A343695
a(n) is the number of preference profiles in the stable marriage problem with n men and n women, where men prefer different women and women prefer different men as their first choices.
Original entry on oeis.org
1, 4, 2304, 967458816, 913008685901414400, 4622106472375910400000000000000, 255573619105709190896159859671040000000000000000, 281792629748570725486109522755987396047015304495104000000000000000000, 10444688389799535672440661668710965357968392730721066975209656086980827545600000000000000000000
Offset: 1
When n = 3, there are 3! ways for men to pick their first choices and 2!^3 ways to complete their lists of preferences. The same calculation works for women's preferences. As the preferences of different genders are independent, we have a total of 3!^2 * 2!^6 = 2304 such preference profiles for n = 3.
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.
Original entry on oeis.org
1, 2, 96, 746496, 1284211998720, 2427160677580800000000, 6166762687851449045483520000000000, 45287412266290145430585597857888710164480000000000, 1555956528335898586085189699733983238252540690603399394099200000000000, 395245501240598487865502317687285665641954608158944047815164739503046322343116800000000000000
Offset: 1
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.
- 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.
A351413
a(n) is the maximum number of stable matchings in the Latin Stable Marriage Problem of order n.
Original entry on oeis.org
1, 2, 3, 10, 9, 48, 61
Offset: 1
Maximal instance of order 2 with 2 stable matchings:
12
21
Maximal instance of order 3 with 3 stable matchings:
123
231
312
Maximal instance of order 4 with 10 stable matchings (group C2xC2):
1234
2143
3412
4321
Maximal instance of order 5 with 9 stable matchings:
12345
21453
34512
45231
53124
Maximal instance of order 6 with 48 stable matchings (Dihedral group):
123456
214365
365214
456123
541632
632541
Maximal instance of order 7 with 61 stable matchings:
1234567
2316745
3125476
4657312
5743621
6471253
7562134
- C. Converse, Lower bounds for the maximum number of stable pairings for the general marriage problem based on the latin marriage problem, Ph. D. Thesis, Claremont Graduate School, Claremont, CA (1992) [Examples are from 69-70].
- A. T. Benjamin, C. Converse, and H. A. Krieger, Note. How do I marry thee? Let me count the ways, Discrete Appl. Math. 59 (1995) 285-292.
- 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 [Sections 3.7 and 4.2].
- J. S. Hwang, Complete stable marriages and systems of I-M preferences, In: McAvaney K.L. (eds) Combinatorial Mathematics VIII. Lecture Notes in Mathematics, vol 884. Springer, Berlin, Heidelberg (1981) 49-63.
- E. G. Thurber, Concerning the maximum number of stable matchings in the stable marriage problem, Discrete Math., 248 (2002), 195-219.
A351781
a(n) = (n-1)^n*(n-1)!^n.
Original entry on oeis.org
0, 1, 64, 104976, 8153726976, 46656000000000000, 28079296819683655680000000, 2400095991902688012207233433600000000, 37800243186554601452585666030525214621696000000000
Offset: 1
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