First some definitions:
To let $ A $ be an abbreviation for an infinite sequence
$ (a (1), a (2), a (3), points) $
positive integers as well $ B $ is an infinite series of positive integers
$ (b (1), b (2), b (3), points) $,
and C is an increasing infinite series of positive integers
$ (c (0), c (1), c (2), c (3), points) $, Call $ c (0) $ the initial value of $ C $,
We say the couple $ (A, B) $ is a recursion pattern for $ C $ if for everyone
we have positive integers n
$ c (n) = a (n) * c (n-1) + b (n) $, Given a recursion pattern $ (A, B) $, and a
Initial value for $ C $We can reconstruct the entire sequence $ C $,
Of course, for every increasing infinite sequence $ C $ We can find a couple $ (A, B) $
Which is a recursion pattern for $ C $simply by taking $ a (n) $ to
be equal $ 1 $and define that $ b (n) $ upon need. If the order $ C $ is
will increase rapidly, there will be many couples $ (A, B) $ what will be
Recursion pattern for $ C $,
In particular, we can find for every increasing sequence of primes
Recursion patterns, many of them when the sequence grows fast.
Now the question: Can a couple $ (A, B) $ to be a recursion pattern for more than
a sequence of primes? In other words, given $ A $ and $ B $ as recursion
Pattern, there is more than one initial value for $ c (0) $
what leads to an infinite sequence of primes?
If the answer is yes and we can prove it, I would expect the proof
quite difficult Maybe a conditional proof would be possible.
On the other hand, it can be fundamental to show that the answer is no.
This question was written by Moshe Newman.