Amicable numbers are two different numbers so related that the sum of the proper divisors of each is equal to the other number.

Is there a name for two different numbers who sum of prime factors is equal to the other number?


I suspect not, for the following reason: The sum of the prime factors of any composite number greater than $5$ is always strictly less than that number. I believe a proof by induction works here, but trying it out with a number of small examples should indicate that this sum grows more slowly than the product. So no such pairs would exist.

Here is the proof:

Let $f(n)$ be the sum of the prime factors of $n$. The claim can then be stated to be, for all $n \geq 5$, if $n$ is composite, then $f(n) < n$.

Let $n \geq 5$ be composite, and suppose (IH) that any composite number at most $n-1$ satisfies the above property. Let $p$ be the smallest prime factor of $n$; it suffices to show that $p + f(n/p) < n$. But $n/p \geq p$ (otherwise, $n$ would have a smaller prime factor), so we have: $$ p + f\left(\frac{n}{p}\right) < p + \frac{n}{p} \leq 2\left( \frac{n}{p}\right) \leq p \left( \frac{n}{p}\right) = n $$ where the last inequality comes from the fact that $p$ is prime (and therefore at least $2$).

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