Find the smallest natural number n such that n^n has at least 720 divisors, comparable to ARML

This problem is smaller constant version of ARML 2010 T-5. Divisor counting function gives the incremented exponent product of prime factorization as the total number divisors. Math people King Squirrel and José Carlos Santos explored the origin/proof of this important number theoretic combinatorics result August 2018

[@DavidCorneth]

Maybe look at prime signatures. Have you tried n = 20?

[@MyOneFiftiethOfADollar]

20^20 = (2^80)(5^20) which has 81x21 divisors >> 720.
18^18 = (2^18)(3^36) which has19x37=703divisors < 720.
Oh I see! Is 20 the smallest you could find? I was just using single occurrences of two primes. My mistake. thanks!
What do you mean by prime signatures? prime factorization.

[@DavidCorneth]

I get 20^20 = 2^40 * 5^20 so is has 41*21 = 861 divisors. And 20 is the smallest positive integer with that property.
Prime signatures are found by listing the exponents with multiplicity in a prime factorization. So primes have signature (1) an 20 has signature (2,1). But maybe 20 is small enough to check positive integers until 20 and don't use prime signatures.

[@DavidCorneth]

Maybe also see OEIS sequence A062319 to verify n = 20

[@MyOneFiftiethOfADollar]

@@DavidCorneth right another mistake on my end. Thx for explanation regarding signatures. I had heard of omega function which just counts the distinct primes in in the prime factorization . It seems like it might be related to this……

[@DavidCorneth]

You're welcome! Maybe omega could be used for this, like one could know for any number k with omega(k) >= 3 we have k^k has more than 720 divisors. But it might get tricky when it produces false negatives.