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حَوّامتي مُمْتِلئة بِأَنْقَلَيْسون
If I am interpreting the problem correctly,
when \(n\) is prime, is the interval \((-n+1, ~2n-1)\) ?
I don't know, I have found a different answer so I'm not sure how you've interpreted it. I guess the way I was imagining it, for k=2 there will be 3 points including n:
where a and b can be any positive or negative integer, as long as it minimizes the size of the set containing n.
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My answer could be totally wrong, how did you come up with your answer? This is just a question that sorta came up while I was doing my own personal research into some stuff haha.
Maybe I should cut to the chase and say what I'm specifically looking for:
I want to, given any number n, be able to find as many numbers that are as small of distance as possible but all contain at least a different prime that the others don't have.
I think this is kind of strange, but if you think of exponents on primes as forming a vector space with each prime a different dimension, then every prime is a linearly independent dimension.
From this I am planning on using Gram-Schmidt orthogonalization with the gcd to remove any common factors they might share.
So I think my phrasing of the question is not quite right earlier since although 3 and 6 are not relatively prime, they are 'linearly independent' in the sense that we can form any multiple of 2 or 3 using these through multiplication and division.
so for any prime p, there is k=p-2,(numbers which are relatively prime with p) right ? this is the trivial case
for a number n! there is 0 numbers that relatively prime with it
im considering the interval (1,n!)
and in p case the interval is (1,p)
is that what ur asking for ?