Here a challenge for all the GoIIT members,Experts and Administrators!!I am sick and tired

Hari Shankar

An attempt:

It is obvious that the f(n) has to be greater than the number of even integers lying in the set which is p = left[rac{n}{2} ight] or left[rac{n}{2} ight]+1

Consider the subset containing all the even numbers.

Statement 1: Now I claim that it is now enough to be able to select two consecutive odd integers and we have three numbers that are mutually prime to each other.

Proof:

If we have the sequence 2n-1,2n,2n+1 it satisfies the condition that all three are mutually prime to each other.

If we have the sequence as 2n,2n+1,2n+2,2n+3, from the previous argument the sub-sequence 2n+1,2n+2,2n+3 satisfies the given conditions.

Statement 2: From the sequence $a_1,a_2,...,a_{2n}$ it is enough to choose n+1 members to ensure that we always have at least two consecutive members

Proof: Arrange the numbers as $(a_1,a_2), (a_3,a_4),...,(a_{2n-1}, a_{2n})$

If we choose n+1 numbers we are forced to choose some two that are from the same set, i.e. two consecutive numbers

If the last index of the sequence is 2n+1, we obviously need to choose n+2.

We can summarise this as: from the sequence a_1,a_2,...,a_n it is enough to choose left[rac{n+1}{2} ight]+1 members to ensure that we always have at least two consecutive members.

Now the number of odd numbers is n-p from which we have to ensure that some two consecutive odd numbers are selected. From the above result this nothing but left[rac{m+1}{2} ight]+1 where m=n-p

Hence we need to select a subset of the size p+left[rac{m+1}{2} ight]+1 where p and m are as above

and this is the expression for f(n).

PS: the m here obviously has no relation to the m appearing in the problem statement.

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