Some thoughts on the space between consecutive prime numbers
There is a property of exponential series a^x for a ≥ 2 where sum of two elements of that series can be ordered as it is shown on the picture below. 1 1 would indicate 2 * first element of the series and 1 2 would indicate sum of first and second element in the series. Arrows on the picture indicate increasing order of that sum series.
It can be shown that 2 is the minimal integer value of a for it to have the property on the picture, and that for all a>2 property persists. It is shown below.
Let’s assume that n is one element from this kind of series and that m is the next element. Let’s now take the critical step on the picture, a step from X X to 1 X+1 (which is for example step from 4 4 to 1 5). So for this step it must be true this:
(1) 2*n ≤ m +1
I have put ≤ and no < because let’s permit that sum X X can be equal to 1 X+1. I don’t think it will change anything when we come to prime numbers. Let’s call x the space between m and n, so x=m — n. Therefore, m = n + x. Let’s put that into (1) and we have.
(2) 2*n ≤ n + x + 1
With some arithmetic it becomes:
(3) x ≥ n — 1
This shows that space between two consecutive elements must be at least as big as the smaller element of the two for the series to have always increasing order. Let’s test this for a=2.
(4) x =2^(n+1) — 2^n
With some arithmetic it becomes:
(5) x = 2^n
So, it is true, and from this (6) it can be obvious that a=2 is the smallest.
(6) 2^n {x}≥ 2^n — 1 {n-1}
For all a>2 space between consecutive elements is 2 * 3^n, 3 * 4^n etc. It just becomes bigger and bigger relatively to the smaller element.
So, space between consecutive elements is also raising exponentially. It is at core same to statement that derivation of exponential function is also exponential function. This is just a switch from continuous to discrete case.
Having this settled, let’s jump to prime numbers. I will be juggling with some of the theorems about prime numbers, so let’s state them.
Goldbach’s conjecture — it states that every integer bigger that 2 can be written as sum of 2 prime numbers
Twin prime conjecture — it states that there is an infinite number of primes that differ by 2
Prime number theorem — it states that average space between consecutive prime numbers increases, and there is a specific formula for the way they sparse, but it won’t be very important here
So, by combining Prime number theorem and Twin prime conjecture we can state that space between two consecutive primes will never really start to always increase from some point because it will always at some point fall to 2.
Let’s suppose that there is an interval where space between consecutive primes starts to increase in exponential way. So, if it start to increase in the exponential way, and if it stays that way for long enough it will start to have the property that is shown on the picture at the beginning. But combining this with Goldbach’s conjecture would state that sums 1 N, 2 N, 3 N etc. will have to be integers that differ by one. The consecutive sums will have to cover consecutive integers now if they are about to cover all integers. But they won’t be able to cover consecutive integers as consecutive primes don’t differ by one mostly. They only differ by one in (1, 2) and (2,3) case. It comes to that that either Goldbach’s conjecture is true, or space between consecutive primes increases in the exponential way for very long. If it does increase in the exponential way for the long enough, Goldbach’s conjecture would be false. By Goldbach’s conjecture being nearly proven I would presume that space between consecutive primes never increases in the exponential way for very long (very long is relative, it could also mean very very long).
What is left to say is that ordered sum of pairs of elements in then series from the picture in the beginning has the biggest density. That, all other series of sums of pairs of elements from exponential series that have increasing property would be less dense than one shown and would derive their property from the series on the picture. I think that proof for that is in the a=2 exponential series, because it barely passes the condition x≥n — 1. The fact that it barely passes the condition and that 2 is the lowest integer (after 1, but 1 is not included) with this property is the reason this sum is the densest. I think that this maximum density between different pair sum series is needed for formal proof of the last sentence from last paragraph. (Less dense pair sum series would be for example 1 1, 2 2, 3 3, 4 4.) But, I think that it becomes obvious and this is not really needed for comprehension.
(from October 19, 2021)
