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.

A069726 Number of rooted planar bi-Eulerian maps with 2n edges. Bi-Eulerian: all its vertices and faces are of even valency.

Original entry on oeis.org

1, 1, 6, 54, 594, 7371, 99144, 1412802, 21025818, 323686935, 5120138790, 82812679560, 1364498150904, 22839100002036, 387477144862128, 6651170184185802, 115346229450879978, 2018559015390399615, 35610482089433479410, 632770874050702595670, 11317118106279639106530
Offset: 0

Views

Author

Valery A. Liskovets, Apr 07 2002

Keywords

Comments

Also counts rooted planar 3-constellations with n triangles: rooted planar maps with bicolored faces having n black triangular faces and an arbitrary number of white faces of degrees multiple to 3. - Valery A. Liskovets, Dec 01 2003

Crossrefs

Programs

  • Maple
    s := 4*(4-81*z)^(1/2): u := 36*I*z^(1/2): a := (s+u)^(1/3): b := (s-u)^(1/3):
    gf := 1 + ((b+a)*s + 108*I*z^(1/2)*(b-a) - 32*(9*z+1))/(432*z):
    simplify(series(gf, z, 22)): seq(coeff(%, z, n), n = 0..20);
    # Peter Luschny, May 19 2024
  • Mathematica
    Join[{1},Table[3^(n-1) Binomial[3n,n+1]/(n(2n+1)),{n,20}]] (* Harvey P. Dale, Oct 18 2013 *)
  • PARI
    A069726(n)=if(n,3^(n-1)*binomial(3*n,n+1)/n/(2*n+1),1)  \\ M. F. Hasler, Mar 26 2012

Formula

a(n) = 3^(n-1)*A000139(n).
a(0)=1, a(n) = 3^(n-1)*binomial(3n, n+1)/(n(2n+1)) for n >= 1.
G.f.: A(x) = (1 + 3*y - y^2)/3 where 3*x^2*y^3 - y + 1 = 0.
G.f. satisfies A(z) = 1 -47*z +3*z^2 +3*z*(22-9*z)*A(z) +9*z*(9*z-2)*A(z)^2 -81*z^2*A(z)^3.
a(n) ~ 2^(-2*n-1)*3^(4*n-1/2)/(sqrt(Pi)*n^(5/2)). - Ilya Gutkovskiy, Dec 04 2016
D-finite with recurrence 2*(n+1)*(2*n+1)*a(n) -9*(3*n-1)*(3*n-2)*a(n-1)=0. - R. J. Mathar, Mar 29 2023
G.f. 1/3 - 2/(27*z) + sqrt(4 - 81*z)*((sqrt(4 - 81*z)/2 + 9*i*sqrt(z)/2)^(1/3) + (sqrt(4 - 81*z)/2 - 9*i*sqrt(z)/2)^(1/3))/(54*z) - (((sqrt(4 - 81*z)/2 + 9*i*sqrt(z)/2)^(1/3) - (sqrt(4 - 81*z)/2 - 9*i*sqrt(z)/2)^(1/3))*i)/(2*sqrt(z)), where i = sqrt(-1). - Karol A. Penson, May 19 2024

Extensions

Entry revised by Editors of the OEIS, Mar 26 - 27 2012