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, 14, 5184, 429981696, 39627113103360000, 11555266180939776000000000000, 24157228657754148059243505254400000000000000, 709983949983801273585561911705687568775548764160000000000000000, 520402602329775972199889472492375107519949414596673059590723457777664000000000000000000

Offset: 1

Author

Tanya Khovanova and MIT PRIMES STEP Senior group, May 30 2021

Keywords

• nonn

Comments

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).

Examples

			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.
		

Links

• 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.

Crossrefs

Cf. A185141, A343474.

Programs

Mathematica

{1, 14}~Join~Table[n^4 (n - 1)!^(2 n), {n, 3, 10}] (* corrected by Michael De Vlieger, Feb 11 2022 *)

Formula

a(n) = n^4*(n-1)!^(2n) for n != 2; a(2) = 14.