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 Classical BigOperators namespace 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.

@[category test, AMS 5 11] theorem erdos_1192.f_r_zero_zero (A : Set ℕ) : f_r A 0 0 = 1 := A:Set ℕ⊢ f_r A 0 0 = 1 A:Set ℕ⊢ {v | (∀ (i : Fin 0), v i ∈ A) ∧ ∑ i, v i = 0}.ncard = 1 have h : { v : Fin 0 → ℕ | (∀ i, v i ∈ A) ∧ ∑ i, v i = 0 } = { default } := A:Set ℕ⊢ f_r A 0 0 = 1 A:Set ℕv:Fin 0 → ℕ⊢ v ∈ {v | (∀ (i : Fin 0), v i ∈ A) ∧ ∑ i, v i = 0} ↔ v ∈ {default} All goals completed! 🐙 All goals completed! 🐙

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

@[category test, AMS 5 11] theorem erdos_1192.f_r_empty (r n : ℕ) : f_r ∅ (r + 1) n = 0 := r:ℕn:ℕ⊢ f_r ∅ (r + 1) n = 0 r:ℕn:ℕ⊢ {v | (∀ (i : Fin (r + 1)), v i ∈ ∅) ∧ ∑ i, v i = n}.ncard = 0 have h : { v : Fin (r + 1) → ℕ | (∀ i, v i ∈ (∅ : Set ℕ)) ∧ ∑ i, v i = n } = ∅ := r:ℕn:ℕ⊢ f_r ∅ (r + 1) n = 0 r:ℕn:ℕv:Fin (r + 1) → ℕ⊢ v ∈ {v | (∀ (i : Fin (r + 1)), v i ∈ ∅) ∧ ∑ i, v i = n} ↔ v ∈ ∅ All goals completed! 🐙 All goals completed! 🐙

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

@[category test, AMS 5 11] theorem erdos_1192.f_r_singleton_self (n : ℕ) : f_r {n} 1 n = 1 := n:ℕ⊢ f_r {n} 1 n = 1 n:ℕ⊢ {v | (∀ (i : Fin 1), v i ∈ {n}) ∧ ∑ i, v i = n}.ncard = 1 have h : { v : Fin 1 → ℕ | (∀ i, v i ∈ ({n} : Set ℕ)) ∧ ∑ i, v i = n } = { fun _ ↦ n } := n:ℕ⊢ f_r {n} 1 n = 1 n:ℕv:Fin 1 → ℕ⊢ v ∈ {v | (∀ (i : Fin 1), v i ∈ {n}) ∧ ∑ i, v i = n} ↔ v ∈ {fun x => n} n:ℕv:Fin 1 → ℕ⊢ (∀ (i : Fin 1), v i = n) ∧ ∑ i, v i = n ↔ v = fun x => n n:ℕv:Fin 1 → ℕ⊢ (∀ (i : Fin 1), v i = n) ∧ ∑ i, v i = n → v = fun x => nn:ℕv:Fin 1 → ℕ⊢ (v = fun x => n) → (∀ (i : Fin 1), v i = n) ∧ ∑ i, v i = n n:ℕv:Fin 1 → ℕ⊢ (∀ (i : Fin 1), v i = n) ∧ ∑ i, v i = n → v = fun x => n n:ℕv:Fin 1 → ℕh1:∀ (i : Fin 1), v i = nright✝:∑ i, v i = n⊢ v = fun x => n n:ℕv:Fin 1 → ℕh1:∀ (i : Fin 1), v i = nright✝:∑ i, v i = ni:Fin 1⊢ v i = n All goals completed! 🐙 n:ℕv:Fin 1 → ℕ⊢ (v = fun x => n) → (∀ (i : Fin 1), v i = n) ∧ ∑ i, v i = n n:ℕ⊢ (∀ (i : Fin 1), (fun x => n) i = n) ∧ ∑ i, (fun x => n) i = n n:ℕ⊢ ∑ i, (fun x => n) i = n All goals completed! 🐙 All goals completed! 🐙

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

@[category test, AMS 5 11] theorem erdos_1192.f_r_no_rep : f_r {0} 1 1 = 0 := ⊢ f_r {0} 1 1 = 0 ⊢ {v | (∀ (i : Fin 1), v i ∈ {0}) ∧ ∑ i, v i = 1}.ncard = 0 have h : { v : Fin 1 → ℕ | (∀ i, v i ∈ ({0} : Set ℕ)) ∧ ∑ i, v i = 1 } = ∅ := ⊢ f_r {0} 1 1 = 0 v:Fin 1 → ℕ⊢ v ∈ {v | (∀ (i : Fin 1), v i ∈ {0}) ∧ ∑ i, v i = 1} ↔ v ∈ ∅ v:Fin 1 → ℕ⊢ ¬((∀ (i : Fin 1), v i = 0) ∧ ∑ i, v i = 1) v:Fin 1 → ℕh1:∀ (i : Fin 1), v i = 0h2:∑ i, v i = 1⊢ False have h3 : v = fun _ ↦ 0 := ⊢ f_r {0} 1 1 = 0 v:Fin 1 → ℕh1:∀ (i : Fin 1), v i = 0h2:∑ i, v i = 1i:Fin 1⊢ v i = 0; All goals completed! 🐙 v:Fin 1 → ℕh1:∀ (i : Fin 1), v i = 0h2:∑ i, (fun x => 0) i = 1h3:v = fun x => 0⊢ False v:Fin 1 → ℕh1:∀ (i : Fin 1), v i = 0h3:v = fun x => 0⊢ ∑ i, (fun x => 0) i = 1 → False All goals completed! 🐙 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 declaration uses 'sorry'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! 🐙

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 declaration uses 'sorry'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 A x) All goals completed! 🐙

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

@[category research solved, AMS 5] theorem declaration uses 'sorry'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