
Re: Largest primeproduct with 2 as factor?
Posted:
Aug 16, 2007 1:06 PM


In article <1187278549.329963.136200@x35g2000prf.googlegroups.com>, Randy Poe <poespamtrap@yahoo.com> 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, 2^41 or 2^42 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^n2 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.

