# Some interesting experimental results about the distribution of primes

Let's consider the following metric of the gap between consecutive primes $$m(k)=\frac {p_k^2-p_{k-1}^2} {24}\;\;\;\;\;(k\ge4)$$

Now, let's define the function

$$\delta(k)=m(k)\;\;\;\;$$ if $$\,m(k)\,$$ is prime $$\;\;\;\;(1)$$

$$\delta(k)=0\;\;\;\;\;\;\;\;\;\;$$ otherwise $$\,\;\;\;\;\;\;\;\;\;\;\;\;\;(2)$$

If we plot the graph of $$\,m(k)\,$$ we obtain (up to $$\,k=10^4$$): But, if we plot the graph of $$\,\delta(k)$$, we obtain something completely different (up to $$\,k=10^4$$): In the second graph the three different increasing curves seem to have the following asymptotic behaviors:

$$\sim \frac {p_k} {2}$$

$$\sim \frac {p_k} {3}$$

$$\sim \frac {p_k} {6}$$

Further, the prime numbers are not uniformly distributed among the three curves. The number of primes decreases passing from the highest to the lowest curve (I'm trying to estimate the number of primes that fall on each curve).

I do not have adequate knowledge of analytic number theory to explain the phenomenon, therefore I ask for your help.

Many thanks.

• The asymptotic behaviors you got is just derived from the Cramer conjecture, and you have multiplied the equality of Cramer conjecture by even integer which is p_k+p_(k-1) , Could you deduce some thing from the plot of Cramer for your graph plot ? – zeraoulia rafik Apr 1 at 0:01
• Your question is not of research level. If $p_k-p_{k-1}=2$, then your expression equals $(p_k+p_{k-1})/12$, which is asymptotically $p_k/6$. Similarly, if $p_k-p_{k-1}=4$, then your expression is asymptotically $p_k/3$. Finally, if $p_k-p_{k-1}=6$, then your expression is asymptotically $p_k/2$. So what you observe follows easily from well-known conjectures on gaps between primes. – GH from MO Apr 1 at 1:45
• I read experimental evidence this way: if $\,m(k)\,$ is prime, then it will be $\,\sim p_k/2\,$ or $\,\sim p_k/3\,$ or $\,\sim p_k/6$, no other possibility given. The question is to estimate the probability of the three cases. – Augusto Santi Apr 1 at 2:05
• See my response below. It answers your question in full. – GH from MO Apr 1 at 2:46

Assume that $$k\geq 5$$ and $$m(k)$$ is prime. Observe the factorization $$\frac{p_k-p_{k-1}}{2}\cdot\frac{p_k+p_{k-1}}{2}=6m(k).$$ On the left hand side, the second fraction is greater than $$6$$, hence the first fraction is smaller than $$m(k)$$. So the first factor is relatively prime to $$m(k)$$, whence it divides $$6$$. So $$p_k-p_{k-1}$$ equals $$2$$ or $$4$$ or $$6$$ or $$12$$.
• If $$p_k-p_{k-1}=2$$, then $$m(k)=(p_k+p_{k-1})/12\sim p_k/6$$.
• If $$p_k-p_{k-1}=4$$, then $$m(k)=(p_k+p_{k-1})/6\sim p_k/3$$.
• If $$p_k-p_{k-1}=6$$, then $$m(k)=(p_k+p_{k-1})/4\sim p_k/2$$.
• If $$p_k-p_{k-1}=12$$, then $$m(k)=(p_k+p_{k-1})/2$$, which is a contradiction, because there is no prime between $$p_k$$ and $$p_{k-1}$$.
This explains the "three increasing curves" $$p_k/6$$, $$p_k/3$$, $$p_k/2$$. The density of these three cases (i.e. the validity of two linear equations in the three primes $$p_k$$, $$p_{k-1}$$, $$m(k)$$) is heuristically known, see e.g. Conjecture 1.2 in Green-Tao: Linear equations in primes.