OFFSET
0,8
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)
LINKS
Peter Luschny, Extensions of the binomial
Eric Weisstein's World of Mathematics, Poisson-Charlier Polynomial
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
EXAMPLE
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.
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 *)
CROSSREFS
KEYWORD
sign,tabl
AUTHOR
Peter Luschny, Apr 12 2016
STATUS
approved