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A060726 For n >= 1, a(n) is the number of permutations in the symmetric group S_n such that their cycle decomposition contains no 6-cycle.

%I A060726 #17 Jul 02 2025 16:02:01
%S A060726 1,1,2,6,24,120,600,4200,33600,302400,3024000,33264000,405820800,
%T A060726 5275670400,73859385600,1107890784000,17726252544000,301346293248000,
%U A060726 5419293175296000,102966570330624000,2059331406612480000
%N A060726 For n >= 1, a(n) is the number of permutations in the symmetric group S_n such that their cycle decomposition contains no 6-cycle.
%C A060726 This is the expansion of exp ((-x^6)/6) /(1-x).
%D A060726 R. P. Stanley, Enumerative Combinatorics, Wadsworth, Vol. 1, 1986, page 93, problem 7.
%H A060726 Harry J. Smith, <a href="/A060726/b060726.txt">Table of n, a(n) for n = 0..100</a>
%H A060726 Plouffe, Simon, <a href="http://www.plouffe.fr/simon/exact.htm">Exact formulas for integer sequences</a>
%F A060726 The formula for a(n) is: a(n) = n! * Sum_{i=0..floor(n/6)} ((-1)^i /(i! * 6^i)) by this formula we have as n -> infinity: a(n)/n! ~ Sum_{i>= 0} (-1)^i /(i! * 6^i) = e^(-1/6) or a(n) ~ e^(-1/6) * n! and using Stirling's formula in A000142: a(n) ~ e^(-1/6) * (n/e)^n * sqrt(2 * Pi * n)
%F A060726 a(n,k) = n!*floor(floor(n/k)!*k^floor(n/k)/exp(1/k) + 1/2)/(floor(n/k)!*k^floor(n/k)), k=6, n >= 0. - _Simon Plouffe_, Feb 18 2011
%e A060726 a(6) = 600 because in S_6 the permutations with no 6-cycle are the complement of the 120 6-cycles so a(6) = 6! - 120 = 600.
%p A060726 for n from 0 to 30 do printf(`%d,`, n! * sum(( (-1)^i /(i! * 6^i)), i=0..floor(n/6))) od:
%o A060726 (PARI) a(n)={n! * sum(i=0, n\6, (-1)^i / (i! * 6^i))} \\ _Harry J. Smith_, Jul 10 2009
%Y A060726 Cf. A000142, A000266, A000090, A000138, A060725, A060726, A060727.
%K A060726 nonn
%O A060726 0,3
%A A060726 Avi Peretz (njk(AT)netvision.net.il), Apr 22 2001
%E A060726 More terms from _James Sellers_, Apr 24 2001