A divisor diversion

Suppose $n$ is an element of $S$ . Its smallest divisor greater than $1$ will be some prime $p$ , and so $n$ will have $p$ divisors. But if the prime factorization of $n$ is $p_{1}^{a_{1}} p_{2}^{a_{2}} \cdot\cdot\cdot p_{k}^{a_{k}}$ then the number of divisors is $(a_{1} + 1)(a_{2} + 1)\cdot\cdot\cdot(a_{k} + 1)$ . For this to equal $p$ only one of these factors can exceed $1$ , and thus $n$ can only have one prime factor, namely $p$ .

We therefore have that if an integer $n$ is an element of $S$ then it must be that $n = p^{p-1}$ for some prime $p$ . Conversely, any integer of the form $p^{p-1}$ for some prime $p$ will be an element of $S$ . The elements of $S$ that are less than $1,000,000$ are then $2^{2-1} = 2, 3^{3-1} = 9, 5^{5-1} = 625$ and $7^{7-1} = 117649$ , and the sum of these four values is $\boxed{118285}.$