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A350558 a(n) = (n-1)!^(2n). 2

%I #34 Feb 18 2022 21:08:36

%S 1,1,64,1679616,63403380965376,8916100448256000000000000,

%T 10061319724179153710638694400000000000000,

%U 173335925289013982808975076100021379095592960000000000000000,79317573895713454077105543742169655162315106629579798748776628224000000000000000000

%N a(n) = (n-1)!^(2n).

%C a(n) is the number of ways to arrange the remaining preferences in the stable marriage problem of order n after the first choice of each participant has been determined. The first choices are often treated separately in order to avoid mutual first choices, or to avoid multiple participants with the same first choice.

%H Matvey Borodin, Eric Chen, Aidan Duncan, Tanya Khovanova, Boyan Litchev, Jiahe Liu, Veronika Moroz, Matthew Qian, Rohith Raghavan, Garima Rastogi, and Michael Voigt, <a href="https://arxiv.org/abs/2201.00645">Sequences of the Stable Matching Problem</a>, arXiv:2201.00645 [math.HO], 2021.

%F a(n) = (n-1)!^(2n).

%F a(n) = A343700(n)/A284458(n).

%F a(n) = A343698(n)/A000142(n).

%F a(n) = A343699(n)/((n-1)!*n^2*(n^2-1)).

%F a(n) = A343699(n)/(A000142(n)*n*(n^2-1)).

%t Table[(n-1)!^(2n),{n,1,9}]

%Y Cf. A000142, A284458, A343698, A343699, A343700.

%K nonn

%O 1,3

%A _Dan Eilers_, Feb 15 2022

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Last modified September 17 11:15 EDT 2024. Contains 375987 sequences. (Running on oeis4.)