A069736 Total number of Eulerian circuits in labeled multigraphs with n edges.

1, 2, 13, 150, 2541, 57330, 1623105, 55405350, 2216439225, 101738006370, 5271938032725, 304455567165750, 19391501988260325, 1350480167457671250, 102096314890336391625, 8327231070135771543750, 728877648485930118800625

Offset

0,2

Links

Table of n, a(n) for n=0..16.

B. Lass, Démonstration combinatoire de la formule de Harer-Zagier, (A combinatorial proof of the Harer-Zagier formula) C. R. Acad. Sci. Paris, Serie I, 333 (2001) No 3, 155-160.

B. Lass, Démonstration combinatoire de la formule de Harer-Zagier, (A combinatorial proof of the Harer-Zagier formula) C. R. Acad. Sci. Paris, Serie I, 333 (2001) No 3, 155-160.

Valery Liskovets, A Note on the Total Number of Double Eulerian Circuits in Multigraphs , Journal of Integer Sequences, Vol. 5 (2002), Article 02.2.5

Formula

a(n) = (2*n)!/(2^n*n!)(3^(n+1)-1)/(2*(n+1)).

E.g.f.: (sqrt(1-2*x)-sqrt(1-6*x))/(2*x).

From Sergei N. Gladkovskii, Jul 25 2012: (Start)

G.f.: 1 + 8*x/(G(0)-8*x); where G(k) = x*(k+1)*(2*k+1)*(9*3^k-1) + (k+2)*(3*3^k-1) - x*((k+2)^2)*(3*3^k-1)*(2*k+3)*(27*3^k-1)/G(k+1); (continued fraction, Euler's 1st kind, 1-step).

G.f.: (3/2)*G(0) where G(k) = 1 - 1/(3*3^k - 27*x*(k+1)*(2*k+1)*9^k/(9*x*(2*k+1)*(k+1)*3^k - (k+2)/Q2)); (continued fraction, 3rd kind, 3-step).

E.g.f.: (sqrt(1-2*x) - sqrt(1-6*x))/(2*x) = G(0)/(2*x); where G(k) = 1 - 3^k/(1 - x*(2*k-1)/(x*(2*k-1) - 3^k*(k+1)/G(k+1))); (continued fraction, 3rd kind, 3-step).

(End)

Mathematica

Table[(2n)!/(2^n n!)(3^(n+1)-1)/(2(n+1)), {n, 0, 20}] (* Harvey P. Dale, Aug 24 2019 *)

Prog

(PARI) x=xx+O(xx^33); Vec(serlaplace((sqrt(1-2*x)-sqrt(1-6*x))/(2*x))) \\ Michel Marcus, Dec 11 2014

Crossrefs

Cf. A011781.

Sequence in context: A373179 A364339 A204554 * A363846 A058192 A367820

Adjacent sequences: A069733 A069734 A069735 * A069737 A069738 A069739

Keyword

easy,nonn

Author

Valery A. Liskovets, Apr 07 2002

Status

approved