Erdős Problem 891

References:

• erdosproblems.com/891
• [Po18] Pólya, Georg, Zur arithmetischen {U}ntersuchung der {P}olynome. Math. Z. (1918), 143--148.
• [Wikipedia] https://en.wikipedia.org/wiki/Dickson%27s_conjecture

theorem Erdos891.erdos_891 : sorry ↔ ∀ k ≥ 2, ∀ᶠ (n : ℕ) in Filter.atTop, ∃ m ∈ Finset.Ico n (n + ∏ i ∈ Finset.range k, Nat.nth Nat.Prime i), k < ArithmeticFunction.cardDistinctFactors m

Let $2=p_1 < p_2 < \cdots$ be the primes and $k\geq 2$. Is it true that, for all sufficiently large $n$, there must exist an integer in $[n,n+p_1\cdots p_k)$ with $>k$ many prime factors?

theorem Erdos891.erdos_891.variants.schinzel (k : ℕ) : k ≥ 2 → ∀ᶠ (n : ℕ) in Filter.atTop, ∃ m ∈ Finset.Ico n (n + (∏ i ∈ Finset.range (k - 1), Nat.nth Nat.Prime i) * Nat.nth Nat.Prime k), k < ArithmeticFunction.cardDistinctFactors m

Schinzel deduced from Pólya's theorem [Po18] (that the sequence of $k$-smooth integers has unbounded gaps) that this is true with $p_1\cdots p_k$ replaced by $p_1\cdots p_{k-1}p_{k+1}$.

theorem Erdos891.erdos_891.variants.case_k_2 : sorry ↔ ∀ᶠ (n : ℕ) in Filter.atTop, ∃ m ∈ Finset.Ico n (n + 6), 3 ≤ ArithmeticFunction.cardDistinctFactors m

This is unknown even for $k=2$ - that is, is it true that in every interval of $6$ (sufficiently large) consecutive integers there must exist one with at least $3$ prime factors?

theorem Erdos891.erdos_891.variants.weisenberg (k : ℕ) (hk : k ≥ 2) : ∃ᶠ (n : ℕ) in Filter.atTop, ∀ m ∈ Finset.Ico n (n + ∏ i ∈ Finset.range k, Nat.nth Nat.Prime i - 1), ArithmeticFunction.cardDistinctFactors m ≤ k

Weisenberg has observed that Dickson's conjecture implies the answer is no if we replace $p_1\cdots p_k$ with $p_1\cdots p_k-1$. Indeed, let $L_k$ be the lowest common multiple of all integers at most $p_1\cdots p_k$. By Dickson's conjecture [Wikipedia], there are infinitely many $n'$ such that $\frac{L_k}{m}n'+1$ is prime for all $1\leq m < p_1\cdots p_k$. It follows that, if $n=L_kn'+1$, then all integers in $[n,n+p_1\cdots p_k-1)$ have at most $k$ prime factors.