For example:
$6$ has this property since proper divisors of $6$ are: $2$ and $3.

From this thread: What does the product of all proper divisors equal to?

My attempt was:
If $n = p_1^{a_1} \times p_2^{a_2} \times ... \times p_k^{a_k}$
Then $n = n^{\frac{\tau(n)}{2} - 1}$. Where $\tau(n) = (a_1 + 1) \times (a_2 + 1) \times ... \times (a_k + 1)$

So is it good enough to stop here, or we can express $n$ in a better formula?

  • $\begingroup$ You missed a factor 2 in the exponent from Arturo Magidin's answer to the earlier question. This leads naturally to Sivaram Ambikasaran's answer to this one. $\endgroup$ – Ross Millikan Feb 28 '11 at 2:14
  • $\begingroup$ @Ross Millikan: Opp! My bad. Thanks for pointing out. $\endgroup$ – Chan Feb 28 '11 at 2:16

$n = n^{\tau(n)/2 - 1}$ implies $\tau(n)=4$ and so $n$ is a product of two primes or the cube of a prime.

  • $\begingroup$ Nice thought! Thanks for this clever hint. $\endgroup$ – Chan Feb 28 '11 at 2:18

Any number of the form $n = p_1 \times p_2$ where $p_1,p_2$ are primes.

  • $\begingroup$ Ambikasaran: Thank you. $\endgroup$ – Chan Feb 28 '11 at 2:19
  • 1
    $\begingroup$ You've missed $n=p^3$. $\endgroup$ – lhf Feb 28 '11 at 14:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.