cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A026418 Number of ordered trees with n edges and having no branches of length 1.

Original entry on oeis.org

1, 0, 1, 1, 2, 3, 6, 11, 22, 43, 87, 176, 362, 748, 1560, 3270, 6897, 14613, 31104, 66459, 142518, 306603, 661572, 1431363, 3104619, 6749390, 14704387, 32098729, 70198656, 153785705, 337443785, 741551614, 1631910081, 3596083215
Offset: 0

Views

Author

Emeric Deutsch, Dec 04 2003

Keywords

Comments

Hankel transform is A166446(n+2). - Paul Barry, Mar 29 2011

Examples

			a(6)=6 because we have the following six ordered trees with 6 edges and no branches of length 1 (hanging from the root): (i) a path of length 6, (ii) a path of length 2 and a path of length 4, (iii) two paths of length 3, (iv) a path of length 4 and a path of length 2, (v) three paths of length 2 and (vi) a path of length 2 at the end of which two paths of length 2 are hanging.
		

Crossrefs

Cf. A001006.

Programs

  • Mathematica
    CoefficientList[Series[(1-x+x^2-Sqrt[1-2*x-x^2+2*x^3-3*x^4])/(2*x^2), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 12 2014 *)

Formula

a(n) = Sum_{j=0..floor(n/2)-1} binomial(n-2-j, j)*m(j), where m(j) = Sum_{k=0..floor(j/2)} binomial(j, 2*k)*binomial(2*k, k)/(k+1) is the Motzkin number A001006(j) for n > 2.
G.f.: g=g(z) satisfies z^2g^2-(1-z+z^2)g+1-z+z^2=0
G.f.: [1,1,2,3,...] has g.f. 1/(1-x-x^2-x^4/(1-x-x^2-x^4/(1-x-x^2-x^4/(1... (continued fraction). - Paul Barry, Jul 16 2009
G.f.: c(x^2/(1-x+x^2)) where c(x) is the g.f. of A000108.
G.f.: g(x)=1/(1-x^2/(1-x+x^2-x^2*g(x)))=1/(1-x^2/(1-x+x^2-x^2/(1-x^2/(1-x+x^2-x^2/(1-... (continued fraction). - Paul Barry, Mar 29 2011
D-finite with recurrence (n+2)*a(n) +(-2*n-1)*a(n-1) +(-n+1)*a(n-2) +(2*n-5)*a(n-3) +3*(-n+4)*a(n-4)=0. - R. J. Mathar, Nov 24 2012
G.f.: (1 - x + x^2 - sqrt(1- 2*x - x^2 + 2*x^3 - 3*x^4))/(2*x^2). - Sergei N. Gladkovskii, Oct 04 2013
a(n) ~ sqrt(26+2*sqrt(13)) * ((1+sqrt(13))/2)^n / (4 * sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Feb 12 2014