There is no limit to this! The Green Tao theorem tells you that the sequence of prime numbers contains arbitrarily long arithmetic progressions. This means that using $a+b\cdot n$ you can get as many primes as you want for some $a,b$ and consecutive values of $n$.
But the theorem does not tell you how to find $a,b$. The longest known such sequence can be ...
The function with rule
produces distinct primes for $n$ up to $25$.
For proof, see the third bullet point on this list of prime number records. It is valid for $x=0,1,...,23$, so I substitute $n=x+2$ so that the set of valid inputs begins at $2$. The function is clearly strictly increasing and so the primes must be distinct.
It is apparently the longest ...
I'm sure I remember a recent puzzle based on the same idea, but I'm failing to find it. There was
Aha, found it:
I'm not sure whether this should be regarded as a duplicate of that one. It isn't the same question but it's clearly closely related.
Based on my previous answer of a similar question, I created a Python program to brute-force all the possible solutions.
To calculate the number of possible solutions:
That is a big number, but not too big for a computer with some time.
However, there is two caveats:
Here is the code:
from dataclasses import dataclass
from enum import Enum
from typing ...
For sake of completeness, I have done a computer search and have found that 40 $\tau$s is the absolute minimum one can achieve. This can be done using Ian's expressions for 1–13 and 17–20, KSmarts' expression for 15, and two new expressions for 14 and 16:
The only expressions requiring 3 $\tau$s are those for 11, 19, and 20, and exhaustive search has shown ...
Partial answer with quick upper and lower bounds for the range.
I believe the highest number you can get should be:
The lowest should be:
Which means that at most there could be:
That said, the problem states numbers, which means we need to include decimals.
Heading out shortly, but thoughts about an answer that probably won't actually help: