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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?


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 $

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  • $\begingroup$ Hi @swallis, welcome to Puzzling SE! Take the tour if you haven't already! If I'm reading your question correctly, the problem here is to find a function $ f $ composed entirely of basic arithmetic functions such that $ f(n) \neq n, $ $ f(n_1) \neq f(n_2) $ if $ n_1 \neq n_2, $ and $ f(n) $ is prime for $ n = 2, 3, \dots, N, $ and ? $\endgroup$ – HTM Jan 15 at 23:34
  • $\begingroup$ @HTM yes indeed. Cheers. $\endgroup$ – swallis Jan 15 at 23:45
  • $\begingroup$ You said "The goal is to use operations on all of these numbers" but you mean "on a subset of these numbers". Because $(2×4)−1=7$ doesn't use $3$. Also, where did the $4$ come from in $(2×4)−1...$? I think you mean we also get as many (integer or rational) constants as we need, to play with. Could you tighten the wording? Assume everyone comes up with a prime AP formula, how do you compare which answer is 'best'? The one with fewest constants? Smallest maximum constant? Smallest product of constants? etc. $\endgroup$ – smci Jan 16 at 23:54
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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).

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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.

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  • 2
    $\begingroup$ The longest known as of September $2019$ is by $3$ terms longer now. " See Largest known primes in AP". $\endgroup$ – Vepir Jan 16 at 10:31
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For $ n = 2, 3, 4, 5 $, the function

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

produces distinct prime numbers:

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

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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 $n=0,1,2\dots,28$

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  • $\begingroup$ I think non-linear functions are not intended (i.e. hiding exponentiation as multiplication). If this would be allowed, you could interpolate a polynomial and generate any specific list of numbers. $\endgroup$ – Vepir Jan 16 at 22:57
  • $\begingroup$ Yeah, as Vepir said. It is a very interesting take on the problem though! $\endgroup$ – swallis Jan 17 at 2:13

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