Skip to content

Erdős problem 392

Let A(n)A(n) denote the least value of tt such that n!=a1at n! = a_1 \cdots a_t with a1atn2a_1 \leq \cdots \leq a_t\leq n^2. Then A(n)=n2n2logn+o(nlogn). A(n) = \frac{n}{2} - \frac{n}{2\log n} + o\left(\frac{n}{\log n}\right).

Sources

Browse retained paths and inspect the exact material available for this Problem.

3 retained statements2415f78e850a

Open selected source

FormalConjectures/ErdosProblems/

392.lean

Retained formal statement2 of 3

Cambie has observed that a positive answer follows from the result above with atna_t \leq n, simply by pairing variables together, e.g. taking ai=a2i1a2ia'_i = a_{2i-1}a_{2i} (and the lower bound follows from Stirling's approximation).

FormalConjectures/ErdosProblems/392.leanErdos392.erdos_392.variants.implication8 linesExact file
(∀ (A : ℕ → ℕ),    (∀ n > 0,        IsLeast {x | ∃ t, ∃ (_ : ∃ a, n.factorial = ∏ i, a iMonotone aa (Fin.last t) ≤ n ^ 2), t + 1 = x} (A n)) →      (fun n => ↑(A n) - ↑n / 2 + ↑n / (2 * Real.logn)) =o[Filter.atTop] fun n => ↑n / Real.logn) →  ∀ (A : ℕ → ℕ),    (∀ n > 0,        IsLeast {x | ∃ t, ∃ (_ : ∃ a, n.factorial = ∏ i, a iMonotone aa (Fin.last t) ≤ n), t + 1 = x} (A n)) →      (fun n => ↑(A n) - ↑n + ↑n / Real.logn) =o[Filter.atTop] fun n => ↑n / Real.logn
SolvedStatement only, no proof

Search problems.science

Find a Problem, Result, source, or page