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A343697 a(n) is the number of preference profiles in the stable marriage problem with n men and n women such that both the men's and women's profiles form Latin squares. 5
1, 4, 144, 331776, 26011238400, 660727073341440000, 3779719071732351369216000000, 11832225237539469009819996424230666240000, 30522879094287825948996777484664523152536511038095360000, 99649061600109839440372937690884668992908741561885362729330828902400000000 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
Equivalently, these are the profiles where each woman is ranked differently by the n men and each man is ranked differently by the women.
The men-proposing Gale-Shapley algorithm on such a set of preferences ends in one round, since every woman receives one proposal in the first round. Similarly, the women-proposing Gale-Shapley algorithm ends in one round.
LINKS
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) = A002860(n)^2.
EXAMPLE
There are 12 Latin squares of order 3, where 12 = A002860(3). Thus, for n = 3, there are A002860(3) ways to set up the men's profiles and A002860(3) ways to set up the women's profiles, making A002860(3)^2 = 144 ways to set up all the preference profiles.
CROSSREFS
Sequence in context: A203424 A055209 A239350 * A030450 A041629 A278845
KEYWORD
nonn
AUTHOR
Tanya Khovanova and MIT PRIMES STEP Senior group, May 26 2021
STATUS
approved

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