FormalConjectures.ErdosProblems.«422»

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module

public import FormalConjecturesUtil

Erdős Problem 422

References:

erdosproblems.com/422

OEIS A005185

@[expose] public sectionnamespace Erdos422open Filteropen scoped Topology

IsHofstadterQ f means that $f$ is Hofstadter's $Q$-sequence (OEIS A005185): $f(1) = f(2) = 1$ and for $n > 2$ $$ f(n) = f(n - f(n - 1)) + f(n - f(n - 2)), $$ where the recurrence requires $f(n - 1) < n$ and $f(n - 2) < n$, so that both arguments on the right-hand side are positive integers. The sequence begins $1, 1, 2, 3, 3, 4, \ldots$.

At most one function satisfies this predicate. Some function satisfies it if and only if $f(n)$ is well-defined for all $n$, which is not known.

def IsHofstadterQ (f : ℕ+ → ℕ+) : Prop :=
  f 1 = 1 ∧ f 2 = 1 ∧
  ∀ n : ℕ+, 2 < n →
    f (n - 1) < n ∧ f (n - 2) < n ∧ f n = f (n - f (n - 1)) + f (n - f (n - 2))

The third term of Hofstadter's $Q$-sequence is $f(3) = f(2) + f(2) = 2$.


@[category test, AMS 11]
theorem erdos_422.test.f3 : ∀ f : ℕ+ → ℕ+, IsHofstadterQ f → f 3 = 2 := by⊢ ∀ (f : ℕ+ → ℕ+), IsHofstadterQ f → f 3 = 2
  intro f ⟨h1, h2, h⟩f:ℕ+ → ℕ+h1:f 1 = 1h2:f 2 = 1h:∀ (n : ℕ+), 2 < n → f (n - 1) < n ∧ f (n - 2) < n ∧ f n = f (n - f (n - 1)) + f (n - f (n - 2))⊢ f 3 = 2
  obtain ⟨-, -, h3⟩ := h 3 (byf:ℕ+ → ℕ+h1:f 1 = 1h2:f 2 = 1h:∀ (n : ℕ+), 2 < n → f (n - 1) < n ∧ f (n - 2) < n ∧ f n = f (n - f (n - 1)) + f (n - f (n - 2))⊢ 2 < 3f:ℕ+ → ℕ+h1:f 1 = 1h2:f 2 = 1h:∀ (n : ℕ+), 2 < n → f (n - 1) < n ∧ f (n - 2) < n ∧ f n = f (n - f (n - 1)) + f (n - f (n - 2))h3:f 3 = f (3 - f (3 - 1)) + f (3 - f (3 - 2))⊢ f 3 = 2 decideAll goals completed! 🐙f:ℕ+ → ℕ+h1:f 1 = 1h2:f 2 = 1h:∀ (n : ℕ+), 2 < n → f (n - 1) < n ∧ f (n - 2) < n ∧ f n = f (n - f (n - 1)) + f (n - f (n - 2))h3:f 3 = f (3 - f (3 - 1)) + f (3 - f (3 - 2))⊢ f 3 = 2)f:ℕ+ → ℕ+h1:f 1 = 1h2:f 2 = 1h:∀ (n : ℕ+), 2 < n → f (n - 1) < n ∧ f (n - 2) < n ∧ f n = f (n - f (n - 1)) + f (n - f (n - 2))h3:f 3 = f (3 - f (3 - 1)) + f (3 - f (3 - 2))⊢ f 3 = 2
  simp only [h3, show (3 : ℕ+) - 1 = 2 from rfl, show (3 : ℕ+) - 2 = 1 from rfl, h1, h2]f:ℕ+ → ℕ+h1:f 1 = 1h2:f 2 = 1h:∀ (n : ℕ+), 2 < n → f (n - 1) < n ∧ f (n - 2) < n ∧ f n = f (n - f (n - 1)) + f (n - f (n - 2))h3:f 3 = f (3 - f (3 - 1)) + f (3 - f (3 - 2))⊢ 1 + 1 = 2
  rflAll goals completed! 🐙

The fourth term of Hofstadter's $Q$-sequence is $f(4) = f(2) + f(3) = 3$.


