A069729 Number of rooted non-separable bi-Eulerian planar maps with 2n edges. Bi-Eulerian means all its vertices and faces are of even valency.

1, 1, 2, 8, 54, 442, 4032, 39706, 413358, 4487693, 50348500, 579994802, 6827955072, 81854670861, 996529292432, 12293898494952, 153421680489694, 1934041122204318, 24599034335501730, 315369011873625930, 4072021557616191708, 52915860528084306704, 691646518495876375968

Offset: 0

Author

Valery A. Liskovets, Apr 07 2002

Keywords

• nonn

Comments

The formula from the article by Liskovets and Walsh, p. 218, B'ns(n), gives incorrect data {1, 4, 25, 204, 1964, 21070, 243681, ...}. Here is the incorrect formula rewritten into Mathematica: Table[(Sum[(-1)^j*3^(n - j - 1)*Binomial[2*n + j - 1, j] * Sum[(-1)^k*Binomial[j, k]*Binomial[3*n, n - j - k - 1], {k, 0, Min[j, n - j - 1]}], {j, 0, n - 1}] - 2*Sum[(-1)^j*3^(n - j - 1)*Binomial[2*n + j - 1, j] * Sum[(-1)^k*Binomial[j, k]*Binomial[3*n, n - j - k - 2], {k, 0, Min[j, n - j - 2]}], {j, 0, n - 2}])/n, {n, 1, 20}]. - Vaclav Kotesovec, Apr 13 2018

Examples

			A(x) = 1 + x + 2*x^2 + 8*x^3 + 54*x^4 + 442*x^5 + 4032*x^6 + ...
		

Links

• Gheorghe Coserea, Table of n, a(n) for n = 0..500
• V. A. Liskovets and T. R. S. Walsh, Enumeration of Eulerian and unicursal planar maps, Discr. Math., 282 (2004), 209-221.

Crossrefs

Cf. A069726, A005470.

Programs

Mathematica

CoefficientList[1 + InverseSeries[Series[-(1+x)^4*(18*x^2-30*x-1 + (1-12*x)^(3/2))/(6*(3*x+2)^3), {x, 0, 25}], x], x] (* Vaclav Kotesovec, Apr 14 2018, after Gheorghe Coserea *)

PARI

seq(N) = {
  my(x='x+O('x^(2*N-1)), y=1+serreverse(x/(3*(1+x)^3)), f=(1+3*y-y^2)/3,
     g=subst(f, 'x, 'x^2), v=Vec(subst(g, 'x, serreverse(x*g^2))));
  vector((#v+1)\2, n, v[2*n-1]);
};
seq(23) \\ Gheorghe Coserea, Apr 13 2018

Formula

G.f. y=A(x) satisfies 0 = y^9 - y^8 + 18*x*y^6 - 66*x*y^5 + 47*x*y^4 + 81*x^2*y^3 - 81*x^2*y^2 + 27*x^2*y - 3*x^2. - Gheorghe Coserea, Apr 13 2018

a(n) ~ 2^(6*n - 1) * 3^(8*n - 1/2) / (3125 * sqrt(Pi) * 13^(4*n - 5/2) * n^(5/2)). - Vaclav Kotesovec, Apr 13 2018

A(x) = 1 + serreverse(-(1+x)^4*(18*x^2-30*x-1 + (1-12*x)^(3/2))/(6*(3*x+2)^3)); equivalently, it can be rewritten as A(x) = 1 + serreverse((y - 1)*(3*y^2 + y - 1)^4 / (243 * y^6 * (2*y-1)^3)), where y = A000108(3*x). - Gheorghe Coserea, Apr 14 2018

Extensions

More terms from Gheorghe Coserea, Apr 13 2018