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Given a prime $\,p\,$ let's consider the following sequence:

$a_0=p$

$a_{n+1}=(a_n-2)\cdot a_n+2$

Is it possible to determine whether the sequence $\,a_n\,$ will reach, sooner or later, another prime number?

Some examples:

for $\,p=2$, $\;\;a_1=2\;\;$ (prime)

for $\,p=3$, $\;\;a_1=5\;\;$ (prime)

for $\,p=29$, $\;\;a_2=614657\;\;$ (prime)

for $\,p=31$, $\;\;a_5=185302018885184100000000000000000000000000000001\;\;$ (prime)

Many thanks.

[ Added ]

Experimental evidence shows that, up to 100, only for the following prime numbers $\,p\,$ the sequence $\,a_n\,$ should never reach another prime: $$13,19,23,43,53,59,61,71,73,79$$ In order to find out the possible divisors of $\,a_n=(p-1)^{2^n}+1\,$ (see the answer of GH from MO) the following result can be exploited:

the only prime divisors of $\,a_n\,$ are of the form $\,k\cdot2^{n+1}+1$.

Example: $$(13-1)^8+1=17\cdot97\cdot260753=(1\cdot2^4+1)\cdot(6\cdot2^4+1)\cdot(16297\cdot2^4+1)$$

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  • $\begingroup$ Notice that we can't even establish the infinitude of primes of the form $n^2+1$, let alone with the added condition of $n$ being a power of $p-1$. $\endgroup$
    – Gjergji Zaimi
    Commented Mar 13, 2020 at 0:01
  • $\begingroup$ @GjergjiZaimi: I think this question is slightly different. I suspect that there exists a prime $p$ such that $a_n$ is never prime, but probably we will never be able to decide this. $\endgroup$
    – GH from MO
    Commented Mar 13, 2020 at 1:06

1 Answer 1

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Your question can be reformulated as follows.

Question. If $p$ is a prime, does there always exist a positive integer $n$ such that $(p-1)^{2^n}+1$ is also a prime?

I believe that this question is out of reach at present (my guess is that the answer is "no", but we will never know). Similar to the well-known questions on Fermat numbers and its generalizations.

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  • $\begingroup$ Maybe did you want to write $(p-1)^{2n}+1$? $\endgroup$
    – Augusto Santi
    Commented Mar 12, 2020 at 23:52
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    $\begingroup$ @AugustoSanti: For $p=31$, your $a_5$ is $30^{32}+1$, not $30^{10}+1$. In general, your $a_n$ is $(p-1)^{2^n}+1$, as I wrote. The exponents are $2$-powers. $\endgroup$
    – GH from MO
    Commented Mar 12, 2020 at 23:58
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    $\begingroup$ Sorry ... you're absolutely right $\endgroup$
    – Augusto Santi
    Commented Mar 13, 2020 at 0:01
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    $\begingroup$ For example if p is a Fermat prime, the OP asks if there exists a bigger Fermat prime. This is a well known open problem for $p=2^{16}+1$. $\endgroup$
    – user6976
    Commented Mar 13, 2020 at 2:35
  • $\begingroup$ @MarkSapir: Excellent point! $\endgroup$
    – GH from MO
    Commented Mar 13, 2020 at 4:32

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