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A343692 a(n) is the number of men's preference profiles in the stable marriage problem with n men and n women, where every man prefers woman number 1 to woman number 2. 1
1, 27, 20736, 777600000, 2176782336000000, 645362587921121280000000, 27285016590396539545426329600000000, 213106813311662727500673631554480635904000000000, 386661002072680852777222237092449665784217600000000000000000000 (list; graph; refs; listen; history; text; internal format)
OFFSET
2,2
COMMENTS
When implementing the men-proposing Gale-Shapley algorithm on such a preference profile, woman number 1's first engagement comes in an earlier round than the engagement of woman number 2.
This is the same as the number of women's preference profiles in the stable marriage problem with n men and n women, where every woman prefers man number 1 to man number 2.
LINKS
FORMULA
a(n) = n!^(n) / 2^n.
a(n) = A338665(n)/n!^(n) = sqrt(A343693(n)).
EXAMPLE
When n = 2, there is exactly 1 way for each man's profile to be completed such that woman number 1 is before woman number 2. Since we are only looking at men's profiles, this means there are 1^n = 1^2 = 1 preference profiles such that every man prefers woman number 1 to woman number 2.
MATHEMATICA
Table[n!^n/2^n, {n, 2, 10}]
CROSSREFS
Sequence in context: A036534 A167731 A357506 * A201369 A036529 A185882
KEYWORD
nonn
AUTHOR
Tanya Khovanova and MIT PRIMES STEP Senior group, May 23 2021
STATUS
approved

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Last modified May 18 08:45 EDT 2024. Contains 372618 sequences. (Running on oeis4.)