Argthtjtr

First, some definitions:

Let $A$ be a shorthand for an infinite sequence
$( a(1), a(2), a(3), dots )$
of positive integers, likewise $B$ is an infinite sequence of positive integers
$( b(1), b(2), b(3) , dots)$,
and C is an increasing infinite sequence of positive integers
$( c(0), c(1), c(2), c(3), dots)$. Call $c(0)$ the initial value of $C$.

We say that the pair $(A,B)$ is a recursion pattern for $C$ if for each
positive integer n we have
$c(n) = a(n) * c(n-1) + b(n)$. Given a recursion pattern $(A,B)$, and an
initial value for $C$, we can reconstruct all the sequence $C$.

Clearly, for any increasing infinite sequence $C$ we can find a pair $(A,B)$
which is a recursion pattern for $C$, simply by taking all the $a(n)$ to
be equal to $1$, and defining the $b(n)$ as required. If the sequence $C$ is
increasing quickly, there will be many pairs $(A,B)$ which will be
recursion patterns for $C$.

In particular, for any increasing sequence of primes, we can find
recursion patterns, many of them if the sequence grows quickly.

Now the question: can a pair $(A,B)$ be a recursion pattern for more than
one sequence of primes? In other words, given $A$ and $B$ as a recursion
pattern, is it possible there is more than one initial value for $c(0)$
which will lead to an infinite sequence of primes?

Comments:
If the answer is yes, and we can prove that, I would expect the proof
to be quite difficult. Maybe a conditional proof would be possible.
On the other hand, it may be elementary to show that the answer is no.
This question was written up by Moshe Newman.

The answer is yes, there are "recursion patterns" which result in sequences of primes for more than one choice of $c(0)$. In fact, if you are given any two "starting points", it is possible to find such sequences. The proof comes from the pigeonhole principle.

Start with $c(0), c'(0)$. These are two prime numbers with difference $d(1) := c'(0) - c(0)$. Then as the set of prime numbers is infinite, there is some pair $p_1 < p'_1$ such that $p_1 equiv p'_1 (text{mod }d)$. Let $a(1) = frac{p'_1 - p_1}{d(1)}$; let $b(1) = p_1 - a(1) c(0)$. (Note: I've slightly cheated here, and will explain the cheat at the end). Then $c(1) = a(1) c(0) + b(1) = a(1) c(0) + p_1 - a(1) c(0) = p_1$, while $c'(1) = a(1) c(0) + b(1) = frac{p'_1 - p_1}{c'(0) - c(0)} c'(0) + p_1 - frac{p'_1 - p_1}{c'(0) - c(0)} c(0) = p'_1$, as desired.

This is a pattern that can be repeated in the obvious way. We therefore have a recursion pattern that results in sequences of primes for multiple "starts".

Now for the cheat: there's no reason so far why this would make $b(i)$ positive. I don't have a completely elementary reason for that, but the prime number theorem should be enough to guarantee it. Essentially, the $n$th prime number doesn't grow "quickly enough" to escape such pairs.

There are very good reasons to think that for every even $g$ there are infinitely many primes $p$ with $p+g$ also prime. I'll mention them at the end.

For $g=10$ the primes with this property start out

$$3, 7, 13, 19, 31, 37, 43, 61, 73, 79, 97, 103, 127, 139, 157, 163, 181, 223, 229,cdots$$

If this is indeed infinite, then one can pick any sub-sequence and the shift by $10$ such as

$$3,19,73,79,157,229,cdots$$
$$13,29,83,89,167,239,cdots.$$

Then $(A,B)$ is a recursion pattern for both sequences of primes where $A=1,1,1,cdots$ and $B=16,54,6,78,72,cdots$ is the correct $B$ sequence.

There is known to be at least one $g leq 246$ with infinitely many primes $p,p+g.$ (in fact infinitely many with the two primes successive, not that we care here.) So that alone shows that the answer to your question is yes.

Assuming what I said about even $g$ is true, if follows that there is great freedom in finding recursion patterns which describe two prime sequences. Specifically: suppose we play a game where you give me two odd primes $p_0 lt p_0'$ a multiplier $a_1$ and an integer $beta_1.$ Then I need to give you a shift $b_1 gt beta_1$ so that $p_1=a_1p_0+b_1$ and $p'_1=a_1p'_0+b_1$ are both prime. Then, after I tell you this, you can give me any $a_2$ and $beta_2$ and I must give $b_2 gt beta_2$ so that $p_2=a_2p_1+b_2$ and $p'_2=a_2p'_1+b_2$ are both prime. We continue in this way forever or until I get stuck.

If I can always manage the first step of finding $b_1$ then I will never get stuck. It is just the same problem over and over with a different given $p,p',a,beta$ each time. Just pick $g=a_0(p'_0-p_0)$ and, since there are infinitely many pairs $p,p+g,$ just pick one with $p-a_1p_0 gt beta_1.$

A theorem about prime constellations would suggest that one can similarly start with $s=3$ or $s=100$ initial primes and an arbitrary $A$ (revealed one step at a time) and pick $B$ to get $s$ prime sequences all with the same recursion pattern $(A,B)$

Here are details on the claim at the top. Define $pi_g(x)$ to be the number of primes $p lt x$ with $p+g$ also prime. Then there are good simple reasons to expect that $pi_2(x)$ is asymptotic to $C_2frac{x}{(ln x)^2} $ for $C_2=Pi_{p geq 3}(1-frac1{(p-1)^2}).$ Numerical calculations up to high $x$ match this quite well. The same reasoning suggests that there is a similar $C_g$ for each $g$ which depends only on the prime divisors of $g$ So $C_{2^k}=C_2$ and for any $g=2^k3^j$ , $C_g=2C_2$. In general, $C_g=C_2Pi_{p mid g}frac{p-1}{p-2}.$ Again, the numerical evidence fits the prediction.