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Erdős Problem 1084

Reference: erdosproblems.com/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 : }

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 : ) : := (s : Finset (ℝ^ d)) (_ : s.card = n) (_ : IsSeparated' 1 (s : Set (ℝ^ d))), unitDistNum s

It 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 declaration uses 'sorry'erdos_1084.variants.upper_d1 : f 1 n = n - 1 := 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 declaration uses 'sorry'erdos_1084.variants.easy_upper_d2 (hn : n 0) : f 2 n < 3 * n := n:hn:n 0f 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 declaration uses 'sorry'erdos_1084.variants.upper_d2 : c > (0 : ), 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 declaration uses 'sorry'erdos_1084.variants.triangular_optimal_d2 : f 2 (3 * n ^ 2 + 3 * n + 1) = 9 * n ^ 2 + 3 * 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 declaration uses 'sorry'erdos_1084.variants.upper_lower_d3 : c₁ : , c₂ > (0 : ), ∀ᶠ n in atTop, 6 * n - c₁ * n ^ (2 / 3 : ) f 3 n f 3 n 6 * n - c₂ * n ^ (2 / 3 : ) := c₁, c₂ > 0, ∀ᶠ (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