I don't know about the promise problem, but my educated hunch is that computing the sum of the prime factors should indeed be roughly as hard as factoring. Here's why: let $N$ be odd and squarefree. Then if we can compute $sopf(N)$, we'll know the parity of the number of prime factors of $N$, which I've gone on record as believing to be hard. (And nobody's contradicted me, so far...)

Terry Tao's answer to that question, by the way, is incredibly useful in thinking about these types of problems. It also indicates that we can sometimes compute the parity even when we can't do anything else, though; I don't know if that applies to number-of-prime-factors. (I am decidedly not a number theorist...)

I don't know about the promise problem, but my educated hunch is that computing the sum of the prime factors should indeed be roughly as hard as factoring. Here's why: let $N$ be odd and squarefree. Then if we can compute $sopf(N)$, we'll know the parity of the number of prime factors of $N$, which I've gone on record as believing to be hard. (And nobody's contradicted me, so far...)

Terry Tao's answer to that question, by the way, is incredibly useful in thinking about these types of problems. It also indicates that we can sometimes compute the parity even when we can't do anything else, though; I don't know if that applies to number-of-prime-factors. (I am decidedly not a number theorist...)

Let $N$ be odd and squarefree. Then if we can compute $sopf(n)$, we'll know the parity of the number of prime factors of $N$, which I've gone on record as believing to be hard. (And nobody's contradicted me so far, although if anything's possible to extract, it's the parity bit.) So, yeah, computing sopf() should be hard.

Terry Tao's answer to that question, by the way, is infinitely more informative and interesting than mine -- mine's simply more relevant.

I don't know about the promise problem, but my educated hunch is that computing the sum of the prime factors should indeed be roughly as hard as factoring. Here's why: let $N$ be odd and squarefree. Then if we can compute $sopf(N)$, we'll know the parity of the number of prime factors of $N$, which I've gone on record as believing to be hard. (And nobody's contradicted me, so far...)

Terry Tao's answer to that question, by the way, is incredibly useful in thinking about these types of problems. It also indicates that we can sometimes compute the parity even when we can't do anything else, though; I don't know if that applies to number-of-prime-factors. (I am decidedly not a number theorist...)