A344662 a(n) is the number of preference profiles in the stable marriage problem with n men and n women so that they form n pairs of people of different genders who rank each other first, and so that the men's preferences arranged in a matrix form a Latin square.

1, 2, 96, 746496, 1284211998720, 2427160677580800000000, 6166762687851449045483520000000000, 45287412266290145430585597857888710164480000000000, 1555956528335898586085189699733983238252540690603399394099200000000000, 395245501240598487865502317687285665641954608158944047815164739503046322343116800000000000000

Offset: 1

Author

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

Keywords

• nonn

Comments

Two people who rank each other first are called soulmates. The profiles in this sequence are required to have n pairs of soulmates.

The profiles with n pairs of soulmates are counted by sequence A343698. The profiles such that the men's preference form a Latin square are counted by A343696. The profiles in this sequence are the intersection of profiles in A343696 and A343698.

The Gale-Shapley algorithm (both men-proposing and women-proposing) on the preference profiles described by this sequence ends in one round.

Examples

			For n = 3, there are A002860(3) = 12 ways to set up the men's preference profiles, where A002860(n) is the number of Latin squares of order n. The men's first preferences set the women's first preferences, so we only need to complete the women's profiles with other preferences, which can be done in 2!^3 = 8 ways. Thus, A344662(3) = 12 * 8 = 96.
		

Links

• 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.
• Wikipedia, Gale-Shapley algorithm.

Crossrefs

Cf. A002860, A185141, A343694, A343695, A343696, A343697, A343698, A344663.

Formula

a(n) = (n-1)!^n * A002860(n) = A343696(n)/n^n.