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%I #28 Jun 05 2021 04:38:35
%S 1,4,36,424,5716,83544,1288296,20637264,340116276,5730014584,
%T 98241641656,1708602483504,30070563388936,534554579527024,
%U 9584333758817616,173120386421418144,3147337611202622196,57545643875054919864,1057492201661230657176
%N Expansion of 1/(1-4x*c(5x)), where c(x) is the g.f. of A000108.
%C Hankel transform is A135420. - _Paul Barry_, Sep 15 2009
%H Vincenzo Librandi, <a href="/A132864/b132864.txt">Table of n, a(n) for n = 0..200</a>
%F a(n) = Sum_{k=0..n} A039599(n,k)*(-1)^k*5^(n-k). - _Philippe Deléham_, Dec 11 2007
%F Integral representation: a(n) = (2/Pi)*Integral_{x=0..20} x^n*sqrt(x*(20-x))/(x*(16+x)). - _Paul Barry_, Sep 15 2009
%F From _Gary W. Adamson_, Jul 18 2011: (Start)
%F a(n) = upper left term in M^n, M = an infinite square production matrix as follows:
%F 4, 4, 0, 0, 0, 0, ...
%F 5, 5, 5, 0, 0, 0, ...
%F 5, 5, 5, 5, 0, 0, ...
%F 5, 5, 5, 5, 5, 0, ...
%F 5, 5, 5, 5, 5, 5, ...
%F ... (End)
%F Conjecture: n*a(n) + 2*(15-2*n)*a(n-1) + 160*(3-2*n)*a(n-2) = 0. - _R. J. Mathar_, Nov 15 2011
%F a(n) ~ 4^n * 5^(n+1) / (9 * n^(3/2) * sqrt(Pi)). - _Vaclav Kotesovec_, Feb 08 2014
%t CoefficientList[Series[1/(1-4*x*(1-Sqrt[1-20*x])/(10*x)), {x, 0, 20}], x] (* _Vaclav Kotesovec_, Feb 08 2014 *)
%t Table[5^(n + 1) * CatalanNumber[n] * Hypergeometric2F1[1, n + 1/2, n + 2, -5/4]/4, {n, 0, 18}] (* _Vaclav Kotesovec_, Jun 05 2021 *)
%Y Cf. A000108, A039599, A135420.
%K nonn
%O 0,2
%A _Philippe Deléham_, Nov 18 2007
%E More terms added by _Paul Barry_, Sep 15 2009
%E More terms from _Vincenzo Librandi_, Feb 11 2014
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