A090317 Row sums of triangle in A090285.

1, 2, 7, 28, 118, 510, 2235, 9876, 43870, 195556, 873814, 3911168, 17527904, 78622982, 352911939, 1584927828, 7120769526, 32002212252, 143859840114, 646819996008, 2908670252676, 13081556909292, 58839348572574, 264674150692488, 1190649451348908, 5356483791828840, 24098774900561500

Offset

0,2

Comments

Apply the inverse of the Riordan array (1/(1-x^2),x/(1+x)^2) to 2^n. - Paul Barry, Mar 13 2009

Hankel transform is A079935. - Paul Barry, Mar 13 2009

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Formula

a(n+1) = A000108(n+1) + Sum_{k=0..n} a(n-k)*A001700(k); a(0) = 1.

G.f.: (1-x^2*c(x)^4)/(1-2x*c(x)^2), where c(x) is the g.f. of the Catalan numbers A000108. - Paul Barry, Mar 13 2009

Recurrence: 2*(n+1)*(n+3)*a(n) = (17*n^2+56*n-21)*a(n-1) - 18*(n+4)*(2*n-3)*a(n-2). - Vaclav Kotesovec, Oct 14 2012

a(n) ~ 9^n/2^(n+2). - Vaclav Kotesovec, Oct 14 2012

a(n) = 4*C(2*n-1,n)/(n+1)+3*Sum_{k=1..n-1}(k+1)*2^k*C(2*n-1,n-k-1)/(n+k+1), n>0, a(0)=1. - Vladimir Kruchinin, Feb 21 2019

Mathematica

Table[SeriesCoefficient[(1-x^2*((1-Sqrt[1-4*x])/(2*x))^4)/(1-2*x*((1-Sqrt[1-4*x])/(2*x))^2), {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Oct 14 2012 *)

Prog

(PARI) x='x+O('x^66); Vec((1-x^2*((1-sqrt(1-4*x))/(2*x))^4)/(1-2*x*((1-sqrt(1-4*x))/(2*x))^2)) \\ Joerg Arndt, May 11 2013
(Maxima)
a(n):=if n=0 then 1 else 4*binomial(2*n-1, n)/(n+1)+3*sum(((k+1)*2^(k)*binomial(2*n-1, n-k-1))/(n+k+1), k, 1, n-1); /* Vladimir Kruchinin, Feb 21 2019 */

Crossrefs

Cf. A000108, A001700, A090285.

Sequence in context: A150647 A150648 A150649 * A150650 A150651 A151298

Adjacent sequences: A090314 A090315 A090316 * A090318 A090319 A090320

Keyword

easy,nonn

Author

Philippe Deléham, Jan 25 2004

Extensions

Term 15 corrected by Paul Barry, Mar 13 2009

Status

approved