Re: [math-fun] a prime in each row?

asimovd@aol.com wrote:

But obviously, there's a prime in the first row, and there's one in the second row guaranteed by Bertrand's Postulate.

Question: Where does the conjectural territory first begin? [2n+1,3n] ? [3n+1,4n] ? Etc.

It follows from the prime number theorem that, given e>0, there exists N=N(e) such that for all x > N, there is a prime between x and (1+e)x. Thus for n large enough there is a prime between 2n+1 and 3n, and also 3n+1 and 4n, and in fact any kn+1 and (k+1)n for fixed k independent of n. Gary McGuire