A292913 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of e.g.f. exp(exp(k*x)-1).
1, 1, 0, 1, 1, 0, 1, 2, 2, 0, 1, 3, 8, 5, 0, 1, 4, 18, 40, 15, 0, 1, 5, 32, 135, 240, 52, 0, 1, 6, 50, 320, 1215, 1664, 203, 0, 1, 7, 72, 625, 3840, 12636, 12992, 877, 0, 1, 8, 98, 1080, 9375, 53248, 147987, 112256, 4140, 0, 1, 9, 128, 1715, 19440, 162500, 831488, 1917999, 1059840, 21147, 0
Offset: 0
Examples
E.g.f. of column k: A_k(x) = 1 + k*x/1! + 2*k^2*x^2/2! + 5*k^3*x^3/3! + 15*k^4 x^4/4! + 52*k^5*x^5/5! + ... Square array begins: 1, 1, 1, 1, 1, 1, ... 0, 1, 2, 3, 4, 5, ... 0, 2, 8, 18, 32, 50, ... 0, 5, 40, 135, 320, 625, ... 0, 15, 240, 1215, 3840, 9375, ... 0, 52, 1664, 12636, 53248, 162500, ...
Crossrefs
Programs
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Maple
A:= (n, k)-> k^n * combinat[bell](n): seq(seq(A(n, d-n), n=0..d), d=0..12); # Alois P. Heinz, Sep 26 2017
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Mathematica
Table[Function[k, n! SeriesCoefficient[Exp[Exp[k x] - 1], {x, 0, n}]][j - n], {j, 0, 10}, {n, 0, j}] // Flatten Table[Function[k, SeriesCoefficient[1/(1 + ContinuedFractionK[-((-1)^(i + 1) (i - 1) + i + 3) k x/4, 1, {i, 0, n}]), {x, 0, n}]][j - n], {j, 0, 10}, {n, 0, j}] // Flatten
Formula
O.g.f. of column k: 1/(1 - k*x/(1 - k*x/(1 - k*x/(1 - 2*k*x/(1 - k*x/(1 - 3*k*x/(1 - k*x/(1 - 4*k*x/(1 - ...))))))))), a continued fraction.
E.g.f. of column k: exp(exp(k*x)-1).
A(n,k) = exp(-1)*k^n*Sum_{j>=0} j^n/j!.
A(n,k) = k^n * Bell(n). - Alois P. Heinz, Sep 26 2017