A078738 Generalized Bell numbers B_{3,2}(n).

1, 13, 355, 16333, 1121881, 106708921, 13354028563, 2118817455385, 414426460442833, 97746679844312581, 27311169061720393411, 8908525371578726747173, 3350963996380181114090665, 1438463413778071631322236593, 698374517715612292764726380851

Offset: 1

Author

N. J. A. Sloane, Dec 21 2002

Keywords

• nonn
• easy

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 1..240
• P. Blasiak, K. A. Penson and A. I. Solomon, The Boson Normal Ordering Problem and Generalized Bell Numbers, arXiv:quant-ph/0212072, 2002.
• P. Blasiak, K. A. Penson and A. I. Solomon, The general boson normal ordering problem, arXiv:quant-ph/0402027, 2004.
• P. Blasiak, K. A. Penson and A. I. Solomon, The general boson normal ordering problem, Phys. Lett. A 309 (2003) 198-205.
• M. Schork, On the combinatorics of normal ordering bosonic operators and deforming it, J. Phys. A 36 (2003) 4651-4665.

Crossrefs

B_{1, 1} = A000110, B_{2, 1} = A000262, B_{3, 1} = A020556 and B_{3, 3} = A069223. Row sums of A078740.

Alternating row sums A090437.

Programs

Mathematica

a[n_] := (n+1)*n!^2*Sum[(-1)^k*HypergeometricPFQ[{2-k, n+1, n+2}, {2, 3}, 1]/(2*(k-2)!), {k, 2, 2n}]; Array[a, 13] (* Jean-François Alcover, Sep 01 2015 *)
Table[Sum[(n + k)!*(n + k + 1)!/(k!*(k + 1)!*(k + 2)!), {k, 0, Infinity}]/E, {n, 1, 20}] (* Vaclav Kotesovec, Jul 27 2018 *)

PARI

nmax = 20; p = floor(3*nmax*log(nmax)); default(realprecision, p);
for(n=1, nmax, print1(round(exp(-1)*suminf(k=0, (n+k)!*(n+k+1)!/(k!*(k+1)!*(k+2)!))), ", ")) \\ G. C. Greubel and Vaclav Kotesovec, Jul 28 2018

Formula

a(n) = Sum_{k=2..2*n} A078740(n, k) = Sum_{k=1..infinity} (1/k!)*Product_{j=1..n}(fallfac(k+(j-1)*(3-2), 2))/exp(1), n>=1. From eq.(9) of the Blasiak et al. reference with r=3, s=2. fallfac(n, m) := A008279(n, m) (falling factorials triangle). a(0) := 1 may be added.

a(n) = Sum_{k>=0} ((n+k)!*(n+k+1)!/(k!*(k+1)!*(k+2)!))/exp(1), n>=1. From eq.(40) of the Blasiak et al. reference. [corrected by Vaclav Kotesovec, Jul 27 2018]

E.g.f. for a(n)/n! with a(0)=(exp(1)-1)/exp(1) added: Sum_{k>=0} (hypergeom([k+2, k+1], [1], z)/(k+2)!)/exp(1). From eq. (41) of the Blasiak et al. reference.

Extensions

Edited by Wolfdieter Lang, Dec 23 2003