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A344663 a(n) is the number of preference profiles in the stable marriage problem with n men and n women where the men's preferences form a Latin square when arranged in a matrix, and no man and woman rank each other first. 3
0, 2, 768, 60466176, 1315033086689280, 37924385587200000000000000, 1726298879786383239996474654720000000000, 261072919520121696668385285116754694244904468480000000000, 208836950100011929062766575947297434628338701720339215752571230617600000000000, 1378135848291144955393621267341374054991268978878673434553714544944450408726397427961036800000000000000 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
This sequence is in contrast to A344662: a(n) is the number of preference profiles in the stable marriage problem with n men and n women so that they form n pairs of people of different genders who rank each other first, and so that the men's preferences form a Latin square when arranged in a matrix.
Two people who rank each other first are called soulmates. Thus, the profiles in this sequence have no soulmate pairs.
The number of profiles without soulmates is counted by sequence A343700. The number of profiles such that the men's preferences form a Latin square is counted by A343696. The profiles in this sequence are the intersection of profiles in A343696 and A343700.
The men-proposing Gale-Shapley algorithm on the preference profiles described by this sequence 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) * (n-1)^n * (n-1)!^n.
EXAMPLE
For n = 3, there are A002860(3) = 12 ways to set up the men's preference profiles, where A002860(n) is the number of Latin squares of order n. Then, since the women can't rank the men who ranked them first as their first preference, there are 2^3 = 8 ways to set up the women's first preferences, and then 2!^3 = 8 ways to finish the women's profiles. So, A344663(3) = 12 * 8 * 8 = 768 preference profiles.
CROSSREFS
Sequence in context: A145124 A102969 A179960 * A167448 A320481 A078169
KEYWORD
nonn
AUTHOR
Tanya Khovanova and MIT PRIMES STEP Senior group, May 31 2021
STATUS
approved

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