A143632 Table T(n,k), n>=0, k>=0, read by antidiagonals, where the e.g.f. for column k satisfies A_k(x) = exp(x*A_k(((x+1)^k-1)/k)) if k>0 and A_0(x) = exp(x*A_0(0)) = exp(x).
1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 3, 16, 1, 1, 1, 3, 19, 125, 1, 1, 1, 3, 22, 185, 1296, 1, 1, 1, 3, 25, 253, 2541, 16807, 1, 1, 1, 3, 28, 329, 4256, 45787, 262144, 1, 1, 1, 3, 31, 413, 6471, 96727, 1037359, 4782969, 1, 1, 1, 3, 34, 505, 9216, 175747, 2828274, 28649553, 100000000, 1
Offset: 0
Examples
Table begins: 1, 1, 1, 1, 1, 1, ... 1, 1, 1, 1, 1, 1, ... 1, 3, 3, 3, 3, 3, ... 1, 16, 19, 22, 25, 28, ... 1, 125, 185, 253, 329, 413, ... 1, 1296, 2541, 4256, 6471, 9216, ...
Links
- Alois P. Heinz, Antidiagonals n = 0..99, flattened
Crossrefs
Programs
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Maple
A:= proc(n,k) option remember; if n<=0 or k=0 then 1 else A(n-1,k)(((x+1)^k-1)/k) fi; unapply(convert(series(exp(x*%), x,n+1), polynom), x) end: T:= (n,k)-> coeff(A(n,k)(x), x,n)*n!: seq(seq(T(n,d-n), n=0..d), d=0..11);
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Mathematica
a[n_, k_][x_] := Module[{f}, f = If[n <= 0 || k == 0, 1, a[n-1, k][((#+1)^k-1)/k]]&; Normal[Series[Exp[y*f[y]], {y, 0, n+1}]] /. y -> x]; t[n_, k_] := Coefficient[a[n, k][x], x, n]*n!; Table[t[n, d-n], {d, 0, 11}, {n, 0, d}] // Flatten (* Jean-François Alcover, Feb 12 2014, translated from Maple *)
Comments