This is a little algorithm I made today, which may appear to be quite complex so I will start with an example.

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 prime factorisation 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.

I do not know whether there are any *dicyclic* primes; that is, primes that are still cyclic after more than one iteration. This is beyond current computation methods as such primes if any exist will be extremely large.

### Questions

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

I believe so. With a few modifications, @EuxhenH has kindly shared their Python code. From the program, it is found that all primes up to $16903$ are eliminated at some point before overflow.

The following table shows how long it takes ($N$ iterations) for the smallest prime (not yet removed in the sum) $P$ to be removed. Although the value of $N$ fluctuates significantly, the general trend seems to be that it takes more iterations to remove a larger $P$.

```
P: N
3: 2
7: 6
11: 7
23: 28
Found cyclic 269
37: 44
47: 48
71: 3
107: 47
109: 142
127: 232
181: 9
277: 247
431: 8
457: 316
479: 217
509: 969
773: 977
1069: 92
1123: 1327
1451: 2059
1483: 1270
1801: 542
Found cyclic 94793
2281: 3558
```

*Follow-up: What is the asymptotic growth of $N(P)$?*For instance, does it admit a $\log$ or $\log\log$ increase?*Are there infinitely many cyclic prime numbers?*

That is, are there infinitely many $n$ such that $p_{n+1}$ divides $\sum_{p_i\in S} p_i$? As of writing, the only known **cyclic** prime numbers are $269$ and $94793$ so they are rather rare to find.