Let $k$, $l$, $n$ be integers such that $k \ge 3$, $l \ge 2$ and $n + k \ge p^{(k)}$,
where $p^{(k)}$ is the least prime satisfying $p^{(k)} \ge k$.
Then there is a prime $p \ge k$ for which $l$ does not divide
the multiplicity of the prime factor $p$ in $(n + 1) \ldots (n + k)$.
Theorem 2 from [ErSe75].
[ErSe75] Erdős, P. and Selfridge, J. L., The product of consecutive integers is never a power. Illinois J. Math. (1975), 292-301.
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theorem erdos_930.variant.consecutive_strong :
∀ k l n, 3 ≤ k → 2 ≤ l → nextPrime k ≤ n + k →
∃ p, k ≤ p ∧ p.Prime ∧
¬ (l ∣ Nat.factorization (∏ m ∈ Icc (n + 1) (n + k), m) p)
