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Empty
 one year ago
Given a number n, what is the smallest interval containing it with k other numbers all mutually relatively prime?
Empty
 one year ago
Given a number n, what is the smallest interval containing it with k other numbers all mutually relatively prime?

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thadyoung
 one year ago
Best ResponseYou've already chosen the best response.0حَوّامتي مُمْتِلئة بِأَنْقَلَيْسون

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1If I am interpreting the problem correctly, when \(n\) is prime, is the interval \((n+1, ~2n1)\) ?

Empty
 one year ago
Best ResponseYou've already chosen the best response.1I 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: \[\gcd(n,n+a)=1\]\[\gcd(n,n+b)=1\]\[\gcd(n+a,n+b)=1\] where a and b can be any positive or negative integer, as long as it minimizes the size of the set containing n.

Empty
 one year ago
Best ResponseYou've already chosen the best response.1My 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.

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Maybe 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 GramSchmidt 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so for any prime p, there is k=p2,(numbers which are relatively prime with p) right ? this is the trivial case

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for 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 ?
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