Are there any conjectures from which we can infer something about the first occurrences of prime gaps length $n$ and their distribution? I've made an interesting graph of these values to make this problem easier, with gauss's approximate of their size, $\log(p)$ on the $x$-axis and on the $y$-axis the proportional error, i.e. $\frac{g(p)}{\log(p)}$. A prime gap $g(p)$ is $p_{n+1} − p_n$. What on earth is going on with gaps with error from 25 to 30, why are some of these gaps starting so late, does this pattern continue somehow? I needn't say anything about the nature of the large 'line' but does anyone see the vertical 'wave' like pattern with the anomalistic bunch?

Approximately the first 1000 gaps

I hope this question doesn't come across as too wordy or vague but would really appreciate any general comment.

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    $\begingroup$ Could you explain how you created the graph? If we enumerate the primes, then for every pair $(p_n,p_{n+1})$, how do you decide whether to draw a point, and where do you put it? $\endgroup$ Aug 1, 2015 at 18:37
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    $\begingroup$ So lets make a set and call it S = {}. A point is drawn if the gap between a prime pair $p_{n+1} - p_n$ produces a value that is not already in this set. Then, we observe gauss's estimate for the prime gap for $p_n$ and put that on the x axis. On the y axis is the proportional error, so the gap divided by log(p) $\endgroup$ Aug 1, 2015 at 18:45
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    $\begingroup$ Also it's for approximately the first 1000 gaps $\endgroup$ Aug 1, 2015 at 18:54
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    $\begingroup$ Hmm, it looks like the large 'line' peters out around $e^{44}\approx 2^{64}$. It is possible that your numbers got truncated to 64 bits somewhere in your calculations? $\endgroup$ Aug 1, 2015 at 19:06
  • $\begingroup$ Still not quite sure how you formed the graph, but I would refer you to my answer to this question: math.stackexchange.com/questions/2311652/… $\endgroup$
    – user56983
    Aug 13, 2017 at 2:32

1 Answer 1


Graham conjecture We can determine a boundary for the first occurance of certain prime gaps if obeying one or more of the two conditions below. If both conditions are obeyed, then we determine that the first occurance of a prime gap is found following the condition which gives the lowest least bound.

condition 1 -symmetric condition for symmetric type gaps. The first symmetric gap of length $2p_{n}$. It is between consecutive primes which are larger than $p_{n+2}^{2}$ and are equal to $k(\#p_{n+1}!) + p_{n}$ and $k(\#p_{n+1}!) - p_{n}$ such that k is the smallest natural number so that they are both larger than $p_{n+2}^{2}$.

Condition 2 - one directional condition for non symmetric gaps.

The first occurance of a non symmetric gap length $p_{n}-1$ is greater than $p_{n-1}^{2}$ and less than $k(\#p_{n-1}!) $where $k$ is the smallest natural number that doesn't produce a contradiction.

Like i said before, and to clarify; the lowest occurance of a gap, if both prerequisites are met - is the lowest boundary given by them. The lowest gap is either symmetric or not, i don't know how to determine this yet, but one if the two conditions will be correct if both prerequisites are satisfied.

  • $\begingroup$ That conjecture is surely false. $\endgroup$
    – quid
    Aug 2, 2015 at 18:05
  • $\begingroup$ There is one other condition. In the case $p_{n} - 1 = 2p_{m}$ we take the smallest of the two boundary conditions given above. Proof? $\endgroup$ Aug 2, 2015 at 18:07
  • $\begingroup$ If anyone finds a counter example let me know, but i have a proof of it being right in my head :) $\endgroup$ Aug 2, 2015 at 18:14
  • $\begingroup$ What do you mean exactly by "$\#p_{n-1}!$"? $\endgroup$
    – quid
    Aug 2, 2015 at 18:22
  • $\begingroup$ $p_{1} \times p_{2} \times p_{3} \times ... \times p_{n-1}$ $\endgroup$ Aug 2, 2015 at 18:25

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