TITLE: Is there an Infinite increasing sequence of primes with bounded second or larger differences?
QUESTION [7 upvotes]: 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?.

REPLY [5 votes]: If this were known for $k=2$, it would mean that for sufficiently large $n$ and some $A$, there is always a prime between $n^2$ and $(n+A)^2$. I believe that this is still open; see Legendre's conjecture. For comparison, the Riemann Hypothesis would imply there is a prime between $n^2$ and $(n+A \log n)^2$.
It is known that for large enough $n,$ there are primes between $n^3$ and $(n+1)^3$ (and you can lower that exponent from $3$). Perhaps you can use this to construct sequences with some differences bounded, although this is not immediate to me.