A144048 Square array A(n,k), n>=0, k>=0, read by antidiagonals, where column k is Euler transform of (j->j^k).

1, 1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 5, 6, 5, 1, 1, 9, 14, 13, 7, 1, 1, 17, 36, 40, 24, 11, 1, 1, 33, 98, 136, 101, 48, 15, 1, 1, 65, 276, 490, 477, 266, 86, 22, 1, 1, 129, 794, 1828, 2411, 1703, 649, 160, 30, 1, 1, 257, 2316, 6970, 12729, 11940, 5746, 1593, 282, 42, 1, 1, 513

Offset: 0

Author

Alois P. Heinz, Sep 08 2008

Keywords

• nonn
• tabl

Comments

In general, column k > 0 is asymptotic to (Gamma(k+2)*Zeta(k+2))^((1-2*Zeta(-k)) /(2*k+4)) * exp((k+2)/(k+1) * (Gamma(k+2)*Zeta(k+2))^(1/(k+2)) * n^((k+1)/(k+2)) + Zeta'(-k)) / (sqrt(2*Pi*(k+2)) * n^((k+3-2*Zeta(-k))/(2*k+4))). - Vaclav Kotesovec, Mar 01 2015

Examples

			Square array begins:
  1,  1,   1,   1,    1,     1, ...
  1,  1,   1,   1,    1,     1, ...
  2,  3,   5,   9,   17,    33, ...
  3,  6,  14,  36,   98,   276, ...
  5, 13,  40, 136,  490,  1828, ...
  7, 24, 101, 477, 2411, 12729, ...
		

Links

• Alois P. Heinz, Antidiagonals = 0..99, flattened
• Vaclav Kotesovec, A method of finding the asymptotics of q-series based on the convolution of generating functions, arXiv:1509.08708 [math.CO], Sep 30 2015, p. 21.
• N. J. A. Sloane, Transforms

Crossrefs

Columns k=0-9 give: A000041, A000219, A023871, A023872, A023873, A023874, A023875, A023876, A023877, A023878.

Rows give: 0-1: A000012, 2: A000051, A094373, 3: A001550, 4: A283456, 5: A283457.

Main diagonal gives A252782.

Cf. A283272.

Programs

Maple

with(numtheory): etr:= proc(p) local b; b:= proc(n) option remember; `if`(n=0,1, add(add(d*p(d), d=divisors(j)) *b(n-j), j=1..n)/n) end end: A:= (n,k)-> etr(j->j^k)(n); seq(seq(A(n,d-n), n=0..d), d=0..13);

Mathematica

etr[p_] := Module[{ b}, b[n_] := b[n] = If[n == 0, 1, Sum[Sum[d*p[d], {d, Divisors[j]}]*b[n - j], {j, 1, n}]/n]; b]; A[n_, k_] := etr[Function[j, j^k]][n]; Table[Table[A[n, d - n], {n, 0, d}], {d, 0, 13}] // Flatten (* Jean-François Alcover, Dec 27 2013, translated from Maple *)

Formula

G.f. of column k: Product_{j>=1} 1/(1-x^j)^(j^k).