%I #20 Sep 07 2024 01:07:38
%S 1,1,7,55,577,7441,113671,2003527,39971905,889608097,21834577351,
%T 585555975511,17027451783937,533460597334705,17908302027585607,
%U 641152804988733031,24380543011087797121,981149507717921468737,41653436572936172408455,1860174362332664149119607
%N Expansion of exp(x*(1+x)/(1-2*x)).
%C In general, e.g.f. exp(x(1+ax)/(1-bx)) has general term sum{i=0..n, sum{j=0..n, a^j*b^(n-i-j)*C(i,j)C(n-j-1,n-i-j)*n!/i!}}.
%C Let k be a positive integer. The sequence obtained by reducing a(n) modulo k is a purely periodic sequence whose period divides k. For example, taken modulo 14 the sequence becomes [1, 1, 7, 13, 3, 7, 5, 1, 1, 7, 13, 3, 7, 5, ...], a periodic sequence of period 7. Cf. A047974. - _Peter Bala_, Jul 15 2022
%F E.g.f.: exp(x*(1+x)/(1-2*x)).
%F a(n) = Sum_{i = 0..n} Sum_{j = 0..n} 2^(n-i-j)*C(i, j)*C(n-j-1, n-i-j)*n!/i!.
%F a(n) ~ 3^(1/4) * 2^(n-1) * n^(n-1/4) * exp(sqrt(3*n)-n-5/8). - _Vaclav Kotesovec_, Sep 25 2013
%F Conjecture: a(n) +(-4*n+3)*a(n-1) +2*(n-1)*(2*n-5)*a(n-2) +2*(n-1)*(n-2)*a(n-3)=0. - _R. J. Mathar_, Feb 20 2015
%F The e.g.f. A(x) satisfies the differential equation (4*x^2 - 4*x + 1)*A'(x) + (2*x^2 - 2*x - 1)*A(x) = 0 with A(0) = 1. Mathar's conjectured recurrence follows easily from this. - _Peter Bala_, Jul 15 2022
%t With[{nn=20},CoefficientList[Series[Exp[(x(x+1))/(1-2x)],{x,0,nn}], x] Range[0,nn]!] (* _Harvey P. Dale_, Sep 21 2011 *)
%K easy,nonn
%O 0,3
%A _Paul Barry_, Aug 29 2005