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Search: id:A101113
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| A101113 |
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Let t(G) = number of unitary factors of the abelian group G. Then a(n) = sum t(G) over all abelian groups G of order exactly n. |
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2
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| 1, 2, 2, 4, 2, 4, 2, 6, 4, 4, 2, 8, 2, 4, 4, 10, 2, 8, 2, 8, 4, 4, 2, 12, 4, 4, 6, 8, 2, 8, 2, 14, 4, 4, 4, 16, 2, 4, 4, 12, 2, 8, 2, 8, 8, 4, 2, 20, 4, 8, 4, 8, 2, 12, 4, 12, 4, 4, 2, 16, 2, 4, 8, 22, 4, 8, 2, 8, 4, 8, 2, 24, 2, 4, 8, 8, 4, 8, 2, 20, 10, 4, 2, 16, 4, 4, 4, 12, 2, 16, 4, 8, 4, 4, 4, 28, 2, 8
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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From Schmidt paper: Let A denote the set of all abelian groups. Under the operation of direct product, A is a semigroup with identity element E, the group with one element. G_1 and G_2 are relatively prime if the only common direct factor of G_1 and G_2 is E. We say that G_1 and G_2 are unitary factors of G if G=G_1 X G_2 and G_1, G_2 are relatively prime. Let t(G) denote the number of unitary factors of G. This sequence is a(n) = sum_{G in A, |G| = n} t(n).
A101113(n) = A034444(n) * A000688(n).
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REFERENCES
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Schmidt, Peter Georg, Zur Anzahl unitaerer Faktoren abelscher Gruppen. [On the number of unitary factors in abelian groups] Acta Arith., 64 (1993), 237-248.
Wu, J., On the average number of unitary factors of finite abelian groups, Acta Arith. 84 (1998), 17-29.
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FORMULA
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a(n) = 2^(number of distinct prime factors of n) * product of prime powers in factorization of n.
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EXAMPLE
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The only finite abelian group of order 6 is C6=C2xC3. The unitary divisors are C1, C2, C3, and C6. So a(6) = 4.
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MATHEMATICA
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Apply[Times, 2*Map[PartitionsP, Map[Last, FactorInteger[n]]]]
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CROSSREFS
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Cf. A101114.
Cf. A034444, A000688.
Sequence in context: A092520 A129089 A124315 this_sequence A055155 A085191 A061142
Adjacent sequences: A101110 A101111 A101112 this_sequence A101114 A101115 A101116
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KEYWORD
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mult,easy,nonn
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AUTHOR
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Russ Cox (rsc(AT)swtch.com), Dec 01 2004
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