%I #7 Jun 07 2021 09:35:39
%S 1,3,56,17550,225150024,197554684517400,16458566311785642529680,
%T 173358539198065045263504881415600,
%U 300709637734376436340098187751948137677075840,109112041481912234203213339867180762753584908387010487351680
%N a(n) is the number of multisets of size n consisting of permutations of n elements.
%C a(n) is the number of different men's preference profiles in the stable marriage problem with n men and n women up to relabeling the men.
%F a(n) = binomial(n! + n  1, n).
%e Consider n = 3. If all three permutations are the same then there are 6 possibilities from which permutation to choose. If two permutations are the same (6 possibilities for each) and the third permutation is different (5 permutations left), then the number of possibilities is 30. If all three permutations are different, then the number of ways to choose them is 6 * 5 * 4/6 = 20. Thus, a(3) = 20 + 30 + 6 = 56.
%t %Table[Binomial[Factorial[n] + n  1, n], {n, 10}]
%Y Cf. A185141.
%K nonn
%O 1,2
%A _Tanya Khovanova_ and MIT PRIMES STEP Senior group, Jun 04 2021
