A320031 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of the e.g.f. exp(x)/(1 - k*x).

1, 1, 1, 1, 2, 1, 1, 3, 5, 1, 1, 4, 13, 16, 1, 1, 5, 25, 79, 65, 1, 1, 6, 41, 226, 633, 326, 1, 1, 7, 61, 493, 2713, 6331, 1957, 1, 1, 8, 85, 916, 7889, 40696, 75973, 13700, 1, 1, 9, 113, 1531, 18321, 157781, 732529, 1063623, 109601, 1, 1, 10, 145, 2374, 36745, 458026, 3786745, 15383110, 17017969, 986410, 1

Offset: 0

Author

Ilya Gutkovskiy, Oct 03 2018

Keywords

• nonn
• tabl

Examples

			E.g.f. of column k: A_k(x) = 1 + (k + 1)*x/1! + (2*k^2 + 2*k + 1)*x^2/2! + (6*k^3 + 6*k^2 + 3*k + 1)*x^3/3! + (24*k^4 + 24*k^3 + 12*k^2 + 4*k + 1)*x^4/4! + ...
Square array begins:
  1,    1,     1,      1,       1,       1,  ...
  1,    2,     3,      4,       5,       6,  ...
  1,    5,    13,     25,      41,      61,  ...
  1,   16,    79,    226,     493,     916,  ...
  1,   65,   633,   2713,    7889,   18321,  ...
  1,  326,  6331,  40696,  157781,  458026,  ...
		

Links

• Alois P. Heinz, Antidiagonals n = 0..140, flattened

Crossrefs

Columns k=0..6 give A000012, A000522, A010844, A010845, A056545, A056546, A056547.

Main diagonal gives A277452.

Cf. A007318, A320032, A371898.

Programs

Maple

A := (n, k) -> simplify(hypergeom([1, -n], [], -k)):
for n from 0 to 5 do seq(A(n, k), k=0..8) od; # Peter Luschny, Oct 03 2018
# second Maple program:
A:= proc(n, k) option remember;
      1 + `if`(n>0, k*n*A(n-1, k), 0)
    end:
seq(seq(A(n, d-n), n=0..d), d=0..12);  # Alois P. Heinz, May 09 2020

Mathematica

Table[Function[k, n! SeriesCoefficient[Exp[x]/(1 - k x), {x, 0, n}]][j - n], {j, 0, 10}, {n, 0, j}] // Flatten
Table[Function[k, HypergeometricPFQ[{1, -n}, {}, -k]][j - n], {j, 0, 10}, {n, 0, j}] // Flatten

Formula

E.g.f. of column k: exp(x)/(1 - k*x).

A(n,k) = Sum_{j=0..n} binomial(n,j)*j!*k^j.

A(n,k) = hypergeom_2F0([1, -n], [], -k).

A(n,k) = 1 + [n > 0] * k * n * A(n-1,k). - Alois P. Heinz, May 09 2020

A(n,k) = floor(n!*k^n*exp(1/k)), k > 0, n + k > 1. - Peter McNair, Dec 20 2021

From Werner Schulte, Apr 14 2024: (Start)

The LU decomposition of this array is given by the upper triangular matrix U which is the transpose of A007318 and the lower triangular matrix L = A371898, i.e., A(n, k) = Sum_{i=0..k} binomial(k, i) * A371898(n, i).

Conjecture: E.g.f. of row n is exp(x) * (Sum_{k=0..n} A371898(n, k) * x^k / k!). (End)