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A299824
a(n) = (1/e^n)*Sum_{j >= 1} j^n * n^j / (j-1)!.
4
2, 22, 309, 5428, 115155, 2869242, 82187658, 2661876168, 96202473183, 3838516103310, 167606767714397, 7949901069639228, 407048805012563038, 22376916254447538882, 1314573505901491675965, 82188946843192555474704, 5448870914168179374456623, 381819805747937892412056342
OFFSET
1,1
COMMENTS
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).
LINKS
FORMULA
a(n) = A189233(n+1,n). - Omar E. Pol, Feb 24 2018
a(n) ~ exp(n/LambertW(1) - 2*n) * n^(n + 1) / (sqrt(1 + LambertW(1)) * LambertW(1)^(n + 1)). - Vaclav Kotesovec, Mar 08 2018
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
EXAMPLE
a(4) = (1/e^4)*Sum_{j >= 1} j^4 * 4^j / (j-1)! = 5428.
PROG
(PARI) a(n) = round(exp(-n)*suminf(j = 1, (j^n)*(n^j)/(j-1)!)); \\ Michel Marcus, Feb 24 2018
(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
KEYWORD
nonn
AUTHOR
Pedro Caceres, Feb 19 2018
STATUS
approved