Re: smallest positive integer that has exactly k divisors

On 28 Okt., 00:01, "christian.bau" <christian....@xxxxxxxxxxxxxxxxxx>
wrote:
On Oct 26, 6:41 am, hagman <goo...@xxxxxxxxxxxxx> wrote:

Indeed. See my other post, where I show (among others) that all
exponents
except that of 2 must be one less than a prime divisor of k.

That is most certainly wrong.

Take n = 64, so you are claiming that the smallest number with exactly
64 divisors is of the form 2^k * 3 * 5 * 7 ...

The numbers of this form are 2 * 3 * 5 * 7 * 11 * 13, 2^3 * 3 * 5 * 7
* 11, 2^7 * 3 * 5 * 7, 2^15 * 3 * 5, 2^31 * 3 and 2^63. The smallest
of these is 2^3 * 3 * 5 * 7 * 11. However, 2^3 * 3^3 * 5 * 7 also has
64 divisors and is smaller.

Oops, thank you for pointing that out.
It seems I swapped some > and < somewhere in my thoughts.
Stated correctly, my ideas only lead to the following:

If n=a(1)*a(2)*..*a(k), a(i)>1, is the factorization
of n that minimizes p(1)^(a(1)-1) * ... * p(k)^(a(k)-1)
and if q is the smallst prime divisor of a(i),
then
p(i)^(a(i)/q) < p(j)^a(j)
for all j.
(This includes p(i)^(a(i)/q) < p(k+1))

Corollary: p(k) < p(i)^a(i) for all j<=k.

Thus nothing speaks against a factor 3^3 (corresponding
to a(2)=4), as long as p(j)^a(j) > 9 for all primes, i.e.
the first non-occurring prime is at least 11 and 2
occurs with at least third power -- both conditions
are of course satisfied by 2^3 * 3^3 * 5 * 7.

This rule is of course only necessary and not sufficient.
However, it turns out that in this form the
candidate 2^3 * 3 * 5 * 7 * 11 (which was "elected" by
my wrong criterion) is ruled out by the corollary
because 11 > 3^2.

hagman

.