Re: Largest primeproduct with 2 as factor?

In article <1187278549.329963.136200@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>, Randy Poe <poespam-trap@xxxxxxxxx> writes:
Given that definition, the product of two primeproducts
is a number of the form p1*p2*p3*p4. It has at most
four PRIME factors, but it also has as factors p1*p2,
p1*p3, p1*p4, p2*p4, p1*p2*p3, p1*p3*p4, etc. In
fact I think it has 2^4 - 1 factors (one for
every subset of {p1, p2, p3, p4} except the empty
set).

2^4, 2^4-1 or 2^4-2 depending on whether you count all factors,
all factors other than 1, or all factors other than 1 and the number
itself.

[If you use the 2^n-2 interpretation then be sure to restrict your
attention to cases where n>0]

But that's assuming that p1, p2, p3 and p4 are all different.

If one or more are the same then you'd need to use a formula
for the number of subsets of a four element multiset.

If one pair is identical then it's 2*2*3 instead of 2^4.
If two pairs are identical then it's 3*3 instead of 2^4.
If there is one triplet then it's 2*4 instead of 2^4.
If all four are identical then it's 5 instead of 2^4.

In general, I believe that the number of subsets of a multiset is
the product of (one more than the multiplicity of the element) evaluated
over each distinct element in the multiset.

And if we multiply two primeproducts of primeproducts can they only
have eight factors?

Again, no. It has many more than 8 factors. I think
it has 2^8 - 1 factors.

Or the full 2^8 if you count 1 as a factor.
.