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import FormalConjecturesUtilErdős Problem 1084
Let f_2(n) be the maximum number of pairs of points at distance exactly 1
among any set of n points in ℝ², under the condition that all pairwise
distances are at least 1.
Estimate the growth of f_2(n).
Status: open.
open Finset Filter Metric Realopen scoped EuclideanGeometry
namespace Erdos1084variable {n : ℕ}
The maximal number of pairs of points which are distance 1 apart that a set of n 1-separated
points in ℝ^d make.
noncomputable def f (d n : ℕ) : ℕ :=
⨆ (s : Finset (ℝ^ d)) (_ : s.card = n) (_ : IsSeparated' 1 (s : Set (ℝ^ d))), unitDistNum sIt is easy to check that $f_1(n) = n - 1$.
@[category research solved, AMS 52, formal_proof using formal_conjectures at "https://github.com/Sanexxxx777/formal-conjectures/blob/9e4f3845be122a8fa3190d38543ebdd0a6f25605/FormalConjectures/ErdosProblems/1084.lean#L232"]
theorem erdos_1084.variants.upper_d1 : f 1 n = n - 1 := n:ℕ⊢ f 1 n = n - 1
All goals completed! 🐙It is easy to check that $f_2(n) < 3n$.
@[category research solved, AMS 52]
theorem erdos_1084.variants.easy_upper_d2 (hn : n ≠ 0) : f 2 n < 3 * n := n:ℕhn:n ≠ 0⊢ f 2 n < 3 * n
All goals completed! 🐙Erdős showed that there is some constant $c > 0$ such that $f_2(n) < 3n - c n^{1/2}$.
@[category research solved, AMS 52]
theorem erdos_1084.variants.upper_d2 : ∃ c > (0 : ℝ), ∀ n > 0, f 2 n < 3 * n - c * sqrt n := ⊢ ∃ c > 0, ∀ n > 0, ↑(f 2 n) < 3 * ↑n - c * √↑n
All goals completed! 🐙Erdős conjectured that the triangular lattice is best possible in 2D, in particular that $f_2(3n^2 + 3n + 1) < 9n^2 + 3n$.
Note: in [Er75f] is read $9n^2 + 6n$, but this seems to be a typo.
@[category research open, AMS 52]
theorem erdos_1084.variants.triangular_optimal_d2 : f 2 (3 * n ^ 2 + 3 * n + 1) = 9 * n ^ 2 + 3 * n := n:ℕ⊢ f 2 (3 * n ^ 2 + 3 * n + 1) = 9 * n ^ 2 + 3 * n
All goals completed! 🐙Erdős claims the existence of two constants $c_1, c_2 > 0$ such that $6n - c_1 n^{2/3} ≤ f_3(n) \le 6n - c_2 n^{2/3}$.
@[category research solved, AMS 52]
theorem erdos_1084.variants.upper_lower_d3 :
∃ c₁ : ℝ, ∃ c₂ > (0 : ℝ), ∀ᶠ n in atTop,
6 * n - c₁ * n ^ (2 / 3 : ℝ) ≤ f 3 n ∧ f 3 n ≤ 6 * n - c₂ * n ^ (2 / 3 : ℝ) := ⊢ ∃ c₁, ∃ c₂ > 0, ∀ᶠ (n : ℕ) in atTop, 6 * ↑n - c₁ * ↑n ^ (2 / 3) ≤ ↑(f 3 n) ∧ ↑(f 3 n) ≤ 6 * ↑n - c₂ * ↑n ^ (2 / 3)
All goals completed! 🐙
end Erdos1084