
Find a consecutive integer sequence – n, n+1, n+2, …., n+k such that there does not exist a number in the sequence which is coprime to all the other numbers in the sequence.
For example, 2 ,3 ,4 ,5 ,6 ,7 ,8 ,9 ,10 is NOT such a sequence because 7 is coprime to all the other numbers in the sequence.

Hey Vivek, do we need to find just one pair (n, k) or prove that the statement holds true for all large enough k? Looks like a fun problem:)

Hi Artur,
We just need to find one pair (n,k) , preferably the least such pair ( as in dictionary order).
The generalisation for all k larger than certain number is the interesting take !! (And That seems to be the case.)
The original question was this : Prove that in any 10 consecutive numbers, We can always find a number which is coprime to all the other. Just extended it and searched for the sequence that first shows exception to this rule. It is the case with 11,12,13 and so on. But Soon we get the exception. And What is that first exceptional sequence?
We can get it easily with programming, Artur 🙂 But, We can also get it analytically. It is the one good fun search !!

Hey Vivek,
Have you found a nice explicit solution (n, k) for large k? I believe you can prove such sequences exist and then construct them by using ∏(p1)/p → 0 (that’s 1/ζ(1)) and the Chinese Remainders Theorem. However, I don’t like much this proof:)

ha, Interesting Artur. Yes, I think I used the chinese remainder theorem. And the first such sequence is of length 17, I think. There are no such sequences of lesser length.
