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A344665 a(n) is the number of preference profiles in the stable marriage problem with n men and n women, where both the men’s preferences and women’s preferences form a Latin square when arranged in a matrix, with no man and woman pairs who rank each other first. 2
0, 2, 48, 124416, 9537454080, 243184270049280000, 1390396658530114967961600000, 4352862027490648408300099378983469056000, 11228731998377005106060609036300637077741992056717312000, 36658843398022550531624696117934603340895735930389121945136191766528000000 (list; graph; refs; listen; history; text; internal format)
The profiles in this sequence are the intersection of the profiles in A343696 and A343697. The Gale-Shapley algorithm on such a set of preference profiles ends in one round.
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) = A002860(n)^2/n! * Sum_{i=0...n} [(-1)^i * n!/i!] = A344664(n) * A000166(n).
For n = 2, there are A002860(2) = 2 ways to set up the men’s profiles. Since the women don’t want to rank the man who ranked them first as first, there is exactly 1 way to set up the women’s profiles. So, there are 2 * 1 = 2 preference profiles for n = 2.
Sequence in context: A203778 A203305 A191954 * A212170 A057527 A166475
Tanya Khovanova and MIT PRIMES STEP Senior group, Jun 22 2021

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