# Making the whole set into primes

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

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

• 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 ?
– HTM
Jan 15 '20 at 23:34
• @HTM yes indeed. Cheers. Jan 15 '20 at 23:45
• 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.
– smci
Jan 16 '20 at 23:54

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).

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.

• The longest known as of September $2019$ is by $3$ terms longer now. " See Largest known primes in AP". Jan 16 '20 at 10:31

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

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

produces distinct prime numbers:

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

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

• 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. Jan 16 '20 at 22:57
• Yeah, as Vepir said. It is a very interesting take on the problem though! Jan 17 '20 at 2:13