3)*stirling(n, m, 1)); \\ Michel Marcus, Sep 14 2021

Formula

a(n) = A008296(n,3).

a(n) = (-1)^n*(3*H(n-4,1)^2 - 3*H(n-4,2) - 11*H(n-4,1) + 6)*(n-4)! for n >= 4, where H(n,1) = Sum_{j=1..n} 1/j = A001008(n)/A002805(n) is the n-th harmonic number and H(n,2) = Sum_{j=1..n} 1/j^2 = A007406(n)/A007407(n).

a(n) = Sum_{m=3..n} binomial(m,3) * 3^(m-3) * Stirling1(n,m). - Alois P. Heinz, Aug 26 2021