Proof of assertion in A253266.

The condition for membership in the sequence is:

Numbers such that the minimum distance between divisors of n occurs only between composite numbers.

The assertion is:

All members of the sequence are either a square minus a small square, or a multiple of a smaller member of the sequence. The square root of the small square must be less than half the fourth root of the number.

Let n be the number in the sequence, and let a and b be the divisors with minimum difference. Now a and b must be relatively prime; if they had a common prime factor p, the difference p - 1 would be less than b - a. Since a and b both divide n, a*b must divide n. But a*b must be a member of the sequence; its divisors are a subset of the divisors of n, and thus cannot have any pairs with a smaller difference than b-a. So we are left with the case where n = a*b, and a and b are the central divisors of n.

No member of the sequence can be even, since then 2 - 1 = 1 must be the minimum difference, and 1 and 2 are not prime. (A similar argument will show that n cannot be divisible by 3, but we do not need this result here.) So a and b are both odd. We can then take x = (b+a) / 2, and y = (b-a) / 2, giving us n = x*2 - y*2, and 2*y is the difference between a and b. Now let p be the smallest prime divisor of a. Then a >= p^2, since it is composite; and also a < sqrt(n). So p < n^(1/4). But p must be > b-a, else p - 1 would be a smaller difference between divisors. We conclude 2*y = b-a < n^(1/4), and y < n^(1/4) / 2, as claimed.