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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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5
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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
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OFFSET
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1,2
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COMMENTS
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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).
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LINKS
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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]]]]
Table[2^(PrimeNu[n])*(Times @@ PartitionsP /@ Last /@ FactorInteger@n), {n, 1, 100}] (* G. C. Greubel, Dec 27 2015 *)
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PROG
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(PARI) a(n)=my(f=factor(n)[, 2]); prod(i=1, #f, numbpart(f[i]))*2^#f; \\ Michel Marcus, Dec 27 2015
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CROSSREFS
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KEYWORD
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mult,easy,nonn
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AUTHOR
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STATUS
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approved
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