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A101880 Number of arrangements of the partitions of n (e.g., 111 counts for 6).

%I A101880 #46 Jul 25 2025 09:47:47
%S A101880 1,1,3,9,35,161,913,6103,47319,416235,4092155,44424095,527511445,
%T A101880 6798907249,94504286703,1408973416617,22426222745159,379522092608177,
%U A101880 6804315177704869,128828842646944135,2568533750228603835,53788282243854336411,1180357162840624656959
%N A101880 Number of arrangements of the partitions of n (e.g., 111 counts for 6).
%C A101880 All terms are odd. - _Alois P. Heinz_, Jul 10 2023
%H A101880 Alois P. Heinz, <a href="/A101880/b101880.txt">Table of n, a(n) for n = 0..450</a>
%H A101880 Jon Perry, <a href="http://www.users.globalnet.co.uk/~perry/maths/partitiontable/partitiontable.htm">Partition Tables</a> [broken link]
%F A101880 a(n) = Sum_{i=1..n} P(n,i)*i!, where P(n,i) is the number of partitions of n into i parts.
%F A101880 G.f.: Sum_{n=1..infinity} (n!*x^n / Product_{k=1..n} (1-x^k)). - _Vladeta Jovovic_, Jan 29 2005
%F A101880 G.f.: ( 1 - G(0) )/(1-x) where G(k) =  1 - (k+1)/(1-x^(k+1))/(1-x/(x-1/G(k+1) )); (recursively defined continued fraction). - _Sergei N. Gladkovskii_, Jan 22 2013
%F A101880 a(n) ~ n! * (1 + 1/n + 2/n^2 + 5/n^3 + 16/n^4 + 60/n^5 + 253/n^6 + 1180/n^7 + 6023/n^8 + 33306/n^9 + 197719/n^10 + ...), for coefficients see A331826. - _Vaclav Kotesovec_, Jan 28 2020
%e A101880 a(3) = 9 as we have 3, 12 (2) and 111 (6).
%e A101880 a(4) = 35 as 4, 31 (2), 22 (2), 211 (6) and 1111 (24).
%p A101880 b:= proc(n, i, p) option remember; `if`(n=0, p!,
%p A101880       `if`(i<1, 0, add(b(n-i*j, i-1, p+j), j=0..n/i)))
%p A101880     end:
%p A101880 a:= n-> b(n$2, 0):
%p A101880 seq(a(n), n=0..25);  # _Alois P. Heinz_, Apr 06 2016
%t A101880 Rest[ CoefficientList[ Series[ Sum[ n!x^n / Product[1 - x^k, {k, n}], {n, 20}], {x, 0, 20}], x]] (* _Robert G. Wilson v_, Feb 10 2005 *)
%o A101880 (Sage)
%o A101880 from sage.combinat.partition import number_of_partitions_length
%o A101880 def A101880(n):
%o A101880     return sum(number_of_partitions_length(n, k)*factorial(k) for k in (0..n))
%o A101880 print([A101880(n) for n in (0..21)]) # _Peter Luschny_, Aug 01 2015
%Y A101880 Cf. A000142, A008284, A274174.
%K A101880 nonn,nice
%O A101880 0,3
%A A101880 _Jon Perry_, Jan 28 2005
%E A101880 More terms from _Vladeta Jovovic_, Jan 29 2005
%E A101880 a(0)=1 prepended by _Alois P. Heinz_, Apr 06 2016