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%I #39 Jan 25 2020 18:13:22
%S 1,5,30,205,1555,12880,115155,1101705,11202680,120415755,1362057155,
%T 16151603830,200144023805,2584429030505,34691478901030,
%U 483040313859705,6963313750468055,103747357497925880,1595132080103893655
%N Number of ways of placing n labeled balls into n unlabeled (but 5-colored) boxes.
%C a(n) is also the exp transform of A010716. - _Alois P. Heinz_, Oct 09 2008
%C The number of ways of putting n labeled balls into a set of bags and then putting the bags into 5 labeled boxes. - _Peter Bala_, Mar 23 2013
%H Alois P. Heinz, <a href="/A144180/b144180.txt">Table of n, a(n) for n = 0..200</a>
%H H. D. Nguyen, D. Taggart, <a href="http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.391.2522&rep=rep1&type=pdf">Mining the OEIS: Ten Experimental Conjectures</a>, 2013; Mentions this sequence. - From _N. J. A. Sloane_, Mar 16 2014
%H N. J. A. Sloane, <a href="/transforms.txt">Transforms</a>
%F a(n) = Sum_{k=0..n} 5^k * A048993(n,k); A048993: Stirling2 numbers.
%F G.f.: A(x) satisfies 5*(x/(1-x))*A(x/(1-x)) = A(x)-1; five times the binomial transform equals this sequence shifted one place left.
%F E.g.f.: exp(5*(exp(x)-1)).
%F G.f.: (G(0) - 1)/(x-1)/5 where G(k) = 1 - 5/(1-k*x)/(1-x/(x-1/G(k+1) )); (recursively defined continued fraction). - _Sergei N. Gladkovskii_, Jan 16 2013
%F a(n) ~ n^n * exp(n/LambertW(n/5)-5-n) / (sqrt(1+LambertW(n/5)) * LambertW(n/5)^n). - _Vaclav Kotesovec_, Mar 12 2014
%F G.f.: Sum_{j>=0} 5^j*x^j / Product_{k=1..j} (1 - k*x). - _Ilya Gutkovskiy_, Apr 07 2019
%p a:= proc(n) option remember; `if`(n=0, 1,
%p (1+add(binomial(n-1, k-1)*a(n-k), k=1..n-1))*5)
%p end:
%p seq(a(n), n=0..25); # _Alois P. Heinz_, Oct 09 2008
%t Table[BellB[n,5],{n,0,20}] (* _Vaclav Kotesovec_, Mar 12 2014 *)
%o (Sage) expnums(19, 5) # _Zerinvary Lajos_, May 15 2009
%Y Cf. A000110, A001861, A027710, A078944. A144223, A144263, A189233, A221159, A221176.
%K nonn
%O 0,2
%A _Philippe Deléham_, Sep 12 2008
%E More terms from _Alois P. Heinz_, Oct 09 2008
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