Yes, a nice one. Suppose, for a contradiction, that such a sequence does exist, and let $p$ be its first term. If $p_0 = p = 2$ or $3$, then $p_1 = 3$ or $p_1 = 5$. Suppose that $p_1 = 3$, then $p_2 = 5$ or $p_2 = 7$. Thus, in each case one of the terms is 5 or 7. But : - If 5 occurs in the sequence then the sequence must go $5 - 11 - 23 - 47 - 93$ or $95$ none of which is a prime. A contradiction. - If 7 occurs in the sequence then the sequence must go $7 - 13 - 25$ or $27$ none of which is prime. A contradiction. Then, we may assume that $p_0 = p > 7$. It follows that $p = 6k+1$ or $p = 6k+5$. In the first case, note that if $x = 1 \pmod 6$ then $2x+1 = 0 \pmod 3$. Thus, by an easy induction, we must have $p_{k+1} = 2p_k - 1$, from which we deduce that for each $n \geq 0$ we have $p_n = 2^n p - 2^n + 1$. But, $p \neq 2$ so, by Fermat's little theorem, we have $2^{p-1} \equiv 1 \pmod p$. It follows that $p_{p-1} > p$ and $p_{p-1} \equiv p \equiv 0 \pmod p$, which contradicts that $p_{p-1}$ is a prime. In the second case, the same reasoning shows that, for each $k$, we must have $p_{k+1} = 2p_k + 1$, and then for each $n \geq 0$: $p_n = 2 ^n p + 2 ^n - 1$, which leads to the same contradiction. The conclusion follws.

This is rather easy. Take $p>3$ to be a term of such a sequence. If it's $3k+2$, then the next one has to be $2p+1$, and the next one has to be $2(2p+1)+1$, and so on (all the terms are of the form $3k+2$). Something similar happens if $p=3k+1$, so let's assume $p=3k+2$ (the other case is similar). The terms of the sequence are of the form $2^np+2^n-1$. If $n=p-1$, then $p|2^np+2^n-1$, so this term of the sequence is not a prime, which yields a contradiction, so the answer is "no" (no such sequences exist).

thanks for your quick response, grobber

Let $p>3$ Choose $p_i=3k+1$ then $p_{i+1}=2p_i-1 \equiv 1 \pmod 3$ which yeilds $p_{i+k}=2^k p_i-2^k+1$(induction) implying $(p_i-1)(2^k-1)=0$ which is absurd so $p_i \equiv 2 \pmod 3$ The terms of the sequence are of the form $2^np+2^n-1$. If $n=p-1$, then $p|2^np+2^n-1$, so this term of the sequence is not a prime, which yields a contradiction. Hence $p<3$ a contradiction hence no such sequence exists.

if n>3, p_n=1or-1 (mod6) if p_n=1 (mod6) and p_n+1=2p_n+1, 3|p_n+1 so not allowed. So, p_n+1=2p_n-1 for all p_,n Same goes for when p_n=-1 (mod6), and p_n+1=2p_n+1 for all p_n lets let p_3=q (q is prime obviously) p_q+2=2^(q-1)q-(2^(q-1)-1) or 2^(q-1)+(2^q-1)-1) , and it is o (modq) because p_q+2 is prime, it is not allowed. so, it does not exist like in the title very beautiful problem. shout out to Baltic Way!