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This is a little algorithm I made today, which may appear to be quite complex, so I will start with an example. Questions are at the end of the post.

The process goes as follows:

  • Start with the first prime number, $2$.

  • From $2$, add the next prime number ($3$) to get $2+3=5$. There are no non-trivial factors, so we move on.

  • From $2+3$, add the next prime number ($5$) to get $2+3+5=10$. Since $10=2\times5$ and these two numbers appear in the sum, we remove $2$ and $5$.

  • We are left with $3$.

  • From $3$, add the next prime number ($5$) to get $3+5=8$. Now $8=2\times2\times2$, but $2$ does not appear in the sum, so we move on.

  • From $3+5$, add the next prime number ($7$) to get $3+5+7=15$. Since $15=3\times5$ and these two numbers appear in the sum, we remove $3$ and $5$.

  • We are left with $7$.

  • (and so on)

So essentially, we keep on adding consecutive prime numbers until we reach a sum whose product contains some of those primes. We remove those primes and start the process once again. Great, except...

There is one more rule that needs to be added. If we continue doing this, we soon find ourselves in a rather strange scenario.

  • (and so on) a continuation:

  • From $37+47+59+\cdots+241+251+257$, add the next prime number ($263$) to get $$37+47+59+\cdots+251+257+263=5918.$$ Now $5918=2\times11\times269$, but neither of the three primes appear in the sum, so we move on.

  • From $37+47+59+\cdots+251+257+263$, add the next prime number ($269$) to get $$37+47+59+\cdots+251+257+263+269=6187.$$ Since $6187=23\times269$ and $269$ appears in the sum, we remove $269$.

  • We are left with $37+47+59+\cdots+251+257+263$.

This is a cycle! The sequence of $263$ and $269$ will continue forever, if we don't add another rule to this process. Therefore, I call $269$ a cyclic prime, and I propose this new rule.

  • From $37+47+59+\cdots+251+257+263$, add the next non-cyclic prime number ($271$) to get $$37+47+59+\cdots+251+257+263+271=6189.$$ Now $6189=2\times2063$ and these two numbers do not appear in the sum, so we move on.

It is then natural to ask if there are primes that are cyclic only after more than one iteration. However, this is beyond the scope of both my knowledge and computation.


Questions

  1. Will every prime number in the sum eventually be removed? If not, which prime numbers will remain in the sum forever?

I believe the answer to this question is yes. The user @EuxhenH has kindly share their code written in Python. From the program, it is found that all primes up to $16903$ are eliminated at some point before overflow.

  1. Are there infinitely many cyclic prime numbers? That is, are there infinitely many $n$ such that $$p_{n+1}\,\Bigg\vert\,\sum_{p_i\in\text{sum}} p_i\,?$$

I believe the answer to this question is no. As of writing, the only known cyclic prime numbers are $269$ and $94793$ so they are rather rare to find.

Any further input will be much appreciated.

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  • $\begingroup$ How can you possibly know which next prime is non-cyclic? I don't think this part is computable, unless there is some hidden proof or trick. $\endgroup$ – EuxhenH Jul 8 at 12:31
  • $\begingroup$ @EuxhenH If $\require{cancel}p_{n+1}\,\cancel{\Bigg\vert}\,\sum_{p_i\in\text{sum}} p_i$ where $\text{sum}$ is the existing sum, then we know that the next prime cannot be cyclic since the factors of $\text{sum}$ do not include the next prime. $\endgroup$ – TheSimpliFire Jul 8 at 12:37
  • $\begingroup$ But at the same time there are primes which divide the sum and yet are not cyclic. As in your example, $5$ divides $2+3$, but $5$ is not cyclic. Let me add the condition that $p_{n+1}$ is cyclic if it divides $\sum_{p_i\in\text{sum}}p_i$ and such that none of the factors of $\frac{\sum_{p_i\in\text{sum}}p_i}{p_{n+1}}$ are present in the list of factors. Note: This may be incorrect, I didn't prove it, but it is just a guess. $\endgroup$ – EuxhenH Jul 8 at 13:13
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    $\begingroup$ @EuxhenH Good point. We can discuss further in my chatroom :) $\endgroup$ – TheSimpliFire Jul 8 at 13:14
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    $\begingroup$ Can someone explain the downvotes? $\endgroup$ – TheSimpliFire Jul 12 at 6:38

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