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A343475 a(n) is the number of preference profiles for n men and n women, where men prefer distinct women as their first choice. 4
1, 8, 10368, 10319560704, 23776267862016000000, 299512499409958993920000000000000, 41761084325232750832975432403386368000000000000000, 117254360528268768669572531322770730078331396796134195200000000000000000, 11151031424792655208856660513601075282865340493496475667265971777832723603783680000000000000000000 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
This sequence is the number of preference profiles for the Stable Marriage Problem such that the male-proposing Gale-Shapley algorithm terminates in one iteration.
This is the same number of preference profiles as when all men rank the different women at the i-th place, where i can be anywhere from 1 to n.
Note this is the same as the number of preference profiles for n men and n women where the women prefer distinct men as their first choice.
LINKS
Michael De Vlieger, Table of n, a(n) for n = 1..22
Matvey Borodin, Eric Chen, Aidan Duncan, Tanya Khovanova, Boyan Litchev, Jiahe Liu, Veronika Moroz, Matthew Qian, Rohith Raghavan, Garima Rastogi, and Michael Voigt, Sequences of the Stable Matching Problem, arXiv:2201.00645 [math.HO], 2021.
FORMULA
a(n) = n!^(n+1) * (n-1)!^n.
EXAMPLE
When n = 3, there are 3! = 6 ways to order the women as first preferences for the men, 2!^3 = 8 ways to finish the mens' profiles, and then 3!^3 = 216 ways to complete the womens' profiles, making a total of 6 * 8 * 216 = 10368 preference profiles.
MATHEMATICA
Table[n!^(n + 1) (n - 1)!^n, {n, 10}]
CROSSREFS
Sequence in context: A188890 A175881 A165429 * A242852 A079235 A134373
KEYWORD
nonn
AUTHOR
Tanya Khovanova and MIT PRIMES STEP Senior group, Apr 16 2021
STATUS
approved

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