cp's OEIS Frontend

A057599 a(n) = (n^2)!/(n!)^(n+1); number of ways of dividing n^2 labeled items into n unlabeled boxes of n items each.

%I A057599 #35 Apr 29 2020 12:12:56
%S A057599 1,1,3,280,2627625,5194672859376,3708580189773818399040,
%T A057599 1461034854396267778567973305958400,
%U A057599 450538787986875167583433232345723106006796340625,146413934927214422927834111686633731590253260933067148964500000000
%N A057599 a(n) = (n^2)!/(n!)^(n+1); number of ways of dividing n^2 labeled items into n unlabeled boxes of n items each.
%C A057599 Note that if n=p^k for p prime then a(n) is coprime to n (i.e., a(n) is not divisible by p).
%C A057599 a(n) is also the number of labelings for the simple graph K_n X K_n, the graph Cartesian product of the complete graph with itself. - _Geoffrey Critzer_, Oct 16 2016
%C A057599 a(n) is also the number of knockout tournament seedings with 2 rounds and n participants in each match. - _Alexander Karpov_, Dec 15 2017
%H A057599 Seiichi Manyama, <a href="/A057599/b057599.txt">Table of n, a(n) for n = 0..27</a>
%H A057599 Alexander Karpov, <a href="https://wp.hse.ru/data/2017/12/12/1160180715/WP7_2017_03_________.pdf">Generalized knockout tournaments</a>, National Research University Higher School of Economics. WP7/2017/03.
%F A057599 a(n) = A034841(n)/A000142(n).
%F A057599 a(n) ~ exp(n - 1/12) * n^((n-1)*(2*n-1)/2) / (2*Pi)^(n/2). - _Vaclav Kotesovec_, Nov 23 2018
%e A057599 a(2)=3 since the possibilities are {{0,1},{2,3}}; {{0,2},{1,3}}; and {{0,3},{1,2}}.
%p A057599 a:= n-> (n^2)!/(n!)^(n+1):
%p A057599 seq(a(n), n=0..10);  # _Alois P. Heinz_, Apr 29 2020
%t A057599 Table[a[z_] := z^n/n!; (n^2)! Coefficient[Series[a[a[z]], {z, 0, n^2}],z^(n^2)], {n, 1, 10}] (* _Geoffrey Critzer_, Oct 16 2016 *)
%o A057599 (PARI) a(n) = (n^2)!/(n!)^(n+1); \\ _Altug Alkan_, Dec 17 2017
%Y A057599 Main diagonal of A060540.
%Y A057599 Cf. A000142, A034841.
%K A057599 nonn
%O A057599 0,3
%A A057599 _Henry Bottomley_, Oct 06 2000