The opposite question is if for any infinite increasing sequence of primes and any $k$ the sequence of the $k$-th order differences of the elements of the sequence is unbounded.

But if the question is true then there are integers $k$ and $B$ and a sequence of primes $q_1 < q_2 < \dots$ such that $$\lvert \Delta^k q_n \rvert < \le B \quad \quad \forall n=1,2,\dots$$ and it is natural to ask what can be the least value of $k$ for which such a sequence exists, can we have $k=2$?

Does somebody know of any result or heuristic in one or the other sense?

EDIT: I expect now the question to have a negative answer. The reason is that if we start with the first $k$ primes a given bound $B$, and follow every possible chain of primes with $k$th order differences bounded by $B$. From an element $q$ of one of these chains we get at most $2B+1$ possible succesors and as the "probability" of one them being prime is about $(2B+1)/\log q$ the expected number of "active chains" decreases to zero as $q$ increases.

By the way I have computed the size of the longuest chain of primes with bounded second differences starting at 2,3 and for small values of $B=2,4,6,8,10$ it gives $57, 421, 1860, 24661, 380028$, it seems to increase very roughly as $e^{2B}$, is that a reasonable estimate?.

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# Is there an Infinite increasing sequence of primes with bounded second or larger differences?

The opposite question is if for any infinite increasing sequence of primes and any $k$ the sequence of the $k$-th order differences of the elements of the sequence is unbounded.

But if the question is true then there are integers $k$ and $B$ and a sequence of primes $q_1 < q_2 < \dots$ such that $$\lvert \Delta^k q_n \rvert < B \quad \quad \forall n=1,2,\dots$$ and it is natural to ask what can be the least value of $k$ for which such a sequence exists, can we have $k=2$?

Does somebody know of any result or heuristic in one or the other sense?