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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).
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%I #28 Apr 14 2024 11:49:34

%S 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,

%T 1,7,61,493,2713,6331,1957,1,1,8,85,916,7889,40696,75973,13700,1,1,9,

%U 113,1531,18321,157781,732529,1063623,109601,1,1,10,145,2374,36745,458026,3786745,15383110,17017969,986410,1

%N 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).

%H Alois P. Heinz, <a href="/A320031/b320031.txt">Antidiagonals n = 0..140, flattened</a>

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

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

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

%F A(n,k) = 1 + [n > 0] * k * n * A(n-1,k). - _Alois P. Heinz_, May 09 2020

%F A(n,k) = floor(n!*k^n*exp(1/k)), k > 0, n + k > 1. - _Peter McNair_, Dec 20 2021

%F From _Werner Schulte_, April 14 2024: (Start)

%F 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).

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

%e 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! + ...

%e Square array begins:

%e 1, 1, 1, 1, 1, 1, ...

%e 1, 2, 3, 4, 5, 6, ...

%e 1, 5, 13, 25, 41, 61, ...

%e 1, 16, 79, 226, 493, 916, ...

%e 1, 65, 633, 2713, 7889, 18321, ...

%e 1, 326, 6331, 40696, 157781, 458026, ...

%p A := (n, k) -> simplify(hypergeom([1, -n], [], -k)):

%p for n from 0 to 5 do seq(A(n, k), k=0..8) od; # _Peter Luschny_, Oct 03 2018

%p # second Maple program:

%p A:= proc(n, k) option remember;

%p 1 + `if`(n>0, k*n*A(n-1, k), 0)

%p end:

%p seq(seq(A(n, d-n), n=0..d), d=0..12); # _Alois P. Heinz_, May 09 2020

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

%t Table[Function[k, HypergeometricPFQ[{1, -n}, {}, -k]][j - n], {j, 0, 10}, {n, 0, j}] // Flatten

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

%Y Main diagonal gives A277452.

%Y Cf. A007318, A320032, A371898.

%K nonn,tabl

%O 0,5

%A _Ilya Gutkovskiy_, Oct 03 2018