cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-6 of 6 results.

A343697 a(n) is the number of preference profiles in the stable marriage problem with n men and n women such that both the men's and women's profiles form Latin squares.

Original entry on oeis.org

1, 4, 144, 331776, 26011238400, 660727073341440000, 3779719071732351369216000000, 11832225237539469009819996424230666240000, 30522879094287825948996777484664523152536511038095360000, 99649061600109839440372937690884668992908741561885362729330828902400000000
Offset: 1

Views

Author

Tanya Khovanova and MIT PRIMES STEP Senior group, May 26 2021

Keywords

Comments

Equivalently, these are the profiles where each woman is ranked differently by the n men and each man is ranked differently by the women.
The men-proposing Gale-Shapley algorithm on such a set of preferences ends in one round, since every woman receives one proposal in the first round. Similarly, the women-proposing Gale-Shapley algorithm ends in one round.

Examples

			There are 12 Latin squares of order 3, where 12 = A002860(3). Thus, for n = 3, there are A002860(3) ways to set up the men's profiles and A002860(3) ways to set up the women's profiles, making A002860(3)^2 = 144 ways to set up all the preference profiles.

Links

Crossrefs

Cf. A002860, A185141, A343696.

Formula

a(n) = A002860(n)^2.

A344663 a(n) is the number of preference profiles in the stable marriage problem with n men and n women where the men's preferences form a Latin square when arranged in a matrix, and no man and woman rank each other first.

Original entry on oeis.org

0, 2, 768, 60466176, 1315033086689280, 37924385587200000000000000, 1726298879786383239996474654720000000000, 261072919520121696668385285116754694244904468480000000000, 208836950100011929062766575947297434628338701720339215752571230617600000000000, 1378135848291144955393621267341374054991268978878673434553714544944450408726397427961036800000000000000
Offset: 1

Views

Author

Tanya Khovanova and MIT PRIMES STEP Senior group, May 31 2021

Keywords

Comments

This sequence is in contrast to 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 form a Latin square when arranged in a matrix.
Two people who rank each other first are called soulmates. Thus, the profiles in this sequence have no soulmate pairs.
The number of profiles without soulmates is counted by sequence A343700. The number of profiles such that the men's preferences form a Latin square is counted by A343696. The profiles in this sequence are the intersection of profiles in A343696 and A343700.
The men-proposing Gale-Shapley algorithm 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. Then, since the women can't rank the men who ranked them first as their first preference, there are 2^3 = 8 ways to set up the women's first preferences, and then 2!^3 = 8 ways to finish the women's profiles. So, A344663(3) = 12 * 8 * 8 = 768 preference profiles.

Links

Crossrefs

Cf. A002860, A185141, A343696, A344662, A343700.

Formula

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

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

Views

Author

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

Keywords

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

Crossrefs

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

Formula

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

A344664 a(n) is the number of preference profiles in the stable marriage problem with n men and n women where both the men's and the women's preferences form a Latin square when arranged in a matrix. In addition, it is possible to arrange all people into n man-woman couples such that they rank each other first.

Original entry on oeis.org

1, 2, 24, 13824, 216760320, 917676490752000, 749944260264355430400000, 293457967200879687743551498616832000, 84112872283641495670736269523436185936222748672000, 27460610008848610956892895086773773421767179663217968124264448000000
Offset: 1

Views

Author

Tanya Khovanova and MIT PRIMES STEP Senior group, Jun 01 2021

Keywords

Comments

Two people who rank each other first are called soulmates. Thus, the profiles in this sequence have n pairs of soulmates.
The profiles with n pairs of soulmates are counted by sequence A343698. The profiles such that the men's preferences form a Latin square are counted by A343696. The profiles such that both men's and women's preferences form a Latin square are counted by A343697. The profiles in this sequence are the intersection of profiles in A343698 and A343697.
Both the men- and the women-proposing Gale-Shapley algorithm on the preference profiles described by this sequence end in one round.

Examples

			For n = 3, there are A002860(3) = 12 Latin squares of order 3. Thus, there are A002860(3) = 12 ways to set up the men's preference profiles. After that, the women's preference profiles form a Latin square with a fixed first column, as the first column is uniquely defined to generate 3 pairs of soulmates. Thus, there are A002860(3)/3! = 12/6 = 2 ways to set up the women's preference profiles, making a(3) = 12 * 2 = 24 preference profiles.

Links

Crossrefs

Cf. A000479, A002860, A185141, A343696, A343697, A343698, A344665.

Formula

a(n) = A002860(n)^2 / n!.
a(n) = A000479(n) * A002860(n).

Extensions

Corrected by Tanya Khovanova, Aug 17 2021

A344665 a(n) is the number of preference profiles in the stable marriage problem with n men and n women, where both the men's preferences and women's preferences form a Latin square when arranged in a matrix, with no paired man and woman who rank each other first.

Original entry on oeis.org

0, 2, 48, 124416, 9537454080, 243184270049280000, 1390396658530114967961600000, 4352862027490648408300099378983469056000, 11228731998377005106060609036300637077741992056717312000, 36658843398022550531624696117934603340895735930389121945136191766528000000
Offset: 1

Views

Author

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

Keywords

Comments

The profiles in this sequence are the intersection of the profiles in A343696 and A343697. The Gale-Shapley algorithm on such a set of preference profiles ends in one round.

Examples

			For n = 2, there are A002860(2) = 2 ways to set up the men's profiles. Since the women don't want to rank the man who ranked them first as first, there is exactly 1 way to set up the women's profiles. So, there are 2 * 1 = 2 preference profiles for n = 2.

Links

Crossrefs

Cf. A002860, A185141, A343696, A343697, A344664.

Formula

a(n) = A002860(n)^2 * Sum_{i=0..n} (-1)^i/i! = A344664(n) * A000166(n).

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

Views

Author

Dan Eilers, Feb 19 2022

Keywords

Comments

a(n) is the number of women's ranking tables in the stable marriage problem that can be paired with a men's ranking table having no two men with the same first choice, without forming any mutual first choices. It has two terms: (n-1)^n from A065440(n), and (n-1)!^n from A091868(n-1). Such men's ranking tables having no two men with the same first choice arise in A343694, A343475, and A344663.
a(n)*A123234 is a useful alternative to A343696 which combines a Latin men's ranking table with an arbitrary women's table, since it gives fewer instances to consider.

Crossrefs

Cf. A065440, A091868, A343694, A343475, A344663, A123234.

Programs

  • Mathematica
    Table[(n-1)^n*(n-1)!^n,{n,1,9}]

Formula

a(n) = (n-1)^n*(n-1)!^n.
a(n) = A065440(n)*A091868(n-1).
Showing 1-6 of 6 results.

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