Let t(G) = no. of unitary factors of the abelian group G, and let
T(n) = Sum t(G)
taken over all abelian groups of order <= n.
There are several papers giving estimates for T(n), e.g.
T(x) = c_1 x (log x + 2 gamma - 1) + c_2 x + ...
My question is, how does the sequence T(n) begin?
Alex Healy and I think we've worked this out. Let S(n) = Sum t(G) over all abelian groups of order exactly n. Then T(n) is the sequence of partial sums of S. Let n = p1^e1 p2^e2 ... pk^ek and consider any G of order n. For each prime p^e in the factorization, if a unitary factor of G contains a subgroup of order p, then it must contain the entire subgroup of order p^e present in G. So there are 2^k unitary factors -- just choose for each prime whether or not to multiply in the subgroup of order p^e from G. This is no different from the integer unitary divisors of n, which is A034444. The number of abelian groups of order n is given by A000688 and is just p(e1) p(e2) ... p(ek) where p(x) is the number of integer partitions of x. So S(n) = A034444(n) * A000688(n). In slightly sleazy Mathematica (thanks to Alex): S[n] = Apply[Times, 2*Map[PartitionsP, Map[Last, FactorInteger[n]]]] T[n] = Sum[Apply[Times, 2*Map[PartitionsP, Map[Last, FactorInteger[i]]]], {i, n}] I have submitted these as A101113 and A101114. Russ