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Third column (k=4) of array A078739(n,k) ((2,2)-generalized Stirling2).
3

%I #23 Oct 22 2024 14:34:23

%S 1,38,652,9080,116656,1446368,17636032,213311360,2569812736,

%T 30898216448,371141389312,4455873443840,53483541999616,

%U 641880868118528,7703040602324992,92439308337643520,1109288626710839296

%N Third column (k=4) of array A078739(n,k) ((2,2)-generalized Stirling2).

%C The numerator of the g.f. is the n=2 row polynomial of the triangle A089275.

%H Vincenzo Librandi, <a href="/A089271/b089271.txt">Table of n, a(n) for n = 0..500</a>

%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 P. Blasiak, K. A. Penson and A. I. Solomon, <a href="https://arxiv.org/abs/quant-ph/0402027">The general boson normal ordering problem</a>, arXiv:quant-ph/0402027, 2004.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (20, -108, 144).

%F G.f.: (1+18*x)/((1-2*1*x)*(1-3*2*x)*(1-4*3*x)).

%F a(n) = 6*12^n - 6*6^n + 2^n = d(n) + 18*d(n-1), n>=1, a(0)=1, with d(n) := A016309(n) = A071951(n+3, 3) = (24*12^n-15*6^n+2^n)/10.

%t Table[6*12^n -6*6^n +2^n, {n,0,30}] (* _G. C. Greubel_, Feb 07 2018 *)

%t LinearRecurrence[{20,-108,144},{1,38,652},20] (* _Harvey P. Dale_, Oct 22 2024 *)

%o (Magma) [6*12^n-6*6^n+2^n: n in [0..20]]; // _Vincenzo Librandi_, Sep 02 2011

%o (PARI) for(n=0,30, print1(6*12^n -6*6^n +2^n, ", ")) \\ _G. C. Greubel_, Feb 07 2018

%Y Cf. A089272, A071951 (Legendre-Stirling triangle).

%K nonn,easy

%O 0,2

%A _Wolfdieter Lang_, Nov 07 2003