A269953 Triangle read by rows: T(n, k) = Sum_{j=0..n} binomial(-j-1, -n-1)*S1(j, k) where S1 are the Stirling cycle numbers A132393.

1, -1, 1, 1, -1, 1, -1, 2, 0, 1, 1, 0, 5, 2, 1, -1, 9, 15, 15, 5, 1, 1, 35, 94, 85, 40, 9, 1, -1, 230, 595, 609, 315, 91, 14, 1, 1, 1624, 4458, 4844, 2779, 924, 182, 20, 1, -1, 13209, 37590, 43238, 26817, 9975, 2310, 330, 27, 1

Offset: 0

Author

Peter Luschny, Apr 12 2016

Keywords

• sign
• tabl

Comments

Replacing the Stirling cycle numbers in the definition by the Stirling set numbers leads to A105794.

From Wolfdieter Lang, Jun 19 2017: (Start)

The triangle t(n, k) = (-1)^(n-k)*T(n, k) is the matrix product of P = A007318 (Pascal) and s1 = A048994 (signed Stirling1). This is Sheffer (exp(t), log(1+t)).

The present triangle T is therefore the Sheffer triangle (exp(-t), -log(1-t)). Note that P is Sheffer (exp(t), t) (of the Appell type). (End)

The triangle T(n,k) is a representative of the parametric family of triangles T(m,n,k), whose columns are the coefficients of the standard expansion of the function f(x) = (-log(1-x))^(k)*exp(-m*x)/k! for the case m=1. See A381082. - Igor Victorovich Statsenko, Feb 14 2025

Examples

			Triangle starts:
   1;
  -1,  1;
   1, -1,  1;
  -1,  2,  0,  1;
   1,  0,  5,  2,  1;
  -1,  9, 15, 15,  5,  1;
   1, 35, 94, 85, 40,  9,  1.
		

Links

• Peter Luschny, Extensions of the binomial
• Eric Weisstein's World of Mathematics, Poisson-Charlier Polynomial

Crossrefs

Columns k=0..4 give A033999, A002741, A381064, A381065, A381066.

Cf. A000166 (row sums), A080956 (diag n,n-1).

Cf. A105793, A105794, A132393, A007318, A048994.

Cf. A001339, A046716, A082030, A095000, A137346.

KummerU(-n,1-n-x,z): this sequence (z=-1), A094816 (z=1), |A137346| (z=2), A327997 (z=3).

Programs

Maple

A269953 := (n,k) -> add(binomial(-j-1,-n-1)*abs(Stirling1(j,k)), j=0..n):
seq(print(seq(A269953(n, k), k=0..n)), n=0..9);
# Alternative:
egf := exp(-t)*(1-t)^(-x): ser := series(egf, t, 12): p := n -> coeff(ser, t, n):
seq(n!*seq(coeff(p(n), x, k), k=0..n), n=0..9); # Peter Luschny, Oct 28 2019

Mathematica

Flatten[Table[Sum[Binomial[-j-1,-n-1] Abs[StirlingS1[j,k]], {j,0,n}], {n,0,9},{k,0,n}]]
(* Or: *)
p [n_] := HypergeometricU[-n, 1 - n - x, -1];
Table[CoefficientList[p[n], x], {n, 0, 9}] (* Peter Luschny, Oct 28 2019 *)

Formula

From Wolfdieter Lang, Jun 19 2017: (Start)

E.g.f. of row polynomials R(n, x) = Sum_{k=0..n} T(n,k)*x^k: exp(-t)/(1 - t)^x.

E.g.f. of column k sequence: exp(-x)*(-log(1-x))^k/k!, k >= 0. (End)

From Peter Bala, Oct 26 2019: (Start)

Let R(n, x) = (-1)^n*Sum_{k >= 0} binomial(n,k)*k!* binomial(-x,k) the n-th row polynomial of this triangle.

R(n, x) = c_n(-x;-1), where c_n(x;a) denotes the n-th Poisson Charlier polynomial.

The series representation e = Sum_{k >= 0} 1/k! is the case n = 0 of the more general result e = n!*Sum_{k >= 0} 1/(k!*R(n,k)*R(n,k+1)), n = 0,2,3,4,.... (End)

R(n, x) = KummerU(-n, 1-n-x, -1). - Peter Luschny, Oct 28 2019