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A143632
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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).
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10
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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
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OFFSET
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0,9
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LINKS
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EXAMPLE
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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, ...
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MAPLE
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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
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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 *)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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