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Search: id:A074351
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| A074351 |
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Number of elements of S_n having order n. |
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+0
4
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| 1, 1, 2, 6, 24, 240, 720, 5040, 40320, 514080, 3628800, 80166240, 479001600, 6797831040, 93774320640, 1307674368000, 20922789888000, 523845011289600, 6402373705728000, 153101632051630080, 2471368711740364800
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OFFSET
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1,3
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COMMENT
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If n is a prime power then a(n) = (n-1)!. - Vladeta Jovovic (vladeta(AT)eunet.rs), Sep 29 2002
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REFERENCES
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J. Kuzmanovich and A. Pavlichenkov, Finite groups of matrices whose entries are integers, Amer. Math. Monthly, 109 (2002), pp. 173-186
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LINKS
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T. D. Noe, Table of n, a(n) for n=1..100
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FORMULA
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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 (vladeta(AT)eunet.rs), Sep 29 2002
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EXAMPLE
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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
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CROSSREFS
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Cf. A001189, A074859.
Sequence in context: A073973 A034874 A052699 this_sequence A052597 A052632 A052692
Adjacent sequences: A074348 A074349 A074350 this_sequence A074352 A074353 A074354
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KEYWORD
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nice,easy,nonn
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AUTHOR
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K Murray Peebles (m.peebles(AT)sms.ed.ac.uk), Sep 26 2002
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