Erdős Problem 285 #
Reference: erdosproblems.com/285
theorem
erdos_285
(f : ℕ → ℕ)
(S : Set ℕ)
(hS : S = {k : ℕ | ∃ (n : Fin k.succ → ℕ), StrictMono n ∧ 0 ∉ Set.range n ∧ 1 = ∑ i : Fin k.succ, 1 / ↑(n i)})
(h :
∀ k ∈ S,
IsLeast
{x : ℕ | ∃ (n : Fin k.succ → ℕ) (_ : StrictMono n) (_ : 0 ∉ Set.range n) (_ : 1 = ∑ i : Fin k.succ, 1 / ↑(n i)),
n (Fin.last k) = x}
(f k))
:
Let $f(k)$ be the minimal value of $n_k$ such that there exist $n_1 < n_2 < \cdots < n_k$ with $$ 1 = \frac{1}{n_1} + \cdots + \frac{1}{n_k}. $$ Is it true that $$ f(k) = (1 + o(1)) \frac{e}{e - 1} k ? $$
Proved by Martin [Ma00].
[Ma00] Martin, Greg, Denser Egyptian fractions. Acta Arith. (2000), 231-260.
theorem
erdos_285.variants.lb
(f : ℕ → ℕ)
(S : Set ℕ)
(hS : S = {k : ℕ | ∃ (n : Fin k.succ → ℕ), StrictMono n ∧ 0 ∉ Set.range n ∧ 1 = ∑ i : Fin k.succ, 1 / ↑(n i)})
(h :
∀ k ∈ S,
IsLeast
{x : ℕ | ∃ (n : Fin k.succ → ℕ) (_ : StrictMono n) (_ : 0 ∉ Set.range n) (_ : 1 = ∑ i : Fin k.succ, 1 / ↑(n i)),
n (Fin.last k) = x}
(f k))
:
It is trivial that $f(k)\geq (1 + o(1)) \frac{e}{e - 1}k$.