A069726 Number of rooted planar bi-Eulerian maps with 2n edges. Bi-Eulerian: all its vertices and faces are of even valency.
1, 1, 6, 54, 594, 7371, 99144, 1412802, 21025818, 323686935, 5120138790, 82812679560, 1364498150904, 22839100002036, 387477144862128, 6651170184185802, 115346229450879978, 2018559015390399615, 35610482089433479410, 632770874050702595670, 11317118106279639106530
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..650
- M. Bousquet-Mélou and A. Jehanne, Polynomial equations with one catalytic variable, algebraic series and map enumeration, arXiv:math/0504018 [math.CO], 2005.
- M. Bousquet-Mélou and G. Schaeffer, Enumeration of planar constellations, Adv. in Appl. Math. v.24 (2000), 337-368.
- V. A. Kazakov, M. Staudacher and Th. Wynter, Character expansion methods for matrix models of dually weighted graphs, arxiv:hep-th/9502132, 1995; Commun. Math. Phys. 177 (1996), 451-468.
- V. A. Liskovets and T. R. S. Walsh, Enumeration of Eulerian and unicursal planar maps, Discr. Math., 282 (2004), 209-221.
Programs
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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
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Mathematica
Join[{1},Table[3^(n-1) Binomial[3n,n+1]/(n(2n+1)),{n,20}]] (* Harvey P. Dale, Oct 18 2013 *)
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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
Comments