@[category test, AMS 11]
theorem erdos_422.test.f4 : ∀ f : ℕ+ → ℕ+, IsHofstadterQ f → f 4 = 3 := by⊢ ∀ (f : ℕ+ → ℕ+), IsHofstadterQ f → f 4 = 3
  intro f hff:ℕ+ → ℕ+hf:IsHofstadterQ f⊢ f 4 = 3
  have h3 := erdos_422.test.f3 f hff:ℕ+ → ℕ+hf:IsHofstadterQ fh3:f 3 = 2⊢ f 4 = 3
  obtain ⟨h1, h2, h⟩ := hff:ℕ+ → ℕ+h3:f 3 = 2h1:f 1 = 1h2:f 2 = 1h:∀ (n : ℕ+), 2 < n → f (n - 1) < n ∧ f (n - 2) < n ∧ f n = f (n - f (n - 1)) + f (n - f (n - 2))⊢ f 4 = 3
  obtain ⟨-, -, h4⟩ := h 4 (byf:ℕ+ → ℕ+h3:f 3 = 2h1:f 1 = 1h2:f 2 = 1h:∀ (n : ℕ+), 2 < n → f (n - 1) < n ∧ f (n - 2) < n ∧ f n = f (n - f (n - 1)) + f (n - f (n - 2))⊢ 2 < 4f:ℕ+ → ℕ+h3:f 3 = 2h1:f 1 = 1h2:f 2 = 1h:∀ (n : ℕ+), 2 < n → f (n - 1) < n ∧ f (n - 2) < n ∧ f n = f (n - f (n - 1)) + f (n - f (n - 2))h4:f 4 = f (4 - f (4 - 1)) + f (4 - f (4 - 2))⊢ f 4 = 3 decideAll goals completed! 🐙f:ℕ+ → ℕ+h3:f 3 = 2h1:f 1 = 1h2:f 2 = 1h:∀ (n : ℕ+), 2 < n → f (n - 1) < n ∧ f (n - 2) < n ∧ f n = f (n - f (n - 1)) + f (n - f (n - 2))h4:f 4 = f (4 - f (4 - 1)) + f (4 - f (4 - 2))⊢ f 4 = 3)f:ℕ+ → ℕ+h3:f 3 = 2h1:f 1 = 1h2:f 2 = 1h:∀ (n : ℕ+), 2 < n → f (n - 1) < n ∧ f (n - 2) < n ∧ f n = f (n - f (n - 1)) + f (n - f (n - 2))h4:f 4 = f (4 - f (4 - 1)) + f (4 - f (4 - 2))⊢ f 4 = 3
  simp only [h4, show (4 : ℕ+) - 1 = 3 from rfl, show (4 : ℕ+) - 2 = 2 from rfl, h2, h3]f:ℕ+ → ℕ+h3:f 3 = 2h1:f 1 = 1h2:f 2 = 1h:∀ (n : ℕ+), 2 < n → f (n - 1) < n ∧ f (n - 2) < n ∧ f n = f (n - f (n - 1)) + f (n - f (n - 2))h4:f 4 = f (4 - f (4 - 1)) + f (4 - f (4 - 2))⊢ 1 + 2 = 3
  rflAll goals completed! 🐙

Let $f(1) = f(2) = 1$ and for $n > 2$ $$ f(n) = f(n - f(n - 1)) + f(n - f(n - 2)). $$ Does $f(n)$ miss infinitely many integers?

@[category research open, AMS 11]
theorem erdos_422 : answer(sorry) ↔
    ∀ f : ℕ+ → ℕ+, IsHofstadterQ f → Set.Infinite {n | ∀ x, f x ≠ n} := by⊢ True ↔ ∀ (f : ℕ+ → ℕ+), IsHofstadterQ f → {n | ∀ (x : ℕ+), f x ≠ n}.Infinite
  sorryAll goals completed! 🐙

Is $f$ surjective?

@[category research open, AMS 11]
theorem erdos_422.variants.surjective : answer(sorry) ↔
    ∀ f : ℕ+ → ℕ+, IsHofstadterQ f → f.Surjective := by⊢ True ↔ ∀ (f : ℕ+ → ℕ+), IsHofstadterQ f → Function.Surjective f
  sorryAll goals completed! 🐙

How does $f$ grow?

@[category research open, AMS 11]
theorem erdos_422.variants.growth_rate :
    ∀ f : ℕ+ → ℕ+, IsHofstadterQ f →
    (fun n ↦ (f n : ℝ)) =O[atTop] (answer(sorry) : ℕ+ → ℝ) := by⊢ ∀ (f : ℕ+ → ℕ+), IsHofstadterQ f → (fun n ↦ ↑↑(f n)) =O[atTop] sorry
  sorryAll goals completed! 🐙

Does $f$ become stationary at some point?

@[category research open, AMS 11]
theorem erdos_422.variants.eventually_const : answer(sorry) ↔
    ∀ f : ℕ+ → ℕ+, IsHofstadterQ f → EventuallyConst f atTop := by⊢ True ↔ ∀ (f : ℕ+ → ℕ+), IsHofstadterQ f → EventuallyConst f atTop
  sorryAll goals completed! 🐙

end Erdos422