Prime factors of consecutive-integer products in a prescribed interval
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Let n>2k be the least integer for which the product (n−k)⋯(n−1) has no prime factor in (k,2k). Must this least n grow faster than every fixed power of k?
Why it matters and what counts as progress
Why it matters: A superpolynomial lower bound separates this consecutive-product question from the earlier polynomial scale and related binomial-coefficient results.
What counts as progress: Improve the polynomial lower bound to a superpolynomial bound or sharpen the exponential-scale lower estimate.
How the problem developed
Paul Erdős Asked how large the least n>2k can be when the product of k consecutive integers ending at n−1 has no prime factor between k and 2k.
- 18 Jun 2026 · Proof [2]
Wouter van Doorn and Quanyu Tang Confirmed Erdős's superpolynomial-growth conjecture, proving n_k>exp(log²k/(20 log log k)) for all sufficiently large k.
AI contribution
Consecutive integers free of certain prime factors
For all sufficiently large k, the least n>2k for which (n−k)⋯(n−1) has no prime factor in (k,2k) exceeds exp((log k)²/(20 log log k)).
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