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%I #15 Aug 28 2019 16:38:27
%S 1,20,10,1,1120,1040,290,30,1,123200,161920,71320,14040,1340,60,1,
%T 22422400,37452800,22097600,6263040,958720,82800,4000,100,1,
%U 6098892800,12222918400,8928102400,3257116800,675281600,84782880,6625920,322000
%N Generalized Stirling2 array (5,2).
%C The row length sequences for this array is [1,3,5,7,9,11,...]=A005408(n-1), n>=1.
%H P. Blasiak, K. 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 W. Lang, <a href="/A091534/a091534.txt">First 6 rows</a>.
%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 a(n, k)=(((-1)^k)/k!)*sum(((-1)^p)*binomial(k, p)*product(fallfac(p+3*(j-1), 2), j=1..n), p=2..k), n>=1, 2<=k<=2*n, else 0. From eq. (12) of the Blasiak et al. reference with r=5, s=2.
%F Recursion: a(n, k)=sum(binomial(2, p)*fallfac(3*(n-1)+k-p, 2-p)*a(n-1, k-p), p=0..2), n>=2, 2<=k<=2*n, a(1, 2)=1, else 0. Rewritten from eq.(19) of the Schork reference with r=5, s=2. fallfac(n, m) := A008279(n, m) (falling factorials triangle).
%t a[n_, k_] := (-1)^k/k!*Sum[(-1)^p*Binomial[k, p]*Product[FactorialPower[p + 3*(j - 1), 2], {j, 1, n}], {p, 2, k}]; Table[a[n, k], {n, 1, 8}, {k, 2, 2 n}] // Flatten (* _Jean-François Alcover_, Sep 01 2016 *)
%Y Cf. A078740 (3, 2)-Stirling2, A090438 (4, 2)-Stirling2.
%Y Cf. A072019 (row sums), A091537 (alternating row sums).
%K nonn,easy,tabf
%O 1,2
%A _Wolfdieter Lang_, Jan 23 2004
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