A322751 Number of directed graphs of 2*n vertices each having an in-degree and out-degree of 1 such that the graph specifies a pairwise connected gift exchange.

1, 0, 4, 80, 4704, 436992, 58897920, 10880501760, 2640513576960, 814928486400000, 311763576754667520, 144816978675459686400, 80294888451877031116800, 52385405443881146567884800, 39727727942688305214337843200, 34656123210118214086941474816000

Offset: 0

Author

Russell Y. Webb, Dec 25 2018

Keywords

• nonn

Comments

The sequence is the number of unique arrangements of directed graphs connecting 2*n vertices, where vertices occur in pairs, and meeting the following requirements:

1. Each vertex has an out-degree and in-degree of 1.

2. No edge connects vertices that are paired.

3. Starting with any pair, following the edges of paired vertices connects all vertices.

The requirements were chosen to yield a nice gift exchange between a set of couples. Acknowledgement to the additional members of the initial, inspirational gift exchange group: Cat, Brad, Kim, Ada, Graham, Nolan, and Leah.

The fraction of graphs meeting the requirements is approximately 0.12. Starting with n=2, the fractions are (0.166666667, 0.111111111, 0.116666667, 0.12042328, 0.122959756, 0.124807468). Is there a way to compute the percentage of graphs satisfying the condition in the limit as n approaches infinity?

Examples

			For n = 1, there is one pair; a(1) = 0 since requirements 1 and 2 can't be satisfied.
For n = 2, there are two pairs; a(2) = 4 graphs given by these edge destinations:
    ((2, 3), (1, 0))
    ((2, 3), (0, 1))
    ((3, 2), (1, 0))
    ((3, 2), (0, 1)).
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 0..100

Crossrefs

A322750 does not allow reciprocal connections.

A010050 is the number of graphs (2n)!.

Cf. A000316, A003471.

Programs

PARI

\\ Here B(n) gives A003471 as vector.
B(n)={my(v=vector(n+1)); v[1]=1; for(n=4, n, my(m = 2-n%2); v[n+1] = v[n]*(n-1) + 2*(n-m)*v[n-2*m+1]); v}
seq(n)={my(v=B(2*n)); Vec(serlaplace(1+log(sum(k=0, n, v[1+2*k]*x^k/k!, O(x*x^n)))))} \\ Andrew Howroyd, Jan 13 2024

Python

from itertools import permutations as perm
def num_connected_by_pairs(assigned, here=0, seen=None):
    seen = (seen, set())[seen is None]
    for proposed in [(here - 1, here), (here, here + 1)][(here % 2) == 0]:
        if proposed not in seen:
            seen.add(proposed)
            num_connected_by_pairs(assigned, assigned[proposed], seen)
    return len(seen)
def valid(assigned, pairs):
    self_give = [assigned[i] == i for i in range(len(assigned))]
    same_pair = [assigned[i] == i + 1 or assigned[i+1] == i for i in range(0, 2*pairs, 2)]
    if pairs == 0 or True in self_give + same_pair:
        return False
    num_connected = [num_connected_by_pairs(assigned, here) for here in range(2, 2*pairs, 2)]
    return min(num_connected) == 2*pairs
print([len([x for x in perm(range(2*pairs)) if valid(x, pairs)]) for pairs in range(0, 6)])

Formula

E.g.f.: 1 + log(B(x)) where B(x) is the e.g.f. of A000316. - Andrew Howroyd, Jan 13 2024

Extensions

a(0) changed to 1 and a(8) onwards from Andrew Howroyd, Jan 13 2024