A239108 Number of hybrid 5-ary trees with n internal nodes.

1, 2, 19, 253, 3920, 66221, 1183077, 21981764, 420449439, 8223704755, 163727846678, 3307039145618, 67600147666909, 1395822347989531, 29070233296701815, 609950649080323320, 12881240945694949696, 273590092192962485985, 5840400740191969187922

Offset

0,2

Links

Alois P. Heinz, Table of n, a(n) for n = 0..300

SeoungJi Hong and SeungKyung Park, Hybrid d-ary trees and their generalization, Bull. Korean Math. Soc. 51 (2014), No. 1, pp. 229-235. See p. 233.

Sheng-liang Yang and Mei-yang Jiang, Pattern avoiding problems on the hybrid d-trees, J. Lanzhou Univ. Tech., (China, 2023) Vol. 49, No. 2, 144-150. (in Mandarin)

Formula

From Paul D. Hanna, Mar 30 2014: (Start)

G.f. A(x) satisfies:

(1) A(x) = (1 + x*A(x)^4) * (1 + x*A(x)^5).

(2) A(x) = ( (1/x)*Series_Reversion( x*(1-x-x^2)^4/(1+x)^4 ) )^(1/4).

(3) A(x) = exp( Sum_{n>=1} x^n*A(x)^(3*n)/n * Sum_{k=0..n} C(n,k)^2 * A(x)^k ).

(4) A(x) = exp( Sum_{n>=1} x^n*A(x)^(4*n)/n * Sum_{k=0..n} C(n,k)^2 / A(x)^k ).

(5) A(x) = Sum_{n>=0} Fibonacci(n+2) * x^n * A(x)^(4*n).

(6) A(x) = G(x*A(x)^3) where G(x) = A(x/G(x)^3) is the g.f. of A007863 (number of hybrid binary trees with n internal nodes).

The formal inverse of g.f. A(x) is (sqrt(1-2*x+5*x^2) - (1+x))/(2*x^5).

a(n) = [x^n] ( (1+x)/(1-x-x^2) )^(4*n+1) / (4*n+1).

(End)

Mathematica

(1/x InverseSeries[x(1 - x - x^2)^4/(1 + x)^4 + O[x]^20])^(1/4) // CoefficientList[#, x]& (* Jean-François Alcover, Oct 02 2019 *)

Prog

(PARI) a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=(1 + x*A^4)*(1 + x*A^5)); polcoeff(A, n)
for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Mar 30 2014
(PARI) a(n)=polcoeff( ((1/x)*serreverse( x*(1-x-x^2)^4/(1+x +x*O(x^n))^4))^(1/4), n)
for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Mar 30 2014
(PARI) a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=exp(sum(m=1, n, sum(j=0, m, binomial(m, j)^2*A^j)*x^m*A^(3*m)/m))); polcoeff(A, n)
for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Mar 30 2014
(PARI) a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=exp(sum(m=1, n, sum(j=0, m, binomial(m, j)^2/A^j)*x^m*A^(4*m)/m))); polcoeff(A, n)
for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Mar 30 2014
(PARI) a(n)=polcoeff(((1+x)/(1-x-x^2 +x*O(x^n)))^(4*n+1)/(4*n+1), n)
for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Mar 30 2014

Crossrefs

Cf. A000045, A007863, A215654, A239107, A239108, A239109.

Column k=5 of A245049.

Sequence in context: A125632 A124125 A234505 * A191806 A349721 A252710

Adjacent sequences: A239105 A239106 A239107 * A239109 A239110 A239111

Keyword

nonn

Author

N. J. A. Sloane, Mar 26 2014

Status

approved