Let $p,q \in \mathbb{P}$, $p \geq 3$ and $q$ is the next prime to $p$.

For $b \in \mathbb{P}$ Consider : $N_b = \displaystyle{\small \prod_{\substack{a \leq b \\ \text{a prime}}} {\normalsize a}}$

Let $n \in \mathbb{N}$,

$n$ is q-point iff $n = q \alpha$ with $\gcd(\alpha, N_p)=1$

My Conjecture:

There is at least one integer $m$, between two consecutive q-points in which $\gcd(m, N_q)=1$

We can reduce this conjecture to the following problem:

For $b \in \mathbb{P}$ We have :

$$\#\{k \leq N_b \, | \, \gcd(k, N_b)=1\} = \displaystyle{\small \prod_{\substack{a \leq b \\ \text{a prime}}} {\normalsize (a-1)}}$$

Let $\beta_b(i)$ be the i-th number coprime to $N_b$ and less than $N_b$, $i \in \{1,2,3,\cdots,\displaystyle{\small \prod_{\substack{a \leq b \\ \text{a prime}}} {\normalsize (a-1)}}\}$

My Conjecture:

$(\forall i \in [1,\displaystyle{\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}}-1]) (\exists j \in [1,\displaystyle{\small \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize (a-1)}}])$ : $$q \cdot \beta_p(i) < \beta_q(j) < q \cdot \beta_p(i+1)$$

I had hard feeling that this conjecture is true, but all my Attempts to prove it failed.

Many thanks for any help..

  • $\begingroup$ "There exist at least $m$ in $\bf N$" doesn't parse. $\endgroup$ Aug 3, 2019 at 23:41
  • 1
    $\begingroup$ “at least $m$ integers” would be the more natural way to say that. $\endgroup$
    – Matt F.
    Aug 4, 2019 at 2:34
  • $\begingroup$ I reformulate the sentence $\endgroup$ Aug 4, 2019 at 9:39
  • $\begingroup$ Should word "one" replace "an" in the phrase "at least an integer $m$", so that the correct phrase would be: "at least one integer $m$"? $\endgroup$
    – Wlod AA
    Aug 9, 2019 at 22:23

1 Answer 1


This is almost surely false. The size of the largest gap between numbers coprime to $N=N_p$ grows at a faster rate than do the primes. There may even be an example with q less than 1000 where q is the least prime factor of the numbers c and d, and every number in between c and d has a smaller least prime factor. Such c and d would witness no beta(j) of your conjecture.

Update 2019.08.11 At https://mathoverflow.net/a/115879 Aaron Meyerowitz has an example with two integers with least prime factor 73, and all integers in between having smaller least prime factor. This is my (or Aaron's) proffered counterexample. I conjecture that this is the smallest such, and would be glad to be shown otherwise.

End Update 2019.08.11.

Gerhard "There May Be Smaller Examples" Paseman, 2019.08.09.

  • $\begingroup$ how can you prove your claim? The Jacobsthal function (of primorial numbers $h(n)$) for example can't help in this case, because the size of the interval $]q \beta_p(i), q \beta_p(i+1)[$ change.. we know that if the size of the interval is always $2q$ my claim is false ($h(n)$ with $q=p_n$ can take values large than $2q$) $\endgroup$ Aug 10, 2019 at 19:02
  • $\begingroup$ I prove the claim by looking at least prime factor of numbers in an interval. When I find an interval bounded by two numbers with lpf q, and all other numbers in between have a smaller lpf, then I have found a spoiler to your conjecture. Gerhard "A Small Matter Of Programming" Paseman, 2019.08.10. $\endgroup$ Aug 10, 2019 at 19:55
  • $\begingroup$ What is your counter-example ? and i don't know a computer can verify up to $\displaystyle{\small \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a}} = e^{q+o(1)}$ with $q$ near $1000$ $\endgroup$ Aug 10, 2019 at 22:05
  • $\begingroup$ Check this out: stackoverflow.com/a/13811371 . It might be a smaller counterexample. Gerhard "That Does It For Conjecture" Paseman, 2019.08.11. $\endgroup$ Aug 11, 2019 at 20:12

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