# On the existence of odd, squarefree semiperfect numbers

A positive integer $$n$$ is called square-free if no square divides it. Moreover, $$n$$ is called semiperfect (or pseudoperfect) if it is equal to the sum of some (or all) of its divisors. The smallest odd semiperfect number is $$945$$, and it has been proved that infinitely many odd semiperfect numbers exist. I have checked the first few odd semiperfect numbers, and noticed that none of them is square-free. Does there exist any odd semiperfect square-free number? If the answer is affirmative, what is the smallest one?

• In the answer below you find the smallest example and in comment the second smallest and the third smallest. Interesting is whether infinite many such examples exist. Oct 24 '20 at 12:08
• Yes there are infinitely many. You need only one, then multiply that number by any additional odd prime factors and multiply the selected divisors by the same. To put it nontechnucally, there are lots of solutions. Oct 24 '20 at 12:30
• Thank you very much! Oct 24 '20 at 12:35

The ratio of the sum of divisors of $$n$$ (including $$n$$ itself) to $$n$$ must be at least than $$2$$ for a semi perfect number, and if $$n$$ is square-free this ratio is the product of $$(p+1)/p$$ for all its prime factors $$p$$. The smallest set of odd prime factors where the product formula is large enough is $$\{3,5,7,11,13\}$$ leading to
$$15015=3×5×7×11×13=\text{ sum of its divisors except }15015,1155,1001,55,15$$.
• You were a bit faster than my program ! Well done and (+1) ! The next number is $19635$ followed by $21945$ Oct 24 '20 at 12:06
• Your next numbers are $3×5×7×11×17$ and $3×5×7×11×19$, which would be the next candidates by my method. Semiperfection seems to be common. Oct 24 '20 at 12:25