

A340890


a(n) is the number of preference profiles for n men and n women, where all men prefer the same woman.


5



1, 8, 5184, 1719926784, 990677827584000000, 2495937495082991616000000000000, 58001506007267709490243656115814400000000000000, 23264754073069200132851692722771970253637181903994880000000000000000
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OFFSET

1,2


COMMENTS

This is the same number of preference profiles as when all men rank the same woman at the ith place, where i can be anywhere from 1 to n.
Note that we can swap men and women in the definition of the sequence.
The total number of possible profiles is A185141.
a(n) = n!^n A342573(n), where A342573 ignores women's preferences.
a(n) is a subsequence of A001013.


LINKS

Table of n, a(n) for n=1..8.
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.


FORMULA

a(n) = n(n1)!^n * n!^n.


EXAMPLE

When n=2, the total number of profiles is 16, and in half of them, the same woman is ranked 1st by both men.


MATHEMATICA

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


CROSSREFS

Cf. A001013, A185141, A342573.
Sequence in context: A055319 A029736 A206460 * A093937 A318897 A178329
Adjacent sequences: A340887 A340888 A340889 * A340891 A340892 A340893


KEYWORD

nonn


AUTHOR

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


STATUS

approved



