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%I #16 Mar 31 2022 09:24:36
%S 1,-1,14,-1,4,-16,504,-16,4,-34,372,2178,35288,2178,372,-34,496,-5888,
%T 65728,749824,4185760,749824,65728,-5888,496,-11056,154912,-767856,
%U 23350656,230640288,770603712,230640288,23350656,-767856,154912,-11056
%N Expansion of g.f.: 2^(1+floor(n/2))*n!*((1-y)^(n+1)/(1+y))*f(x, y, m), where f(x, y, m) = 2^(m+1)*exp(2^m*t)/((1-y*exp(t))*(1 + (2^(m+1)-1)*exp(2^m*t))), and m = 0.
%H G. C. Greubel, <a href="/A171693/b171693.txt">Rows n = 0..40 of the irregular triangle, flattened</a>
%F G.f.: 2^(1+floor(n/2))*n!*((1-y)^(n+1)/(1+y))*f(x, y, m), where f(x, y, m) = 2^(m+1)*exp(2^m*t)/((1-y*exp(t))*(1 + (2^(m+1)-1)*exp(2^m*t))), and m = 0.
%e Irregular triangle begins as:
%e 1;
%e -1, 14, -1;
%e 4, -16, 504, -16, 4;
%e -34, 372, 2178, 35288, 2178, 372, -34;
%e 496, -5888, 65728, 749824, 4185760, 749824, 65728, -5888, 496;
%t m= 0;
%t f[t_, y_, m_]= 2^(m+1)*Exp[2^m*t]/((1-y*Exp[t])*(1+(2^(m+1)-1)*Exp[2^m*t]));
%t T[n_]:= T[n]= CoefficientList[2^(1+Floor[n/2])*n!*(1-y)^(n+1)/(1 + y)*SeriesCoefficient[Series[f[t, y, m], {t,0,20}], n], y];
%t Table[T[2*n+1], {n,0,12}]//Flatten (* modified by _G. C. Greubel_, Mar 30 2022 *)
%Y Cf. A060187, A159041, A171692, A171694, A171695.
%K sign,tabf
%O 0,3
%A _Roger L. Bagula_, Dec 15 2009
%E Edited by _G. C. Greubel_, Mar 31 2022
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