[seqfan] Arithmetic progressions of primes

[hv at crypt.org]

This is a request for a sanity check.

I am continuing to investigate the recently submitted A165500 and A165501:
maximum length and first common difference respectively of arithmetic
progressions starting at n such that each element has the same number of
divisors.

For prime n = p_i, this resolves to A165501(n)=A088430(i), which assumes
the long held conjecture that A165500(n)=n for all such cases, i.e. that
for all primes p there exists a difference d that yields an arithmetic
progression of p primes starting at p.

My question is: is it reasonable to conjecture further that for all p
there are an infinite number of differences d yielding such a progression?

I believe yes, based on this rather rough calculation:

For given n, k, d, the probability that n+kd is prime is approximately
1 / ln(n+kd).

The probability that a particular n, d yields an AP of primes is thus:
  p(n, d) = prod_{k=0}^{n-1}{ 1 / ln(n+kd) }
assuming the probability in each case is independent of the others
(** this is probably the most dubious assumption **).

If d is much larger than n, this approximates to:
  p(n, d) = 1 / ln^n(d)

So for given n, the expected number of such APs of primes is:
  E(n) = sum_{d=2}^{inf}{ p(n, d) }
       = sum_{d=2}^{inf}{ 1 / ln^n(d) }
       = (1 / ln^n(2)) sum_{i=1}^{inf}{ sum_{d=2^i}^{2^(i+1)-1}{ 1 / lg^n(d) }}
       > (1 / ln^n(2)) sum_{i=1}^{inf}{ 2^i / (i+1)^n }
.. which is clearly infinite.

Is this a reasonable argument to justify the conjecture that every prime p
starts an infinite number of (length p) APs of primes?

Hugo