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A344689 a(n) is the number of preference profiles in the stable marriage problem with n men and n women such that one man and one woman are ranked last by all the people of the opposite gender except each other. 1
1, 14, 5184, 429981696, 39627113103360000, 11555266180939776000000000000, 24157228657754148059243505254400000000000000, 709983949983801273585561911705687568775548764160000000000000000, 520402602329775972199889472492375107519949414596673059590723457777664000000000000000000 (list; graph; refs; listen; history; text; internal format)
The members of such a pair of people are called outcasts. The outcasts must be matched with each other in any stable matching independently of how they rank each other.
For n other than 2, there can be at most one pair of outcasts.
The number of profiles where the pair of outcasts exists and they rank each other last is A343474(n).
Michael De Vlieger, Table of n, a(n) for n = 1..23
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^4*(n-1)!^(2n) for n != 2; a(2) = 14.
Each person makes a ranking list for all members of the opposite gender without ties. The outcasts are ranked n-th (last) by at least n-1 persons of the opposite gender. This is why for n>2 at most one pair of outcasts can exist.
For n>2, we have n^2 ways to pick the two outcasts, then n!^2 ways to complete the outcasts' preference profiles, and finally (n-1)!^(2n-2) ways to complete everyone else's profiles.
{1, 14}~Join~Table[n^4 (n - 1)!^(2 n), {n, 3, 10}] (* corrected by Michael De Vlieger, Feb 11 2022 *)
Sequence in context: A180588 A214441 A241326 * A159372 A060856 A030531
Tanya Khovanova and MIT PRIMES STEP Senior group, May 30 2021

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Last modified May 18 09:54 EDT 2024. Contains 372620 sequences. (Running on oeis4.)