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A074351
Number of elements of S_n having order n.
7
1, 1, 2, 6, 24, 240, 720, 5040, 40320, 514080, 3628800, 80166240, 479001600, 6797831040, 93774320640, 1307674368000, 20922789888000, 523845011289600, 6402373705728000, 153101632051630080, 2471368711740364800, 51182316789956352000, 1124000727777607680000
OFFSET
1,3
COMMENTS
If n is a prime power then a(n) = (n-1)!. - Vladeta Jovovic, Sep 29 2002
LINKS
Alois P. Heinz, Table of n, a(n) for n = 1..450 (first 100 terms from T. D. Noe)
J. Kuzmanovich and A. Pavlichenkov, Finite groups of matrices whose entries are integers, Amer. Math. Monthly, 109 (2002), 173-186.
FORMULA
n!/(a_1!*a_2!*...*a_d!*k_1^a_1*k_2^a_2*...*k_d^a_d) is the number of elements of S_n having order n that are permutations with distinct cycle-lengths k_1, ..., k_d having multiplicities a_1, ..., a_d, where lcm(k_1, ..., k_d)=n. Summing over all permutation types gives the total.
a(n) = n!*coefficient of x^n in expansion of Sum_{i divides n} mu(n/i)*exp(Sum_{j divides i} x^j/j). - Vladeta Jovovic, Sep 29 2002
EXAMPLE
a(10) = 514080 because {10}, {5, 2, 2, 1} and {5, 2, 1, 1, 1} are the unique multisets of cycle lengths summing to 10 whose lcm is 10 and 10!/(1!*10^1) + 10!/(1!*2!*1!*5^1*2^2*1^1) + 10!/(1!*1!*3!*5^1*2^1*1^3) = 514080.
MATHEMATICA
a[n_] := SeriesCoefficient[ Series[ Sum[ MoebiusMu[n/i]*Exp[Sum[x^j/j, {j, Divisors[i]}]], {i, Divisors[n]}], {x, 0, n}], n]*n!; Table[a[n], {n, 1, 21}] (* Jean-François Alcover, May 21 2012, after Vladeta Jovovic *)
PROG
(PARI) a(n)={n!*polcoeff(sumdiv(n, i, moebius(n/i)*exp(sumdiv(i, j, x^j/j) + O(x*x^n))), n)} \\ Andrew Howroyd, Jul 02 2018
CROSSREFS
Cf. A001189, A074859, A290961 (the same for endofunctions).
Main diagonal of A057731.
Sequence in context: A355284 A034874 A052699 * A352060 A346121 A052597
KEYWORD
nice,easy,nonn
AUTHOR
K Murray Peebles (m.peebles(AT)sms.ed.ac.uk), Sep 26 2002
STATUS
approved