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%I #31 Mar 27 2017 03:53:42
%S 1,1,1,1,3,11,31,71,157,397,1141,3301,9087,24311,66067,185771,532121,
%T 1520889,4316233,12255913,35079739,101232419,293236615,849895311,
%U 2465119669,7167636741,20909386941,61162159501,179214613111,525803297743,1544899158331
%N Expansion of series reversion of x/(1 + x + 2*x^4).
%H G. C. Greubel, <a href="/A190590/b190590.txt">Table of n, a(n) for n = 1..1000</a> (terms 1..165 from Vincenzo Librandi)
%H Vladimir Kruchinin, <a href="http://arxiv.org/abs/1211.3244">The method for obtaining expressions for coefficients of reverse generating functions</a>, arXiv:1211.3244 [math.CO], 2012
%F a(n) = Sum_{j=floor((3*n+1)/4)..n} binomial(j,-3*n+4*j-1)*2^(n-j)*binomial(n,j))/n.
%F Recurrence: 3*(n-1)*(3*n-7)*(3*n+1)*a(n) = 3*(2*n-3)*(18*n^2 - 54*n + 29)*a(n-1) - 3*(n-2)*(54*n^2 - 216*n + 209)*a(n-2) + 54*(n-3)*(n-2)*(2*n-5)*a(n-3) + 485*(n-4)*(n-3)*(n-2)*a(n-4). - _Vaclav Kotesovec_, Aug 20 2013
%F a(n) ~ 6^(1/4)*sqrt(2*6^(3/4)+16)*(1+4/3*6^(1/4))^n/(24*sqrt(Pi)*n^(3/2)). - _Vaclav Kotesovec_, Aug 20 2013
%e g.f.: x + x^2 + x^3 + x^4 + 3*x^5 + 11*x^6 + 31*x^7 + ...
%t Rest[CoefficientList[InverseSeries[Series[x/(1+x+2*x^4),{x,0,20}],x],x]] (* _Vaclav Kotesovec_, Aug 20 2013 *)
%o (Maxima) a(n):=sum(binomial(j, -3*n+4*j-1)*2^(n-j)*binomial(n, j), j, floor((3*n+1)/4), n)/n;
%o (PARI) x='x+O('x^66); /* that many terms */
%o Vec(serreverse(x/(1+x+2*x^4))) /* show terms */ /* _Joerg Arndt_, May 27 2011 */
%K nonn
%O 1,5
%A _Vladimir Kruchinin_, May 26 2011
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