A171367 Antidiagonal sums of triangle of Stirling numbers of 2nd kind A048993.
1, 0, 1, 1, 2, 4, 9, 22, 58, 164, 495, 1587, 5379, 19195, 71872, 281571, 1151338, 4902687, 21696505, 99598840, 473466698, 2327173489, 11810472444, 61808852380, 333170844940, 1847741027555, 10532499571707, 61649191750137, 370208647200165, 2278936037262610, 14369780182166215
Offset: 0
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..665
- P. Flajolet, Combinatorial aspects of continued fractions, Discrete Mathematics, Volume 32, Issue 2, 1980, pp. 125-161.
Programs
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Maple
b:= proc(n, m) option remember; `if`(n<=m, `if`(n=m, 1, 0), m*b(n-1, m)+b(n-1, m+1)) end: a:= n-> b(n, 0): seq(a(n), n=0..30); # Alois P. Heinz, May 16 2023 -
Mathematica
Table[Sum[StirlingS2[n-k, k], {k, 0, n}], {n, 0, 30}] (* Vaclav Kotesovec, Oct 18 2016 *) -
Maxima
makelist(sum(stirling2(n-k,k),k,0,n),n,0,60); /* Emanuele Munarini, Jun 01 2012 */
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PARI
a(n) = sum(k=0, n, stirling(n-k, k,2)); /* Joerg Arndt, Jan 16 2013 */
Formula
G.f.: 1/(1-x^2/(1-x/(1-x^2/(1-2x/(1-x^2/1-3x/(1-x^2/(1-4x/(1-x^2/(1-5x/(1-... (continued fraction).
G.f.: (G(0) - 1)/(x-1)/x where G(k) = 1 - x/(1-k*x)/(1-x/(x-1/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 16 2013
G.f.: T(0)/(1-x^2), where T(k) = 1-x^3*(k+1)/(x^3*(k+1)-(1-x*(x+k))*(1-x*(x+1+k))/T(k+1) ); (continued fraction, after P. Flajolet, p. 140). - Sergei N. Gladkovskii, Oct 30 2013
G.f. (alternating signs): Sum_{k>=0} S(x,k)*x^k, where S(x,k)*exp(-x) is the inverse Mellin transform of Gamma(s)*s^k. - Benedict W. J. Irwin, Oct 14 2016