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%I #17 Feb 02 2022 23:34:18
%S 1,4,576,26873856,1585084524134400,320979616137216000000000000,
%T 493004666484778531821296025600000000000000,
%U 11093499218496894899774404870401368262117949440000000000000000
%N a(n) is the number of preference profiles for n men and n women, where all men prefer the same woman and all women prefer the same man.
%C Every preference profile of this type has exactly one pair of people who rank each other first.
%C This is the same number of preference profiles as when all men rank the same woman at only the i-th place, and all women rank the same man at only the j-th place, where i and j can be anywhere from 1 to n.
%C The total number of possible profiles is A185141.
%H Matvey Borodin, Eric Chen, Aidan Duncan, Tanya Khovanova, Boyan Litchev, Jiahe Liu, Veronika Moroz, Matthew Qian, Rohith Raghavan, Garima Rastogi, and Michael Voigt, <a href="https://arxiv.org/abs/2201.00645">Sequences of the Stable Matching Problem</a>, arXiv:2201.00645 [math.HO], 2021.
%F a(n) = n^2 * (n-1)!^(2*n).
%F a(n) = A342573(n)^2, where A342573 ignores women's preferences.
%e When n=2, there are 4 ways to pick a man and woman who are preferred by all people of the opposite gender, and then 1 way to fill in each of the remaining slots in every person's preference profile. So, there are 4 different preference profiles.
%t Table[n^2 (n - 1)!^(2n), {n, 10}]
%Y Cf. A185141, A342573, A340890.
%K nonn
%O 1,2
%A _Tanya Khovanova_ and MIT PRIMES STEP Senior group, Apr 16 2021
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