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%I #4 Mar 30 2012 17:07:15
%S 1,3,5,9,11,15,17,23,27,31,33,41,43,47,51,61,63,71,73,81,85,89,91,103,
%T 107,111,117,125,127,135,137,151,155,159,163,179,181,185,189,201,203,
%U 211,213,221,229,233,235,255,259,267,271,279,281,293,297,309,313,317
%N 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 <= n.
%C 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. Sequence gives T(n) = sum_{G in A, |G| <= n} t(G).
%D Schmidt, Peter Georg, Zur Anzahl unitaerer Faktoren abelscher Gruppen. [On the number of unitary factors in Abelian groups] Acta Arith., 64 (1993), 237-248.
%D Wu, J., On the average number of unitary factors of finite Abelian groups, Acta Arith. 84 (1998), 17-29.
%F a(n) = partial sums of A101113
%e A101113 begins 1, 2, 2, 4, 2. So a(5) = 11.
%t Sum[Apply[Times, 2*Map[PartitionsP, Map[Last, FactorInteger[i]]]], {i, n}]
%Y Cf. A101113.
%K easy,nonn
%O 1,2
%A _Russ Cox_, Dec 01 2004
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