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A343696 a(n) is the number of preference profiles in the stable marriage problem with n men and n women, such that the men's preference profiles form a Latin square. 6
1, 8, 2592, 191102976, 4013162496000000, 113241608573209804800000000, 5078594244241245901264634511360000000000, 759796697672599288560347581750936194390876487680000000000, 602809439070636186475532789128702956081602819845966698324215778508800000000000 (list; graph; refs; listen; history; text; internal format)
Equivalently, these are the profiles where each woman is ranked differently by the n men.
Equivalently, for every rank i, there is exactly one woman who is ranked i by a given man.
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.
Due to symmetry, a(n) is the number of preference profiles in the stable marriage problem with n men and n women, such that the women’s preference profiles form a Latin square.
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.
a(n) = n!^n * A002860(n).
For n = 3, there are 3!^3 ways to set up the women's preference profiles and A002860(3) ways to set up the men's preference profiles, where A002860(3) = 12 (there are 12 different Latin squares of order 3). Thus a(3) = 3!^3 * A002860(3) = 216 * 12 = 2592.
Sequence in context: A268150 A325062 A247733 * A210126 A259440 A172957
Tanya Khovanova and MIT PRIMES STEP Senior group, May 25 2021

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