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A162477 Expansion of (1/(1-x)^2)*c(x/(1-x)^4) where c(x) is the g.f. of A000108.

%I A162477 #25 Jul 30 2023 09:20:04
%S A162477 1,3,11,50,255,1391,7939,46821,283081,1745212,10929625,69338213,
%T A162477 444668749,2877994064,18774736487,123321704739,814930698217,
%U A162477 5413955476648,36138368789601,242252716083298,1630170332414433
%N A162477 Expansion of (1/(1-x)^2)*c(x/(1-x)^4) where c(x) is the g.f. of A000108.
%C A162477 Partial sums of A162476.
%H A162477 Seiichi Manyama, <a href="/A162477/b162477.txt">Table of n, a(n) for n = 0..1000</a>
%F A162477 G.f.: 1/((1 - x)^2 - x/((1 - x)^2 - x/((1 - x)^2 - x/((1 - x)^2 - ... (continued fraction);
%F A162477 a(n) = Sum_{k=0..n} C(n+3k+1,n-k)*A000108(k).
%F A162477 Conjecture: (n+1)*a(n) +4*(1-2*n)*a(n-1) +6*(n-2)*a(n-2) +2*(7-2*n)*a(n-3) +(n-5)*a(n-4) = 0. - _R. J. Mathar_, Nov 17 2011
%F A162477 G.f.: (1 - 2*x + x^2 - sqrt(1 - 8*x + 6*x^2 - 4*x^3 + x^4))/(2*x). Remark: using this form of the g.f., it is straightforward to prove the above conjectured recurrence. - _Emanuele Munarini_, Aug 31 2017
%F A162477 G.f. A(x) satisfies: A(x) = (1 + x*A(x)^2) / (1 - x)^2. - _Ilya Gutkovskiy_, Jun 30 2020
%F A162477 G.f.: 1/G(x), where G(x) = 1 - (3*x - x^2)/(1 - x/G(x)) (continued fraction). - _Nikolaos Pantelidis_, Jan 08 2023
%t A162477 Table[Sum[Binomial[n+3k+1,4k+1]CatalanNumber[k],{k,0,n}],{n,0,100}] (* _Emanuele Munarini_, Aug 31 2017 *)
%o A162477 (Maxima) makelist(sum(binomial(n+3*k+1,4*k+1)*binomial(2*k,k)/(k+1),k,0,n),n,0,12); /* _Emanuele Munarini_, Aug 31 2017 */
%Y A162477 Cf. A000108, A086616, A162476.
%K A162477 easy,nonn
%O A162477 0,2
%A A162477 _Paul Barry_, Jul 04 2009