%I #30 Feb 02 2022 23:34:04
%S 1,8,5184,1719926784,990677827584000000,
%T 2495937495082991616000000000000,
%U 58001506007267709490243656115814400000000000000,23264754073069200132851692722771970253637181903994880000000000000000
%N a(n) is the number of preference profiles for n men and n women, where all men prefer the same woman.
%C This is the same number of preference profiles as when all men rank the same woman at the ith place, where i can be anywhere from 1 to n.
%C Note that we can swap men and women in the definition of the sequence.
%C The total number of possible profiles is A185141.
%C a(n) = n!^n A342573(n), where A342573 ignores women's preferences.
%C a(n) is a subsequence of A001013.
%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(n1)!^n * n!^n.
%e When n=2, the total number of profiles is 16, and in half of them, the same woman is ranked 1st by both men.
%t Table[n (n  1)!^n n!^n, {n, 10}]
%Y Cf. A001013, A185141, A342573.
%K nonn
%O 1,2
%A _Tanya Khovanova_ and MIT PRIMES STEP Senior group, Mar 31 2021
