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

0,2

Comments

Partial sums of A162476.

Links

Seiichi Manyama, Table of n, a(n) for n = 0..1000

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

Mathematica

Table[Sum[Binomial[n+3k+1, 4k+1]CatalanNumber[k], {k, 0, n}], {n, 0, 100}] (* Emanuele Munarini, Aug 31 2017 *)

Prog

(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 */

Crossrefs

Cf. A000108, A086616, A162476.

Sequence in context: A203163 A024333 A024334 * A115081 A323672 A103466

Adjacent sequences: A162474 A162475 A162476 * A162478 A162479 A162480

Keyword

easy,nonn

Author

Paul Barry, Jul 04 2009

Status

approved