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A176826 a(n+1) = Sum_{p=0..n} (a(p)*a(n-p)+k)+l for n>=1, where a(0)=1, a(1)=1, k=1 and l=-1. 1
1, 1, 3, 9, 27, 85, 283, 985, 3539, 13013, 48707, 184921, 710347, 2755669, 10780139, 42477977, 168439619, 671641685, 2691362195, 10832277401, 43771088315, 177504638933, 722178443963, 2946919157081, 12057932335283, 49461067106261, 203355663470307, 837870162610137 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

LINKS

Table of n, a(n) for n=0..27.

FORMULA

G.f.: (1-sqrt(1-4*z*(a(0)-z*a(0)^2+z*a(1)+(k+l)*z^2/(1-z)+k*z^2/(1-z)^2)))/(2*z) (k=1, l=-1).

Conjecture: (n+1)*a(n) +(-7*n+2)*a(n-1) +3*(5*n-7)*a(n-2) +(-17*n+46)*a(n-3) +4*(2*n-7)*a(n-4)=0. - R. J. Mathar, Feb 18 2016

a(n) = Sum_{k=0..n} C(k)*Sum_{i=0..(n-k)/2} binomial(k+1,i)*binomial(n-k-1,n-k-2*i), where C(k) is the k-th Catalan number (A000108). - Vladimir Kruchinin, May 09 2018

a(n) ~ sqrt(3*((32 - 13*sqrt(2))^(1/3) + (32 + 13*sqrt(2))^(1/3))) * ((2 + (sqrt(2) - 1)^(2/3) + (1 + sqrt(2))^(2/3))^n / (2^(11/6) * sqrt(Pi) * n^(3/2))). - Vaclav Kotesovec, May 11 2018

EXAMPLE

a(2)=2*1*1+2-1=3. a(3)=2*1*3+2+1^1+1-1=9. a(4)=2*1*9+2+2*1*3+2-1=27.

MAPLE

l:=-1: : k := 1 : m:=1:d(0):=1:d(1):=m: for n from 1 to 30 do d(n+1):=sum(d(p)*d(n-p)+k, p=0..n)+l:od :

taylor((1-sqrt(1-4*z*(d(0)-z*d(0)^2+z*m+(k+l)*z^2/(1-z)+k*z^2/(1-z)^2)))/(2*z), z=0, 30); seq(d(n), n=0..30);

PROG

(Maxima)

a(n):=sum((binomial(2*k, k)*sum(binomial(k+1, i)*binomial(n-k-1, n-k-2*i), i, 0, (n-k)/2))/(k+1), k, 0, n); /* Vladimir Kruchinin, May 09 2018 */

CROSSREFS

Cf. A000108, A176759.

Sequence in context: A131428 A099787 A308520 * A146786 A151029 A179263

Adjacent sequences:  A176823 A176824 A176825 * A176827 A176828 A176829

KEYWORD

easy,nonn

AUTHOR

Richard Choulet, Apr 27 2010

EXTENSIONS

More terms from Vaclav Kotesovec, May 11 2018

STATUS

approved

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Last modified January 23 16:48 EST 2020. Contains 331173 sequences. (Running on oeis4.)