# In number theory, is there the idea of a "metaprime" -- a number with a prime number of factors?

This could be the definition or perhaps the sum of its factors being prime. There must be many numbers with a prime number of factors, in fact every prime is a metaprime by this definition but I don't know about the sum of all factors or prime factors. 9 would be a metaprime where all (positive) factors add up to 13. I am wondering if this is even an area of study -- I can't find anything when I google it.

• Note that if there are two factors, the number itself must be prime. For an odd number of factors, and therefore for other primes, note that factors normally come in pairs, so the only positive integers with an odd number of factors are square numbers. Dec 11, 2016 at 19:46
• Please clarify, is it a number of factors, number of prime factors or the sum of them?
– z100
Dec 11, 2016 at 19:49
• Dec 11, 2016 at 19:58
• But, ideally, a "meta-prime" should be a prime lying over. Dec 11, 2016 at 20:23
• @EricM.Schmidt: Thanks for this -- looks like an AI (?) came up with this concept?
– Jeff
Dec 11, 2016 at 21:48

Suppose $\tau (n)$ is the number of factors of $n$. We know that $\tau$ is a multiplicative function and that $\tau (n) = 1 \iff n = 1$. That means, that $\tau (n)$ can be prime only in case, when n is a degree of a prime. For prime p and natural m, we have $\tau(p^m) = m+1$. Thus all numbers with prime number of factors have the form $p^{q-1}$ where both $p$ and $q$ are primes.
Suppose $\sigma (n)$ is the sum of factors of n. We know that $\sigma$ is a multiplicative function and that $\sigma (n) = 1 \iff n = 1$. That means, that $\sigma (n)$ can be prime only in case, when n is a degree of a prime. For prime $p$ and natural $m$, we have $\sigma(p^m) = \sum_{k = 0}^{m} p^k$ . Thus all numbers with prime sum of factors have the form $p^m$, where both $p$ and $R_{m+1}^{(p)}$ are primes.