A162477 Expansion of (1/(1-x)^2)*c(x/(1-x)^4) where c(x) is the g.f. of A000108.
1, 3, 11, 50, 255, 1391, 7939, 46821, 283081, 1745212, 10929625, 69338213, 444668749, 2877994064, 18774736487, 123321704739, 814930698217, 5413955476648, 36138368789601, 242252716083298, 1630170332414433
Offset: 0
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..1000
Programs
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Mathematica
Table[Sum[Binomial[n+3k+1,4k+1]CatalanNumber[k],{k,0,n}],{n,0,100}] (* Emanuele Munarini, Aug 31 2017 *) -
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 */
Formula
G.f.: 1/((1 - x)^2 - x/((1 - x)^2 - x/((1 - x)^2 - x/((1 - x)^2 - ... (continued fraction);
a(n) = Sum_{k=0..n} C(n+3k+1,n-k)*A000108(k).
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
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
G.f. A(x) satisfies: A(x) = (1 + x*A(x)^2) / (1 - x)^2. - Ilya Gutkovskiy, Jun 30 2020
G.f.: 1/G(x), where G(x) = 1 - (3*x - x^2)/(1 - x/G(x)) (continued fraction). - Nikolaos Pantelidis, Jan 08 2023
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