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I have been thinking a lot about prime gaps.

It seemed to me that it is much easier to reason about reduced residue systems which can then be used to reason back to prime gaps.

Below is an example of thinking about prime gaps through reasoning about reduced residue systems. Instead of reasoning about $p_{n+1} - p_{n}$, I find it much easier to think about $c_{n,i+1} - c_{n,i}$.

Let:

  • $p_n$ be the $n$th prime
  • $c_{n,i}$ be the $i$th element in the ordered list of integers relatively prime to $p_n$ so that $c_{n,1} = 1$ and $c_{n,p_n-1} = p_n-1$ and $c_{n,p_n} = p_n+1$
  • $D_n(x)$ be defined for $x > n$ to be the least $j-i$ where:
    • $j > i$
    • $p_x | c_{n,j}$
    • $p_x | c_{n,i}$
  • $G_n(x,m)$ be defined for $x > n, m \ge 2$ to be the least $j - i$ where:
    • $j > i$
    • $p_x | c_{n,j}$
    • $p_x | c_{n,i}$
    • $c_{n,j} - c_{n,i}$ is a multiple of $m$

From this perspective, I was wondering if the following propositions are straight forward to show when reasoning with reduced residue systems.

Would the following assumptions be correct for all $n > 0$?

  • If $a > b$, then $D_n(a) \ge D_n(b)$

  • If $s > t$, then $G_n(a,s) \ge G_n(a,t)$

These seem obvious to me but I wanted to confirm since often there are edge cases that serve as counter examples to broad generalizations about primes.

I was working on arguments to confirm each assumption but I found the argument more complicated than I expected. Would there be a simple argument to confirm each assumption?

I did find a way to complete the above argument but I need to spend more time validating it to make sure that it does not contain a mistake.

Edit: I have added context for the question to address the concern that had been raised. Hopefully, my answer addresses the concern raised.

Alex Ravsky
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Larry Freeman
  • 10,189

1 Answers1

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Let $n<x$ be any natural numbers. Then the ordered list of natural numbers relatively prime to $p_n$ is $(1,2,\dots)$ with the multiples of $p_n$ removed. Thus the $c_{n,i}$'s divisible by $p_x$ are exactly the naturals divisible by $p_x$ but not by $p_n$.

The case of $D_n(x)$.

The value $j-i$ required for $D_n(x)$ is minimized when either $c_{n,j}=c_{n,i}+p_x$ or ($c_{n,j}=c_{n,i}+2p_x$ and $p_n|(c_{n,i}+p_x)$).

Suppose first that $p_n>2$. If $p_x\ne 1\pmod {p_n}$ then it is easy to see that the required minimum is attained in the first case, when $c_{n,i}=-1\pmod {p_n}$ so $D_n(x)=\min\{j-i\}=p_x-\left\lceil\frac{p_x}{p_n}\right\rceil$. Similarly, if $p_x=1\pmod {p_n}$ then the required minimum is attained again in the first case, when $c_{n,i}=-2\pmod {p_n}$ and $D_n(x)=\min\{j-i\}=p_x-\left\lceil\frac{p_x-1}{p_n}\right\rceil$. Note that anyway $$D_n(x)\ge p_x-\left\lceil\frac{p_x}{p_n}\right\rceil\ge p_x-\frac{p_x+p_n-1}{p_n}.$$

So to show that the function $D_n(x)$ is nondecreasing it suffices to note that $$p_x-\left\lceil\frac{p_x-1}{p_n}\right\rceil\le p_x-\frac{p_x-1}{p_n}\le p_{x+1}-\frac{p_{x+1}+p_n-1}{p_n}.$$ The latter inequality holds because $$(p_{x+1}-p_x)\left(1-\frac 1{p_n}\right)\ge 2\left(1-\frac 1{2}\right)=1.$$

Suppose now that $p_n=2$. Then the first case is impossible and so the required minimum is attained in the second case. Then $$D_n(x)=\min\{j-i\}=2p_x-\left\lceil\frac{p_x}{2}\right\rceil=\frac 32\cdot p_x-\frac{1}{2}.$$

Clearly, the function $D_n(x)$ is nondecreasing.

The case of $G_n(x,m)$.

This case is similar to the previous.

Indeed, let $m\ge 2$ be any natural number an $M$ be the least common multiple of $m$ and $p_x$. The value $j-i$ required for $G_n(x,m)$ is minimized when either $c_{n,j}=c_{n,i}+M$ or ($c_{n,j}=c_{n,i}+2M$ and $p_n|(c_{n,i}+M)$).

Suppose first that $p_n>2$. If $M\ne 1\pmod {p_n}$ then it is easy to see that the required minimum is attained in the first case, when $c_{n,i}=-1\pmod {p_n}$ so $G_n(x,m)=\min\{j-i\}=M-\left\lceil\frac{M}{p_n}\right\rceil$. Similarly, if $M=1\pmod {p_n}$ then the required minimum is attained again in the first case, when $c_{n,i}=-2\pmod {p_n}$ and $G_n(x,m)=\min\{j-i\}=M-\left\lceil\frac{M-1}{p_n}\right\rceil$.

But if $p_x|(m+1)$ then can happen that $G_n(x,m)>G_n(x,m+1)$. For instance, let $p_n=3$, $p_x=5$, and $m=4$. Then $M=20\ne 1\pmod {p_n}$, so $$G_n(x,m)=M-\left\lceil\frac{M}{p_n}\right\rceil=20-\left\lceil\frac{20}{3}\right\rceil=13.$$ But $$G_n(x,m+1)=5-\left\lceil\frac{5}{3}\right\rceil=3.$$

Alex Ravsky
  • 106,166