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Search: id:A147848
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| A147848 |
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Number (up to isomorphism) of groups of order 2n that have Z/nZ as a subgroup (that is, that have an element of order n). |
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1
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| 1, 2, 2, 4, 2, 4, 2, 6, 2, 4, 2, 8, 2, 4, 4, 6, 2, 4, 2, 8, 4, 4, 2, 12, 2, 4, 2, 8, 2, 8, 2, 6, 4, 4, 4, 8, 2, 4, 4, 12, 2, 8, 2, 8, 4, 4, 2, 12, 2, 4, 4, 8, 2, 4, 4, 12, 4, 4, 2, 16
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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This sequence is related to A060594 : in fact, for every square root of unity modulo n, there are either one or two such groups of order 2n.
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FORMULA
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a(2) = 2; a(4) = 4; a(2^k) = 6 for k >= 3.
a(p^k) = 2 for any odd prime number p and k >= 1.
For other values of n, you can find a(n) by using the fact that the sequence is multiplicative.
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EXAMPLE
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Two such groups that always exist are the cyclic group Z/(2n)Z and the dihedral group Dih_n. If n is prime, these are the only such groups, so the n-th term equals 2.
For even values of n, we also have the direct product Z/nZ x Z/2Z and the dicyclic group Dic_n. If n = 2p with p prime, there are no other groups, so the n-th term equals 4.
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CROSSREFS
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Cf. A060594.
Adjacent sequences: A147845 A147846 A147847 this_sequence A147849 A147850 A147851
Sequence in context: A001223 A118776 A092520 this_sequence A129089 A124315 A101113
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KEYWORD
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easy,nice,nonn,mult
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AUTHOR
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Ilia Smilga (ilia.smilga(AT)ens.fr), Nov 15 2008
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EXTENSIONS
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Extended comments, references and confirmed "mult" keyword. - Ilia Smilga (ilia.smilga(AT)ens.fr), Nov 17 2008
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