The smallest number with given number of divisors

Question

I'm looking for a formula which gives the smallest number such that the number of its divisors is known. Exemple : let $n=2$ is the number of the divsors of $x$, then $x=2$ because 1 and 2 are a divisors of 2. Thanks.

Answer 1

Hint:

For prime $n$ the smallest number is obviously $2^{n-1}$.

For products of two prime numbers: $n=\pi_1\cdot \pi_2$ ($\pi_1\ge \pi_2$) it is $2^{\pi_1-1}3^{\pi_2-1}$ and so on but with growing complications.

The art of complications is the following. Based on the previous examples one can naively assume that the general expression for the number in question is $$ M(n)=\prod_{i=1}^N p_i^{\pi_i-1},\tag{1} $$ where $p_i$ are the prime numbers $2,3,5,\dots$ and $\pi_i$ are (not necessarily distinct) prime divisors of $n$: $$ n=\prod_{i=1}^N\pi_i,\quad\text{with}\quad \pi_{i+1}\le\pi_i. $$

And indeed an application of the expression (1) gives correctly almost all first 42 members of A005179 with notable exceptions for $n=8,16,24,$ and $32$.

It is not hard to understand the reason for the exceptions. Consider the number $M(n)$ with $n$ having more than two prime factors and ask: what would happen if we remove in (1) the last prime $p_N$ and instead increase the power of 2? Then we shall look whether $$ 2^{\pi_1\pi_N-1}<2^{\pi_1-1}p_N^{\pi_N-1}\Rightarrow \pi_1\pi_N-1<\pi_1-1+(\pi_N-1)\log_2p_N\Rightarrow \pi_1<\log_2p_N. $$ And this is exactly what happens in the case of $n=8$: $$ \pi_1=2<\log_25=\log_2 p_3. $$ This is not yet the end of story as besides increasing the power of 2 one can play also with simultaneous increasing it for 2 and 3, for 2, 3 and 5 and so on. It can be however conjectured that the "density" of the exceptions from (1) is generally rather low.