tau(k)/bigomega(k)

[hv at crypt.org]

Emeric Deutsch <deutsch at duke.poly.edu> wrote:
:Dear Seqfans,
:Sequence A109421 gives the numbers k such that tau(k)/bigomega(k)
:is an integer [tau(k)=number of divisors of k; bigomega(k)=number
:of prime divisors of k, counted with multiplicities].
:
:My question: does the set {tau(k)/bigomega(k): k=2,3,...} contain
:all integers n>=2?
:
:We (= Maple and I) have found that the least k such that
:tau(k)/bigomega(k)=n is given by the table:
:n	k
:-	-
:2	2
:3	60
[...]
:17	>11,000,000 if it exists

If n = 2^(k-1) . 3^(k-2) . 5, we have:
  tau(n) = 2k(k-1)
  bigomega(n) = 2(k-1)
so tau(n) / bigomega(n) = k, and this represents a possible candidate
for a(k) for any k >= 2.

It is clear from the results that this doesn't always give the least
solution (ie it is not a constructor for a(k)), but it does show that
a value exists for all k.

All a(k) must be in A025487, and I think it should be possible to
restrict the search for a given k to be reasonably efficient.

Hugo