$p_{n+1}$ is defined as the lowest prime factor of the number $2^{p_n}+1$.
Let’s try something:
Let $p_0 = 2$. Then $p_1$ is the lowest prime factor of $2^2 + 1 = 5$, so 5. Then $p_2$ is the lowest prime factor of $2^5 +!1 = 33$, so 3. Then $p_3$ is the lowest prime factor of $2^3 + 1 = 9$....which is 3 again! Looks like this sequence has an infinite repetition and converges to 3...
The question is,
Is this true for any $p_0$? The fact of the matter is that we have a countable set of $p_n$, because the number of $p_n$ must be less than the number of integers $2^x + 1$ there are — and this is one-to-one with the natural numbers $\mathbb{N}$. It seems logical that at some point in the sequence of $p_n$, you’ll find a number divisible by 3. To avoid this, you’d essentially need to find a series of primes — quasi-Mersenne — where each continuing $2^{p_n} + 1$ is also prime. I don’t think that such a sequence exists...
Sample starting points:
$p_0 = 2$. 2, 5, 3, 3, 3, ...
(My favourite) $p_0 = 3$. 3, 3, 3, 3, 3, ...
$p_0 = 5$. 5, 3, 3, 3, 3, ...
$p_0 = 7$. 7, 3, 3, 3, 3, ... (because of 129)
$p_0 = 11$. 11, 3, 3, 3, 3, ... (because of 2049)....
Hmm....
Does every $p_0 > 2$ converge to 3 within one step? @WeatherVane found that this is at least the case for $3 \leq p_0 \leq 61$. This is essentially saying for prime $p$, does $3 | 2^p + 1$? Further, for any odd natural number 2m+1, does $3 | 2^{2m+1} + 1$? Let’s see...this is true iff $3 | 2^{2m+1} - 2 = 2( 2^{2m} - 1)$ iff $3 | 2^{2m} - 1$ iff $3 | 2^{2m} - 4$.... commences rambling...
Let’s use
Induction. We seem to have exhaustively proved base cases above. Then, assume that for some odd integer $n = 2m + 1$, $3 | 2^n + 1$. Then consider $p$ = 2m + 3, the next odd integer. Does $3 | 2^p + 1 = 2^{2m+3} + 1 = 4(2^{2m+1}) + 1$? Well, if $3 | 2^n + 1 = 2^{2m+1} + 1$, then $3 | 4(2^{2m+1} + 1)$. So $3 | 4(2^{2m+1}) + 4$. It must therefore also divide 3 less than that, and so $3 | 4(2^{2m+1}) + 1 = 2^{2m+3} + 1 = 2^p + 1$. We have therefore shown that the inductive hypothesis does indeed imply the inductive conclusion. By induction, therefore, $3|2^n+1$ for any positive odd integer $n$.
It naturally follows that
Any starting prime number greater than 2 is a positive odd integer, and so for any $p_0 > 2$, we must have $p_n = 3 \forall n \in \mathbb{N}$. Since we have shown that $p_0 = 2$ converges to 3 within 2 steps, for any prime $p_0$, $p_n = 3 \forall n \geq 2$. This is why there is no sequence (not even the hypothetical prime ladder postulated above) which is suitable for puzzling, because every prime sequence (without restriction) converges to 3 infinitely. $\square$