A185141 -id:A185141 - cp's OEIS frontend

A290770 a(n) = Product_{k=1..n} k^(2*k).

1, 1, 16, 11664, 764411904, 7464960000000000, 16249593066946560000000000, 11020848942410302096869949440000000000, 3102093199396597590886754340698424229232640000000000, 465607547420733489126893933985879279492195953053596584509440000000000

Offset: 0

Ilya Gutkovskiy, Aug 10 2017

G. C. Greubel, Table of n, a(n) for n = 0..27
Eric Weisstein's World of Mathematics, Hyperfactorial
Index entries for sequences related to factorial numbers

Cf. A000178, A001044, A001818, A002109, A051675, A055209, A061742, A062206, A074962, A184877, A185141, A260122.

[1] cat [(&*[k^(2*k): k in [1..n]]): n in [1..10]]; // G. C. Greubel, Oct 14 2018

Table[Product[k^(2 k), {k, 1, n}], {n, 0, 9}]
Table[Hyperfactorial[n]^2, {n, 0, 9}]
Table[n!^(2 n)/BarnesG[n + 1]^2, {n, 0, 9}]

a(n) = prod(k=1, n, k^(2*k)) \\ Felix Fröhlich, Aug 10 2017

a(n) = A002109(n)^2.
a(n) = A185141(n)/A000178(n-1)^2 for n > 0.
a(n) = (n!)^(2*n)/G(n+1)^2, where G() is the Barnes G-function.
a(n) ~ A^2*exp(-n^2/2)*n^(n*(n+1))*n^(1/6), where A is the Glaisher-Kinkelin constant (A074962).

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.

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

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

nonn

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.

			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.

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

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

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

nonn

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.

			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.

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.

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

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

a(n) = A000479(n) * A002860(n).

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

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

nonn

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.

			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.

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.

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

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

A344670 a(n) is the number of preference profiles in the stable marriage problem with n men and n women such that there exists a stable matching with one couple where both the man and the woman rank each other last.

Original entry on oeis.org

1, 4, 4536, 5113774080

Offset: 1

Tanya Khovanova and MIT PRIMES STEP Senior group, Jun 02 2021

A man and a woman who rank each other last and end up in a marriage are called a hell-couple. A stable matching cannot have more than one hell-couple.
Given a profile, if there exists a stable matching with a hell-couple, then all the stable matchings for this profile have the same hell-couple.
The Gale-Shapley algorithm (both men-proposing and women-proposing) for such a profile needs at least n rounds to terminate.

			For n = 2, there is a stable matching with a hell-couple if and only if the other two people rank each other first. Now, there are 2 ways to pair the men and women, and 2 ways to choose which couple has a man and woman ranking each other first, making a(2) = 2 * 2 = 4.

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.

Cf. A185141, A344671.

A344671 a(n) is the total number of stable matchings for all possible preference profiles in the stable marriage problem with n men and n women such that there exists a married couple where the woman and the man rank each other last.

Original entry on oeis.org

1, 4, 4608, 5317484544

Offset: 1

Tanya Khovanova and MIT PRIMES STEP Senior group, Jun 05 2021

A man and a woman who rank each other last and end up in a marriage are called a hell-couple. A stable matching cannot have more than one hell-couple.
Given a profile, if there exists a stable matching with a hell-couple, then all the stable matchings for this profile have the same hell-couple.
The Gale-Shapley algorithm (both men-proposing and women-proposing) for such a profile needs at least n rounds to terminate.
A344670(n) is the number of preference profiles such that there exists a stable matching with a hell-couple.
This sequence is distinct from A344670 because in this sequence profiles are counted with their respective multiplicity if they yield multiple stable matchings with a hell-couple.

			For n = 2, each preference profile that has a hell-couple has exactly one stable matching, thus a(2) = A344670(2) = 4. For n > 2, this is no longer the case and a(n) > A344670(n).

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.

Cf. A185141, A344670.

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

A290770 a(n) = Product_{k=1..n} k^(2*k).

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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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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.

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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.

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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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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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A344670 a(n) is the number of preference profiles in the stable marriage problem with n men and n women such that there exists a stable matching with one couple where both the man and the woman rank each other last.

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Author

Keywords

Comments

Examples

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