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A101113 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. 5
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; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

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).

REFERENCES

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.

LINKS

Table of n, a(n) for n=1..98.

FORMULA

a(n) = 2^(number of distinct prime factors of n) * product of prime powers in factorization of n.

EXAMPLE

The only finite Abelian group of order 6 is C6=C2xC3. The unitary divisors are C1, C2, C3 and C6. So a(6) = 4.

MATHEMATICA

Apply[Times, 2*Map[PartitionsP, Map[Last, FactorInteger[n]]]]

CROSSREFS

Cf. A101114.

Cf. A034444, A000688.

Sequence in context: A129089 A169594 A124315 * A055155 A085191 A188581

Adjacent sequences:  A101110 A101111 A101112 * A101114 A101115 A101116

KEYWORD

mult,easy,nonn

AUTHOR

Russ Cox, Dec 01 2004

STATUS

approved

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Last modified November 29 01:24 EST 2014. Contains 250479 sequences.