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%I #6 Dec 06 2021 03:16:54
%S 1,0,1,2,0,0,1,0,6,0,0,7,24,0,0,12,0,0,25,0,120,0,0,260,0,0,61,720,0,
%T 0,360,0,0,1470,0,0,841,0,5040,0,0,15960,0,0,5082,0,0,5251,40320,0,0,
%U 20160,0,0,122640,0,0,134456,0,0,20497,0,362880,0,0,1512000
%N Expansion of e.g.f.: exp(t*x)/(1 - x^2/t - t^3*x^3).
%H G. C. Greubel, <a href="/A158785/b158785.txt">Rows n = 0..50 of the irregular triangle, flattened</a>
%F T(n, k) = coefficients of e.g.f.: t^floor(n/2)*exp(t*x)/(1 - x^2/t - t^3*x^3).
%F From _G. C. Greubel_, Dec 05 2021: (Start)
%F T(n, floor(n/2) + n) = A330044(n).
%F T(n, 0) = A005359(n).
%F T(n, 1) = A005212(n). (End)
%e Irregular triangle begins as:
%e 1;
%e 0, 1;
%e 2, 0, 0, 1;
%e 0, 6, 0, 0, 7;
%e 24, 0, 0, 12, 0, 0, 25;
%e 0, 120, 0, 0, 260, 0, 0, 61;
%e 720, 0, 0, 360, 0, 0, 1470, 0, 0, 841;
%e 0, 5040, 0, 0, 15960, 0, 0, 5082, 0, 0, 5251;
%e 40320, 0, 0, 20160, 0, 0, 122640, 0, 0, 134456, 0, 0, 20497;
%t Table[CoefficientList[Expand[t^Floor[n/2]*n!*SeriesCoefficient[Series[Exp[t*x]/(1 - x^2/t - t^3*x^3), {x, 0, 20}], n]], t], {n, 0, 10}]//Flatten
%o (Sage)
%o f(x, t) = exp(t*x)/(1 - x^2/t - t^3*x^3)
%o def A158785(n, k): return ( factorial(n)*t^(n//2)*( f(x, t) ).series(x, 20).list()[n] ).series(t, 2*n+1).list()[k]
%o flatten([[A158785(n, k) for k in (0..n+(n//2))] for n in (0..10)]) # _G. C. Greubel_, Dec 05 2021
%Y Cf. A005212, A005359, A158706, A158757, A330044.
%K nonn,tabl
%O 0,4
%A _Roger L. Bagula_, Mar 26 2009
%E Edited by _G. C. Greubel_, Dec 05 2021
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