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A213048 Number of preferential arrangements of n labeled elements with repetitions allowed.

%I A213048 #22 Nov 21 2020 16:15:28
%S A213048 1,1,5,31,229,1981,19775,224589,2864901,40591255,632760105,
%T A213048 10765616885,198543617119,3945765358653,84070841065937,
%U A213048 1911864488674531,46222718892288645,1183919151676806177,32025836905529003471,912372909851608715945,27304698509111141688969
%N A213048 Number of preferential arrangements of n labeled elements with repetitions allowed.
%H A213048 Alois P. Heinz, <a href="/A213048/b213048.txt">Table of n, a(n) for n = 0..250</a>
%F A213048 a(n) = Sum_{k=1..n} C(n+k-1,k)*a(n-k) for n>0, a(0) = 1.
%F A213048 a(n) = Sum_c(n) C(n+k1-1,k1) C(n-k1+k2-1,k2) C(n-k1-k2+k3-1,k3) ..., where Sum_c(n) denotes the sum over all compositions (ordered partitions) of n = k1 + k2 + ... .
%F A213048 a(n) ~ c * n! * n^(log(2)) / (log(2))^n, where c = 0.9387523255426859866752735339706007723805611... . - _Vaclav Kotesovec_, May 03 2015
%e A213048 For n=2 the a(2) = 5 solutions are (1,2), (1|2), (2|1), (1|1), (2|2).
%p A213048 a:= proc(n) option remember;
%p A213048       `if`(n=0, 1, add(binomial(n+k-1, k)*a(n-k), k=1..n))
%p A213048     end:
%p A213048 seq(a(n), n=0..25);
%t A213048 a[n_] := a[n] = If[n==0, 1, Sum[Binomial[n+k-1, k] a[n-k], {k, 1, n}]];
%t A213048 a /@ Range[0, 25] (* _Jean-François Alcover_, Nov 21 2020 *)
%Y A213048 Cf. A000670.
%K A213048 nonn
%O A213048 0,3
%A A213048 _Thomas Wieder_, Jun 03 2012