A033452 "STIRLING" transform of squares A000290.
0, 1, 5, 22, 99, 471, 2386, 12867, 73681, 446620, 2856457, 19217243, 135610448, 1001159901, 7714225057, 61904585510, 516347066551, 4468588592739, 40058673825258, 371421499686007, 3556976106133821, 35138574378189700, 357654857584636597, 3746672593640388775
Offset: 0
Keywords
Examples
G.f. = x + 5*x^2 + 22*x^3 + 99*x^4 + 471*x^5 + 2386*x^6 + 12867*x^7 + 73681*x^8 + ...
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..574
Programs
-
Maple
a := n -> add(Stirling2(n, j)*j^2, j=0..n): seq(a(n), n=0..20); # Zerinvary Lajos, Apr 18 2007 # second Maple program: b:= proc(n, m) option remember; `if`(n=0, m^2, m*b(n-1, m)+b(n-1, m+1)) end: a:= n-> b(n, 0): seq(a(n), n=0..23); # Alois P. Heinz, Aug 04 2021
-
Mathematica
max = 20; Clear[g]; g[max + 2] = 1; g[k_] := g[k] = 2 - 1/(1 - k*x)/(1 - x/(x - 1/g[k + 1])); gf = 1/x + 1/x^2 - g[0]/x^2; CoefficientList[ Series[gf, {x, 0, max}], x] (* Jean-François Alcover, Jan 24 2013, after Sergei N. Gladkovskii *)
-
PARI
{a(n) = if( n<0, 0, n! * polcoeff( (exp(x + x * O(x^n)) - 1) * exp( exp(x + x * O(x^n)) - 1 + x), n))}; /* Michael Somos, Mar 28 2012 */
Formula
Representation as an infinite series: a(n) = (Sum_{k>=1} k^n*k*(k-2)/k!)/exp(1), n >= 1. This is a Dobinski-type summation formula. - Karol A. Penson, Mar 21 2002
a(n) = A005493(n) - A000110(n+1). - Floor van Lamoen and Christian Bower, Oct 16 2002. (n^2 has e.g.f.: e^x * (x^2+x), a(n) thus has e.g.f: e^(e^x-1) * ( (e^x-1)^2 + (e^x-1) ) which simplifies to e^(e^x-1) * (e^2x - e^x). A005493 has e.g.f.: e^(e^x+2x-1), A000110 has e.g.f.: e^(e^x-1), A000110(n+1) has as e.g.f.: derivative of A000110 which is e^(e^x+x-1).) [corrected by Georg Fischer, Jun 17 2020]
a(n) = Bell(n+2) - 2*Bell(n+1). - Vladeta Jovovic, Jul 28 2003
G.f.: sum{k>=0, k^2*x^k/prod[l=1..k, 1-lx]}. - Ralf Stephan, Apr 18 2004
E.g.f.: exp( exp(x) - 1 + x) * (exp(x) - 1). - Michael Somos, Mar 28 2012
a(n) = A123158(n,3). - Philippe Deléham, Oct 06 2006
G.f.: G(0)/x -1/x, where G(k) = 1 - x^2*(k+1)/( x^2*(k+1) - (2*x+x*k-1)*(3*x+x*k-1)/G(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Feb 25 2014
Comments