[go: up one dir, main page]

login
A007082
Number of Eulerian circuits on the complete graph K_{2n+1}, divided by (n-1)!^(2n+1).
(Formerly M2183)
10
2, 264, 1015440, 90449251200, 169107043478365440, 6267416821165079203599360, 4435711276305905572695127676467200, 58393052751308545653929138771580386824519680, 14021772793551297695593332913856884153315254190271692800, 60498832138791357698014788383803842810832836262245623803123983974400
OFFSET
1,1
REFERENCES
D. E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, Section 7.2.1.3, p. 745, Problem 107.
B. D. McKay, Applications of a technique for labeled enumeration, Congress. Numerantium, 40 (1983), 207-221.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
Brendan D. McKay and Robert W. Robinson, Asymptotic enumeration of Eulerian circuits in the complete graph, Combinatorics, Probability and Computing, 7 (1998), 437-449. (gives terms up to n=10, i.e., up through K_{21})
FORMULA
a(n) = A135388(n) / (n-1)!^(2n+1) = A350028(2n+1) / (n-1)!^(2n+1) = A357887(2n+1,n(2n+1)) / (n-1)!^(2n+1). - Max Alekseyev, Oct 19 2022
EXAMPLE
From Günter Rote, Dec 09 2021: (Start)
For n=2, in the graph K5, if we fix the Euler tour to start with the edge 12, we get 132 Euler tours. Here are the first and the last few in lexicographic order.
12314253451
12314254351
12314352451
12314354251
12314524351
...
12543153241
12543241351
12543241531
12543513241
12543514231.
To get all 264*1!^5 = 264 Euler tours, the number must be multiplied by 2 to include the reversed tours. (End)
PROG
(Python)
# for n <= 4
def A(n, w="12"):
if len(w) > (2*n+1)*n: return 2
return sum(A(n, w+t) for t in "123456789"[:2*n+1]
if t!=w[-1] and t+w[-1] not in w and w[-1]+t not in w)
KEYWORD
nonn,nice,walk
EXTENSIONS
More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Mar 28 2003
STATUS
approved