A343474 -id:A343474 - cp's OEIS frontend

A343694 a(n) is the number of men's preference profiles in the stable marriage problem with n men and n women, such that all men prefer different women as their first choices.

1, 2, 48, 31104, 955514880, 2149908480000000, 505542895416115200000000, 16786680128857246009393152000000000, 102199258264429373853760111996211036160000000000, 143679021498505654124337567125614729953051527872512000000000000

Offset: 1

Tanya Khovanova and MIT PRIMES STEP Senior group, May 25 2021

nonn

For such a preference profile, the men-proposing Gale-Shapley algorithm ends in one round since each woman receives exactly one proposal in the first round.

This sequence also counts the preference profiles for n men who rank n women such that all men prefer different women as their i-th choice, for an i between 1 and n, inclusive.

			For n=2, there are two ways to pick men's first preferences, and then one way to complete the preference profile, making a total of 2 preference profiles.

Wikipedia, Gale-Shapley algorithm.

Cf. A001013, A185141, A343474, A343695, A342573, A340890.

Mathematica
```
Table[n! (n - 1)!^n, {n, 10}]
```

a(n) = n! * (n-1)!^n.

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

Tanya Khovanova and MIT PRIMES STEP Senior group, May 25 2021

For these profiles both men-proposing and women-proposing Gale-Shapley algorithms end in one round.

This is a subsequence of A001013.

			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.

Wikipedia, Gale-Shapley algorithm.

Cf. A001013, A185141, A343474, A343694, A342573, A340890.

Mathematica
```
Table[n!^2 (n - 1)!^(2 n), {n, 10}]
```

a(n) = n!^2 * (n-1)!^(2*n).

a(n) = A343694(n)^2.

A338665 a(n) is the number of preference profiles for n men and n women where every man prefers woman number 1 to woman number 2.

Original entry on oeis.org

4, 5832, 6879707136, 19349176320000000000, 303256405652583481344000000000000, 53311087345695615264200592956011315200000000000000, 190584865366582887488321066784947980317795794157526056960000000000000000

Offset: 2

Tanya Khovanova and MIT PRIMES STEP Senior group, Apr 22 2021

nonn

This is also the number of preference profiles for n men and n women where every woman prefers man number 1 to man number 2.

When implementing the men-proposing Gale-Shapley algorithm on such a preference profile, woman number 1 gets her first engagement in an earlier round than woman number 2.

			When n = 2, we have exactly 1 way to arrange each man's profiles such that woman number 1 is ranked before woman number 2. Each woman's profile can be set in 2! = 2 ways, so the total number of preference profiles such that every man prefers woman number 1 to woman number 2 is 1^2 * 2^2 = 4.

Wikipedia, Gale-Shapley algorithm.

Cf. A001013, A185141, A342573, A340890, A343474, A343475.

Mathematica
```
Table[n!^(2 n)/2^n, {n, 2, 10}]
```

a(n) = n!^(2n) / 2^n.

A343692 a(n) is the number of men's preference profiles in the stable marriage problem with n men and n women, where every man prefers woman number 1 to woman number 2.

Original entry on oeis.org

1, 27, 20736, 777600000, 2176782336000000, 645362587921121280000000, 27285016590396539545426329600000000, 213106813311662727500673631554480635904000000000, 386661002072680852777222237092449665784217600000000000000000000

Offset: 2

Tanya Khovanova and MIT PRIMES STEP Senior group, May 23 2021

nonn

When implementing the men-proposing Gale-Shapley algorithm on such a preference profile, woman number 1's first engagement comes in an earlier round than the engagement of woman number 2.

This is the same as the number of women's preference profiles in the stable marriage problem with n men and n women, where every woman prefers man number 1 to man number 2.

			When n = 2, there is exactly 1 way for each man's profile to be completed such that woman number 1 is before woman number 2. Since we are only looking at men's profiles, this means there are 1^n = 1^2 = 1 preference profiles such that every man prefers woman number 1 to woman number 2.

Wikipedia, Gale-Shapley algorithm.

Cf. A185141, A338665, A343474, A343693.

Mathematica
```
Table[n!^n/2^n, {n, 2, 10}]
```

a(n) = n!^(n) / 2^n.

a(n) = A338665(n)/n!^(n) = sqrt(A343693(n)).

A343693 a(n) is the number of preference profiles in the stable marriage problem with n men and n women, where every man prefers woman number 1 to woman number 2 and every woman prefers man number 1 to man number 2.

Original entry on oeis.org

1, 729, 429981696, 604661760000000000, 4738381338321616896000000000000, 416492869888246994251567132468838400000000000000, 744472130338214404251254167128703048116389820927836160000000000000000, 45414513879851870274245681660582356320629081347021328317938070440504213897216000000000000000000

Offset: 2

Tanya Khovanova and MIT PRIMES STEP Senior group, May 24 2021

nonn

When implementing the men-proposing Gale-Shapley algorithm on such a preference profile, woman number 1's first engagement comes in an earlier round than the first engagement of woman number 2. Similarly, when implementing the women-proposing Gale-Shapley algorithm on such a preference profile, man number 1's first engagement comes in an earlier round than the first engagement for man number 2.

			When n = 2, each man and each woman have fixed preferences, so every person has exactly 1 way to set their personal preferences, yielding 1 total preference profile.

Wikipedia, Gale-Shapley algorithm.

Cf. A185141, A338665, A343474, A343692.

Mathematica
```
Table[n!^(2 n)/4^n, {n, 2, 10}]
```

a(n) = n!^(2*n) / 4^n.

a(n) = A338665(n)/2^n = A343692(n)^2.

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.

Original entry on oeis.org

1, 14, 5184, 429981696, 39627113103360000, 11555266180939776000000000000, 24157228657754148059243505254400000000000000, 709983949983801273585561911705687568775548764160000000000000000, 520402602329775972199889472492375107519949414596673059590723457777664000000000000000000

Offset: 1

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

nonn

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).

			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.

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.

Cf. A185141, A343474.

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

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

cp's OEIS Frontend

A343694 a(n) is the number of men's preference profiles in the stable marriage problem with n men and n women, such that all men prefer different women as their first choices.

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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.

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A338665 a(n) is the number of preference profiles for n men and n women where every man prefers woman number 1 to woman number 2.

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Author

Keywords

Comments

Examples

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Programs

Mathematica

Formula

A343692 a(n) is the number of men's preference profiles in the stable marriage problem with n men and n women, where every man prefers woman number 1 to woman number 2.

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Keywords

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Examples

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Crossrefs

Programs

Mathematica

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A343693 a(n) is the number of preference profiles in the stable marriage problem with n men and n women, where every man prefers woman number 1 to woman number 2 and every woman prefers man number 1 to man number 2.

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Author

Keywords

Comments

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Crossrefs

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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.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula