This is follow up from previous posts. In the last post I’ve claimed (and I think the claim stands on its own without elaboration):
“… if we can guarantee for all n that there must be at least one candidate twin prime pair less than P(n+1)^2 remaining after sieving to the n-th prime then the twins are infinite.”
So I’ve been exploring the distribution of twin primes and twin prime candidates to see if there is a way to make that guarantee. This post doesn’t make that connection, but I try to illustrate some aspects of their distribution to help me think about it a little further.
First, the following graph shows the distribution of actual twin prime pairs less than any given n-th prime, Pn, or less than Pn^2.
You can see that, like the primes, the distribution of twin primes has convexity, or the density decreases as n, and Pn, increase.
In previous posts I’ve shown that after n rounds of sieving to Pn, then within the n-th primorial we are left with Kn candidate k-tuples, which may or may not be sieved out by further sieving by higher primes, where Kn can be generalized as follows:
And for various k-tuples including primes (Nn) and twin primes (Tn) we have the following specific relationships:
Now we can also recognize that after n rounds of sieving to Pn, we have “fixed” all primes and k-tuples to Pn^2, in other words further rounds of sieving cannot remove any further primes (nor eliminate any further twins or k-tuples) and hence any candidate k-tuples between Pn and Pn^2 are actual primes, twins or k-tuples.
I propose that since the step-wise sieving process leaves a symmetric pattern of candidate k-tuples within the set of natural numbers, that the density of candidate primes and k-tuples should be relatively constant to infinity, and therefore the density of actual primes and k-tuples to Pn^2 should be approximately the same as the overall density of candidates after n rounds of sieving.
Hence we would expect that (Kn/Pn#)*Pn^2 might equal the actual number of k-tuples to Pn^2. The following two graphs explore this idea.
This graph shows the actual counts of primes and twin primes less than Pn^2, in comparison with the expected values as calculated as previously suggested. One can see that the actual numbers of primes and twins are slightly less than the estimate. We also show, for the primes, the curve for (Pn^2)/ln(Pn^2), which gives a lower estimate than the actual number of primes.
I would have expected this estimate to give a closer approximation to the actual values, or at least to converge toward the correct values. This does not seem to be the case, unfortunately. I would extend the curves, but the numbers get too big to work with practically.
This final graph compares the estimate for primes with actual count of primes for some much larger numbers. The value of the prime count for the very high numbers was obtained from Wolfram Mathematica.
Still, my estimate exceeds the actual count, if perhaps not by a large margin.
I’ve been puzzling over why these estimates aren’t better than they are and haven’t come up with anything yet. Still, it is clear that the estimate trends the same as the actual count, and trends upward with increasing Pn. So it seems hopeful to be able to demonstrate the validity of my first claim, which requires only ONE candidate twin less than Pn^2 (actually, only one less than P(n+1)^2).
More to follow as anything interesting develops. 🙂