

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
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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), 237248.
Wu, J., On the average number of unitary factors of finite Abelian groups, Acta Arith. 84 (1998), 1729.


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



