Start with any prime number $p$.

If $mod(p,6)=1$ apply this: $$ x_{n+1} = 2x_n-1 $$

Otherwise $mod(p,6)=5$ and apply this: $$ x_{n+1} = 2x_n+1 $$

We then form a sequence starting with $p$ and if we continue this process indefinitely, $mod(x,6)$ stays the same, ergo if the starting prime is 1 below a multiple of 6, all numbers of the sequence will be 1 below a multiple of 6, the same applies for primes 1 above a multiple of 6. This can be proven algebraically.

The sequences look like this:

$mod(p,6)=1 : 6n+1; 12n+1; 24n+1; 48n+1; ... $

$mod(p,6)=5 : 6n-1; 12n-1; 24n-1; 48n-1; ... $

Can we guarantee that these sequences have infinitely many composite numbers in them (for all starting prime values)?

Or in another way: How do we prove, that we don't get stuck into a sequence of purely prime numbers?

  • 1
    $\begingroup$ The exponential growth should guarantee infinite many composites , but I do not see a proof. Modulo-calculation with some integer $m\ge 2$ does not help since in the case $m\mid n$, every member of the sequences is coprime to $m$. $\endgroup$
    – Peter
    Nov 20, 2021 at 17:27
  • $\begingroup$ I object. The exponential growth per se has nothing to do with the number of composites. $\endgroup$ Nov 20, 2021 at 17:38
  • $\begingroup$ @IvanNeretin Of course not , but this is an iterative process and we know none producing infinite many primes without a gap. I also have not claimed that this is a proof. $\endgroup$
    – Peter
    Nov 20, 2021 at 17:42
  • $\begingroup$ @Peter In that sense, I agree with you. $\endgroup$ Nov 20, 2021 at 17:43

1 Answer 1


Suppose $p$ is the starting prime of the form $6n-1$. Then the sequence goes as follows: $$p,\;2p+1,\;4p+3,\;\dots,2^kp+(2^k-1)\dots$$ As soon as $2^k-1$ is divisible by $p$ (which is going to happen at $k=p-1$ and multiples thereof, according to the Fermat's little theorem), the whole expression will also turn out divisible by $p$.

The same reasoning with opposite signs applies to the other case.


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