A299824 - OEIS

%I #64 Mar 17 2020 22:46:29

%S 2,22,309,5428,115155,2869242,82187658,2661876168,96202473183,

%T 3838516103310,167606767714397,7949901069639228,407048805012563038,

%U 22376916254447538882,1314573505901491675965,82188946843192555474704,5448870914168179374456623,381819805747937892412056342

%N a(n) = (1/e^n)*Sum_{j >= 1} j^n * n^j / (j-1)!.

%C For m>1, A242817(m) and a(m-1) are also the m-th and (m+1)-st terms of the sequences "Number of ways of placing X labeled balls into X unlabeled (but (m-1)-colored) boxes". For instance, sequence A144180 for 5-colored boxes (m = 6), has A144180(6) = 12880, and A144180(7) = 115155, which are A242817(6) and a(5) respectively. Same pattern can be observed for A027710, A144223, A144263 (comment added after _Omar E. Pol_'s formula).

%H Alois P. Heinz, <a href="/A299824/b299824.txt">Table of n, a(n) for n = 1..368</a>

%F a(n) = A189233(n+1,n). - _Omar E. Pol_, Feb 24 2018

%F a(n) ~ exp(n/LambertW(1) - 2*n) * n^(n + 1) / (sqrt(1 + LambertW(1)) * LambertW(1)^(n + 1)). - _Vaclav Kotesovec_, Mar 08 2018

%F Or: a(n) ~ (1/sqrt(1+w)) * exp(1/w-2)^n * (n/w)^(n+1), with w = LambertW(1) ~ 0.56714329... The relative error decreases from 10^-2 for a(2) to 10^-3 for a(15), but reaches 10^-3.5 only at a(45). - _M. F. Hasler_, Mar 09 2018

%e a(4) = (1/e^4)*Sum_{j >= 1} j^4 * 4^j / (j-1)! = 5428.

%o (PARI) a(n) = round(exp(-n)*suminf(j = 1, (j^n)*(n^j)/(j-1)!)); \\ _Michel Marcus_, Feb 24 2018

%o (PARI) A299824(n,f=exp(n),S=n/f,t)=for(j=2,oo,S+=(t=j^n*n^j)/(f*=j-1);t<f&&j>n&&return(ceil(S))) \\ For n > 23, use \p## with some ## >= 2n. - _M. F. Hasler_, Mar 09 2018

%Y Cf. A027710, A144180, A144223, A144263, A189233, A242817, A292860.

%K nonn

%O 1,1

%A _Pedro Caceres_, Feb 19 2018