A158785 - OEIS

%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