0,3

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

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.

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

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

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

Cf. A069726, A005470.

Sequence in context: A296788 A079503 A052694 * A346647 A354690 A052599

Adjacent sequences: A069726 A069727 A069728 * A069730 A069731 A069732

Valery A. Liskovets, Apr 07 2002

More terms from Gheorghe Coserea, Apr 13 2018

approved