1. FormalConjectures
  2. ErdosProblems
  3. «1192»
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Erdős Problem 1192

References:

    erdosproblems.com/1192

    [Ru90] Ruzsa, Imre Z., A just basis. Monatsh. Math. (1990), 145--151.

    [Er80] Erdős, Paul, A survey of problems in combinatorial number theory. Ann. Discrete Math. 6 (1980), 89--115.

open Nat Filter Finset Setopen scoped Asymptotics BigOperatorsnamespace Erdos1192

For $A\subset \mathbb{N}$ let $f_r(n)$ count the number of solutions to $n=a_1+\cdots+a_r$ with $a_i\in A$.

noncomputable def f_r (A : Set ℕ) (r n : ℕ) : ℕ := { v : Fin r → ℕ | (∀ i, v i ∈ A) ∧ ∑ i, v i = n }.ncard

The empty sum ($r = 0$) gives exactly one representation of $0$: the empty tuple.

All goals completed! 🐙

With an empty set, there are no valid $r$-tuples for $r \geq 1$.

All goals completed! 🐙

For $r = 1$, $f_1({n}, n) = 1$: the only $1$-tuple from ${n}$ summing to $n$ is $(n)$.

All goals completed! 🐙

$f_1({0}, 1) = 0$: can't represent $1$ with a single element from ${0}$.

All goals completed! 🐙

Does there exist, for all $r\geq 2$, a basis $A$ of order $r$ (so that $f_r(n)>0$ for all large $n$) such that $$\sum_{n\leq x}f_r(n)^2 \ll x$$ for all $x$?

@[category research open, AMS 5] theorem erdos_1192 : answer(sorry) ↔ ∀ r ≥ 2, ∃ A : Set ℕ, (∀ᶠ n in atTop, f_r A r n > 0) ∧ (fun (x : ℕ) ↦ ∑ n ∈ range (x + 1), (f_r A r n : ℝ) ^ 2) =O[atTop] (fun (x : ℕ) ↦ (x : ℝ)) := ⊢ True ↔ ∀ r ≥ 2, ∃ A, (∀ᶠ (n : ℕ) in atTop, f_r A r n > 0) ∧ (fun x ↦ ∑ n ∈ Finset.range (x + 1), ↑(f_r A r n) ^ 2) =O[atTop] fun x ↦ ↑x All goals completed! 🐙open scoped Classical in

Erdős and Rényi proved by the probabilistic method that there exists a set $A$ such that $$\sum_{n\leq x}f_r(n)^2 \ll x$$ and $$\lvert A\cap [1,x]\rvert\gg x^{1/r}$$ for all $x$.

@[category research solved, AMS 5] theorem erdos_1192.variants.renyi : ∀ r ≥ 2, ∃ A : Set ℕ, ((fun (x : ℕ) ↦ ∑ n ∈ range (x + 1), (f_r A r n : ℝ) ^ 2) =O[atTop] (fun (x : ℕ) ↦ (x : ℝ))) ∧ ((fun (x : ℕ) ↦ (x : ℝ) ^ (1 / (r : ℝ))) =O[atTop] (fun (x : ℕ) ↦ (count (· ∈ A) x : ℝ))) := ⊢ ∀ r ≥ 2, ∃ A, ((fun x ↦ ∑ n ∈ Finset.range (x + 1), ↑(f_r A r n) ^ 2) =O[atTop] fun x ↦ ↑x) ∧ (fun x ↦ ↑x ^ (1 / ↑r)) =O[atTop] fun x ↦ ↑(count (fun x ↦ x ∈ A) x) All goals completed! 🐙

Ruzsa [Ru90] proved that the answer is yes for $r=2$.

@[category research solved, AMS 5] theorem erdos_1192.variants.ruzsa : ∃ A : Set ℕ, (∀ᶠ n in atTop, f_r A 2 n > 0) ∧ (fun (x : ℕ) ↦ ∑ n ∈ range (x + 1), (f_r A 2 n : ℝ) ^ 2) =O[atTop] (fun (x : ℕ) ↦ (x : ℝ)) := ⊢ ∃ A, (∀ᶠ (n : ℕ) in atTop, f_r A 2 n > 0) ∧ (fun x ↦ ∑ n ∈ Finset.range (x + 1), ↑(f_r A 2 n) ^ 2) =O[atTop] fun x ↦ ↑x All goals completed! 🐙end Erdos1192