[math-fun] A missing sequence from group theory?
Russ Cox
russcox at gmail.com
Thu Dec 2 02:26:02 CET 2004
> 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
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