Making the whole set into primes

Question

Let's say you start with a set of sequential integers starting from 2, so: $ 2, 3, 4, 5, \dots, N $ for some $ N > 2. $

The goal is to use identical basic arithmetic operations ($ +, -, \times, \div $) to all of these numbers to have them end up them all being different prime numbers.

Is 2, 3, 4, 5 possible? How much higher can you go? Is there a limit to what can be achieved here?

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Rules

• For a given number, the end number must be different than the starting number (for example, you can't have 3 become 3, but you could have 2 become 3).
• The end primes must be all different.
• The numbers must remain whole and non-negative at all times, but you may multiply/divide them by a non-whole number, such as 1.5, if this doesn't cause the result to break this pattern.

Examples

2, 3

$ (2 \times 4) - 1 = 7 $
$ (3 \times 4) - 1 = 11 $

2, 3, 4

$ (((2 \times 2) - 2) \times 2) - 1 = 3 $
$ (((3 \times 2) - 2) \times 2) - 1 = 7 $
$ (((4 \times 2) - 2) \times 2) - 1 = 11 $

Answer 1

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 found at "Largest known primes in AP". As of $2019$, the longest known is $n=0,1,2\dots,26$ which can be rewritten to fit your problem for $N=2,3,4\dots,28$.

That is, the following function gives distinct primes for $n=2,3,4\dots,28$:

$$ 18135696597948930\cdot n+188313212743640051 $$

And is the best known AP of primes at this time (according to linked wikipedia article).

Answer 2

The function with rule

$$f(n)= 45872132836530n + 376651396831763$$

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 known arithmetic sequence composed of entirely primes.

Answer 3

For $ n = 2, 3, 4, 5 $, the function

$$ f(n) = 6 \times n - 1 $$

produces distinct prime numbers:

$$ \begin{equation*} f(2) = 6 \times 2 - 1 = 11 \\ f(3) = 6 \times 3 - 1 = 17 \\ f(4) = 6 \times 4 - 1 = 23 \\ f(5) = 6 \times 5 - 1 = 29 \end{equation*} $$

Answer 4

If I understood everything correctly, this solution is also acceptable. If not - it's still very interesting and simple one.

$$ 2 \times n \times n + 29 $$

It's valid for $n=0,1,2\dots,28$