A071879 G.f. satisfies: A(x) = 1 + x*A(x) + x^3*A(x)^3.
1, 1, 1, 2, 5, 11, 24, 57, 141, 349, 871, 2212, 5688, 14730, 38403, 100829, 266333, 706997, 1885165, 5047522, 13565203, 36578497, 98934826, 268342933, 729709432, 1989021256, 5433518806, 14873285506, 40790118487, 112064912455, 308390452661, 849969894794, 2346045295997
Offset: 0
Keywords
Examples
G.f.: A(x) = 1 + x + x^2 + 2*x^3 + 5*x^4 + 11*x^5 + 24*x^6 + ... The first-order differences of the terms form the coefficients of A(x)^3: A(x)^3 = 1 + 3*x + 6*x^2 + 13*x^3 + 33*x^4 + 84*x^5 + 208*x^6 + 522*x^7 + ...
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..1000
- Paul Barry, Riordan arrays, generalized Narayana triangles, and series reversion, Linear Algebra and its Applications, 491 (2016) 343-385.
- Olivier Bodini, Matthieu Dien, Antoine Genitrini, and Frédéric Peschanski, The Ordered and Colored Products in Analytic Combinatorics: Application to the Quantitative Study of Synchronizations in Concurrent Processes. In 2017 Proceedings of the Fourteenth Workshop on Analytic Algorithmics and Combinatorics (ANALCO).
- Isaac DeJager, Madeleine Naquin, and Frank Seidl, Colored Motzkin Paths of Higher Order, VERUM 2019.
Programs
-
Maple
a:= n-> add(binomial(n+1,1+2*i)*binomial(n-2*i,i), i=0..floor(n/3))/(n+1): seq(a(n), n=0..29);
-
Mathematica
a[n_] := Sum[Binomial[n+1, 1+2i]*Binomial[n-2i, i], {i, 0, Floor[n/3]}]/(n+1);
-
PARI
a(n)=if(n<0,0,polcoeff(serreverse(x^2/(x+x^2+x^4+x^2*O(x^n))),n+1))
-
PARI
Vec(serreverse(x/(1+x+x^3)+O(x^66))/x) /* Joerg Arndt, Aug 19 2012 */
-
PARI
{a(n)=local(A=1); for(i=1,n,A=sum(m=0, n, x^m*sum(k=0, m, binomial(m, k)*(x*A)^(2*k)) +x*O(x^n))); polcoeff(A, n)} for(n=0, 40, print1(a(n), ", ")) \\ Paul D. Hanna, Sep 05 2014
Formula
a(n) = (Sum_{i=0..floor(n/3)} C(n+1, 1+2i)*C(n-2i, i))/(n+1). - Emeric Deutsch, Nov 03 2002
a(n) = Sum_{k=0..floor(n/3)} C(n,3k)*C(3k,k)/(2k+1). - Paul Barry, Sep 07 2006
D-finite with recurrence: 2*n*(2*n+3)*a(n) + 2*(1-6*n^2)*a(n-1) + 6*(2*n-1)*(n-1)*a(n-2) - 31*(n-1)*(n-2)*a(n-3) = 0. - R. J. Mathar, Nov 13 2012
a(n) ~ (2+3*2^(1/3))^(3/2) * (1+3*2^(-2/3))^n/(4*sqrt(6*Pi)*n^(3/2)). - Vaclav Kotesovec, Aug 19 2013
G.f. satisfies: A(x) = Sum_{n>=0} x^n * Sum_{k=0..n} C(n,k) * (x*A(x))^(2*k). - Paul D. Hanna, Sep 05 2014
Extensions
Name changed by Paul D. Hanna, Nov 15 2012
Comments