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I started with this question about linking at least one twin prime with each prime. My approach is:

  1. Give each prime two buckets.
  2. Put each odd 3n in the second bucket of it's largest prime divisor.
  3. Put it in the first bucket if (3n-4, 3n-2) are twin primes.
  4. Prove that no first bucket is empty, or that the density drops slow enough.

Edit: Let
$d(p, 3n) = \begin{cases} 1 & p|n \wedge p \text{ is largest prime divisor}\\ 0 & \text{ otherwise } \end{cases} \\ t(p, 3n) = \begin{cases} 1 & d(p, n) = 1 \wedge (3n-4, 3n-2) \text{ are twin primes}\\ 0 & \text{ otherwise } \end{cases} \\ \\ f(p) = \sum_{3n}^N {t(p, 3n)} \\ g(p) = \sum_{3n}^N {d(p, 3n)} \\ h(p) = \frac {f(p)}{g(p)} \\ $

When I graph $f(p)$, it's about what you'd expect:
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And the graph of $g(p)$ is no surprise either:
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However, when I graph $h(p)$ for $N = 20M, 30M, ..$ I get this:
(y = 5%, 10%, 15%; x = primes < 20000; color = N +10M)

Ratio of 3pn twins to total 3pn
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Are these lines a result of my small sample size and ratio-similarity? Or is this a pattern in how twin primes are distributed?

maybeso
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    Impossible to understand. Try to simplify your question as possible, explaining your plot with a simple latex formula. – reuns Jul 02 '17 at 01:36
  • I added the math and brief description. click on the link to view the graph. – maybeso Jul 02 '17 at 04:19
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    It is not clear what is $\sum^N$. There are infinitely many integers having $p$ as its largest prime factor ie. $\sum_n d(p,3n) = \infty$ – reuns Jul 02 '17 at 05:16
  • N is the limit, why did you remove the limit in your example?? so the absolute maximum possible would be N/3, not $\infty$. – maybeso Jul 02 '17 at 06:35
  • You meant $f(p) = \displaystyle\sum_{n=1}^N d(p,3n)$ ..? In that case $f(p),g(p)$ depend on $N$. Also do you know the twin prime constant ? It comes from a conjectured model and asymptotic distribution for the primes and twin primes. – reuns Jul 02 '17 at 08:07
  • Please read the question again, and either answer or don't. – maybeso Jul 02 '17 at 09:23

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