%I #21 Apr 04 2024 11:00:17

%S 1,0,1,2,12,64,470,3828,36456,387840,4603392,60061440,855664656,

%T 13207470912,219609303888,3912940891104,74377769483520,

%U 1502277409668096,32130095812624512,725400731911792896,17240044059713320704,430231117562438446080,11248105572520779755520

%N Number of alignments of n points with no singleton cycles

%C a(n) is the number of labeled sequences of cycles, where no cycle has size 1.

%D P. Flajolet and R. Segdewick, Analytic Combinatorics, Cambridge University Press, 2009, page 119

%H Vincenzo Librandi, <a href="/A226226/b226226.txt">Table of n, a(n) for n = 0..200</a>

%F E.g.f.: 1/(1+x-log(1/(1-x)))

%F a(n) ~ n!*c/(1-c)^(n+2), where c = -LambertW(-exp(-2)) = 0.158594339563... - _Vaclav Kotesovec_, Jun 02 2013

%F a(0) = 1; a(n) = Sum_{k=0..n-2} binomial(n,k) * (n-k-1)! * a(k). - _Ilya Gutkovskiy_, Apr 26 2021

%e For n=4, the a(4)=12 alignments with no singletons are: 1234, 1243, 1324, 1342, 1423, 1432, 12|34, 13|24, 14|23, 23|14, 24|13, 34|12.

%t Range[0, 50]! CoefficientList[ Series[(1 + z - Log[1/(1 - z)])^(-1), {z, 0, 50}], z]

%o (PARI) x='x+O('x^66); Vec(serlaplace(1/(1+x-log(1/(1-x))))) \\ _Joerg Arndt_, Jun 01 2013

%Y The alignments with singletons included are given by A007840.

%K nonn

%O 0,4

%A _Ricardo Bittencourt_, May 31 2013