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a(n) = denominator of (1/e)^n * Sum_{k>=0}(n^k*k^n)/(n!*k!).
1

%I #25 May 27 2022 15:34:53

%S 1,1,1,2,2,3,120,720,1008,40320,362880,45360,39916800,68428800,

%T 6227020800,87178291200,1307674368000,1046139494400,355687428096000,

%U 376610217984000,40548366802944000,2432902008176640000,5676771352412160000,40142883134914560000,25852016738884976640000

%N a(n) = denominator of (1/e)^n * Sum_{k>=0}(n^k*k^n)/(n!*k!).

%F A339401(n)/a(n) = A242817(n)/n!. - _Pontus von Brömssen_, Dec 03 2020

%F a(n) = denominator([x^n] exp(n*(exp(x)-1))). - _Alois P. Heinz_, Dec 07 2020

%p A:= proc(n, k) option remember; `if`(n=0, 1, (1+

%p add(binomial(n-1, j-1)*A(n-j, k), j=1..n-1))*k)

%p end:

%p a:= n-> denom(A(n$2)/n!):

%p seq(a(n), n=0..30); # _Alois P. Heinz_, Dec 07 2020

%t a[n_] := BellB[n, n]/n! // Denominator;

%t Table[a[n], {n, 0, 30}] (* _Jean-François Alcover_, May 27 2022 *)

%Y Cf. A339401 for numerators and relation to A242817.

%K nonn,frac

%O 0,4

%A _William C. Laursen_, Dec 03 2020