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%I #28 Jan 09 2025 04:24:09
%S 1,73,16333,8030353,7209986401,10541813012041,23227377813664333,
%T 72925401604382826913,312727862321385812968033,
%U 1772004571987390827615327241,12917715377912025572750844722221,118521774439119390334062953438350513,1343761301099219856651740487814621053313
%N Generalized Bell numbers B_{4,3}.
%H G. C. Greubel, <a href="/A070531/b070531.txt">Table of n, a(n) for n = 1..200</a>
%H P. Blasiak, Karol A. Penson and A. I. Solomon, <a href="http://www.arXiv.org/abs/quant-ph/0402027">The general boson normal ordering problem</a>, arXiv:quant-ph/0402027, 2004.
%H P. Blasiak, Karol A. Penson and A. I. Solomon, <a href="http://dx.doi.org/10.1016/S0375-9601(03)00194-4">The general boson normal ordering problem</a>, Phys. Lett. A 309 (2003) 198-205.
%H M. Schork, <a href="http://dx.doi.org/10.1088/0305-4470/36/16/314">On the combinatorics of normal ordering bosonic operators and deforming it</a>, J. Phys. A 36 (2003) 4651-4665.
%F In Maple notation, a(n) = (1/12)*n!*(n+1)!*(n+2)!*hypergeom([n+1, n+2, n+3], [2, 3, 4], 1)/exp(1).
%F a(n) = Sum_{k=3..3*n} A090440(n, k) = (Sum_{k>=3} (1/k!)*Product_{j=1..n} fallfac(k+(j-1)*(4-3), 3))/exp(1), n>=1. From eq.(9) of the Blasiak et al. reference with r=4, s=3. fallfac(n, m) := A008279(n, m) (falling factorials triangle). a(0) := 1 may be added.
%t ff[n_, k_] = Pochhammer[n - k + 1, k]; a[1, 3] = 1; a[n_, k_] := a[n, k] = Sum[Binomial[3, p]*ff[(n - 1 - p + k), 3 - p]*a[n - 1, k - p], {p, 0, 3} ]; a[n_ /; n < 2, _] = 0; Table[Sum[a[n, k] , {k, 3, 3 n}], {n, 1, 9}] (* _Jean-François Alcover_, Sep 01 2011 *)
%Y Cf. A091028 (alternating row sums of A090440).
%K nonn,changed
%O 1,2
%A _Karol A. Penson_, May 02 2002
%E Edited by _Wolfdieter Lang_, Dec 23 2003