I'm curious to see if this conjecture is true: if $ m > 4 $ is a positive integer not divisible by $ 2 $ or $ 3 $, it's possible to find a positive integer $ n $ such that the difference of the $ 2 $ Fibonacci numbers $ F_{m+n}-F_n $ is a prime number.
Various example of $ [m, n] $, with the smallest n, are $[5,3],[7,3], [11,4], [13,5], [17,3],..., [619,1353]...$
I don't know if this could be of any help: I converted the primes I found that way in the Zeckendorf representation and they are an alternating sequence of $ 1 $ and $ 0 $ ending with two adjacent $ 0 $ and $ 1 $ (and eventually just various $ 0 $)
Edit: I posted this problem on Mathoverflow too
I can't say for sure, but intuitively, since primes are truly random and infinitely many exist, then it's very likely the conjecture is true: that it's always possible.
I was able to come up with this semi-proof: $F(x)$ is the x-th number in the Fibonacci sequence
In the interval that lies between $F(A+x)-2(F(x))$ up to $F(A+x)$, there exists just around ${2F(x)\over \ln(F(A+x))}$ primes. And, we know that F(A+x)--F(x) is in this interval.
Since the interval holds 2(F(x)) integer numbers, then the chance of randomly selecting a Prime, if you choose a number in that interval, is $1\over ln(F(A+x))$. We will always choose the number F(A+x)-F(x).
The chance of failure, is, 1 - $1\over ln(F(A+x))$ for the first attempt. If every time F(A+x)-F(x) failed to hit a prime number, you tried again, but increased x by 1, forever, then the chance of hitting at least one prime number is 1 -- every attempt has failed, or $1 - \prod_{x=0}^{\infty} \left(1 - \frac{1}{ln(F(A+x))}\right).$
Regardless of what A is, the chance eventually converges to 100%.
Here that is in desmos. https://www.desmos.com/calculator/29bkcvb5mt
In all seriousness though, this assumes that selecting F(A+x)--F(x) every time as our lottery numbers, is as arbitrary as selecting any other number or function of x. If we waited for truly random selection to land upon F(A+x)-F(x), it'd take increasingly forever, as A is larger, and then on top of that we'd have just a $1/\ln(F(A+x))$ chance that it's prime.
I suppose though, that since we could do THAT forever too, we can randomly land upon a F(A+x)--F(x) infinitely many times. And so eventually one of the F(A+x)--F(x) would be landed upon randomly and on top of that also be prime. (regardless of A)