OFFSET
0,4
COMMENTS
From Bob Beals: Let P[n] = probability that a random permutation in S_n has odd order. Then P[n] = sum_k P[random perm in S_n has odd order | n is in a cycle of length k] * P[n is in a cycle of length k]. Now P[n is in a cycle of length k] = 1/n; P[random perm in S_n has odd order | k is even] = 0; P[random perm in S_n has odd order | k is odd] = P[ random perm in S_{n-k} has odd order]. So P[n] = (1/n) * sum_{k odd} P[n-k] = (1/n) P[n-1] + (1/n) sum_{k odd and >=3} P[n-k] = (1/n)*P[n-1] + ((n-2)/n)*P[n-2] and P[1] = 1, P[2] = 1/2. The solution is: P[n] = (1 - 1/2) (1 - 1/4) ... (1-1/(2*[n/2])).
LINKS
T. D. Noe, Table of n, a(n) for n=0..100
FORMULA
E.g.f.: (1-sqrt(1-x^2))/(1-x).
a(2n) = (2n-1)! + (2n-1)a(2n-1), a(2n+1) = (2n+1)a(2n).
a(n) = n! - A000246(n). - Victor S. Miller
EXAMPLE
A permutation in S_4 has even order iff it is a transposition, a product of two disjoint transpositions or a 4 cycle so a(4) = C(4,2)+ C(4,2)/2 + 3! = 15.
MAPLE
s := series((1-sqrt(1-x^2))/(1-x), x, 21): for i from 0 to 20 do printf(`%d, `, i!*coeff(s, x, i)) od:
MATHEMATICA
a[n_] := a[n] = n! - ((n-1)! - a[n-1]) * (n+Mod[n, 2]-1); a[0] = 0; Table[a[n], {n, 0, 20}](* Jean-François Alcover, Nov 21 2011, after Pari *)
With[{nn=20}, CoefficientList[Series[(1-Sqrt[1-x^2])/(1-x), {x, 0, nn}], x] Range[0, nn]!] (* Harvey P. Dale, Aug 05 2015 *)
PROG
(PARI) a(n)=if(n<1, 0, n!-((n-1)!-a(n-1))*(n+n%2-1))
(GAP) List([1..9], n->Length(Filtered(SymmetricGroup(n), x->(Order(x) mod 2)=0)));
CROSSREFS
KEYWORD
nonn,nice
AUTHOR
Avi Peretz (njk(AT)netvision.net.il), Feb 25 2001
EXTENSIONS
Additional comments and more terms from Victor S. Miller, Feb 25 2001
Further terms and e.g.f. from Vladeta Jovovic, Feb 28 2001
STATUS
approved