Smallest index $k > n$ such that $(p_k+p_{k+1})/(p_n+p_{n+1})$ is an integer $\ge 2$

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import FormalConjecturesUtil

/-!
# Smallest index $k > n$ such that $(p_k+p_{k+1})/(p_n+p_{n+1})$ is an integer $\ge 2$

*References:*
- [A167918](https://oeis.org/A167918)-/

namespace OeisA167918

/-- $P(i)$ is the $i$-th prime, 1-indexed. -/
noncomputable def P (i : ℕ) : ℕ := Nat.nth Nat.Prime (i - 1)

/-- $S(i) = p_i + p_{i+1}$. -/
noncomputable def S (i : ℕ) : ℕ := P i + P (i + 1)

/-- Smallest index $k > n$ such that $S(n) \mid S(k)$. -/
noncomputable def a (n : ℕ) : ℕ :=
  if n = 0 then 0
  else sInf { k : ℕ | k > n ∧ S n ∣ S k }

/-- Value of the sequence `a` at 0. -/
@[category test, AMS 11]
theorem a_0 : a 0 = 0 := by rfl

@[category API, AMS 11]
lemma P_1 : P 1 = 2 := Nat.nth_prime_zero_eq_two

@[category API, AMS 11]
lemma P_2 : P 2 = 3 := Nat.nth_prime_one_eq_three

@[category API, AMS 11]
lemma P_3 : P 3 = 5 := Nat.nth_prime_two_eq_five

@[category API, AMS 11]
lemma P_4 : P 4 = 7 := Nat.nth_prime_three_eq_seven

@[category API, AMS 11]
lemma P_5 : P 5 = 11 := Nat.nth_prime_four_eq_eleven

@[category API, AMS 11]
lemma P_6 : P 6 = 13 := by
  dsimp [P]
  have h : (13 : ℕ).Prime := by decide
  exact Nat.nth_count h

@[category API, AMS 11]
lemma P_7 : P 7 = 17 := by
  dsimp [P]
  have h : (17 : ℕ).Prime := by decide
  exact Nat.nth_count h

@[category API, AMS 11]
lemma P_8 : P 8 = 19 := by
  dsimp [P]
  have h : (19 : ℕ).Prime := by decide
  exact Nat.nth_count h

@[category API, AMS 11]
lemma S_1 : S 1 = 5 := by dsimp [S]; rw [P_1, P_2]

@[category API, AMS 11]
lemma S_2 : S 2 = 8 := by dsimp [S]; rw [P_2, P_3]

@[category API, AMS 11]
lemma S_3 : S 3 = 12 := by dsimp [S]; rw [P_3, P_4]

@[category API, AMS 11]
lemma S_4 : S 4 = 18 := by dsimp [S]; rw [P_4, P_5]

@[category API, AMS 11]
lemma S_5 : S 5 = 24 := by dsimp [S]; rw [P_5, P_6]

@[category API, AMS 11]
lemma S_6 : S 6 = 30 := by dsimp [S]; rw [P_6, P_7]

@[category API, AMS 11]
lemma S_7 : S 7 = 36 := by dsimp [S]; rw [P_7, P_8]

/-- Value of the sequence `a` at 1. -/
@[category test, AMS 11]
theorem a_1 : a 1 = 6 := by
  have h_least : IsLeast { k : ℕ | k > 1 ∧ S 1 ∣ S k } 6 := by
    constructor
    · simp only [Set.mem_ofPred_eq]
      refine ⟨by decide, ?_⟩
      rw [S_1, S_6]
      decide
    · intro k hk
      simp only [Set.mem_ofPred_eq] at hk
      by_contra! h
      have hk_gt := hk.1
      interval_cases k
      · have hdiv := hk.2
        rw [S_1, S_2] at hdiv
        revert hdiv; decide
      · have hdiv := hk.2
        rw [S_1, S_3] at hdiv
        revert hdiv; decide
      · have hdiv := hk.2
        rw [S_1, S_4] at hdiv
        revert hdiv; decide
      · have hdiv := hk.2
        rw [S_1, S_5] at hdiv
        revert hdiv; decide
  have ha1 : a 1 = sInf { k : ℕ | k > 1 ∧ S 1 ∣ S k } := by
    unfold a; split <;> [omega; rfl]
  rw [ha1, h_least.csInf_eq]

/-- Value of the sequence `a` at 2. -/
@[category test, AMS 11]
theorem a_2 : a 2 = 5 := by
  have h_least : IsLeast { k : ℕ | k > 2 ∧ S 2 ∣ S k } 5 := by
    constructor
    · simp only [Set.mem_ofPred_eq]
      refine ⟨by decide, ?_⟩
      rw [S_2, S_5]
      decide
    · intro k hk
      simp only [Set.mem_ofPred_eq] at hk
      by_contra! h
      have hk_gt := hk.1
      interval_cases k
      · have hdiv := hk.2
        rw [S_2, S_3] at hdiv
        revert hdiv; decide
      · have hdiv := hk.2
        rw [S_2, S_4] at hdiv
        revert hdiv; decide
  have ha2 : a 2 = sInf { k : ℕ | k > 2 ∧ S 2 ∣ S k } := by
    unfold a; split <;> [omega; rfl]
  rw [ha2, h_least.csInf_eq]

/-- Value of the sequence `a` at 3. -/
@[category test, AMS 11]
theorem a_3 : a 3 = 5 := by
  have h_least : IsLeast { k : ℕ | k > 3 ∧ S 3 ∣ S k } 5 := by
    constructor
    · simp only [Set.mem_ofPred_eq]
      refine ⟨by decide, ?_⟩
      rw [S_3, S_5]
      decide
    · intro k hk
      simp only [Set.mem_ofPred_eq] at hk
      by_contra! h
      have hk_gt := hk.1
      interval_cases k
      · have hdiv := hk.2
        rw [S_3, S_4] at hdiv
        revert hdiv; decide
  have ha3 : a 3 = sInf { k : ℕ | k > 3 ∧ S 3 ∣ S k } := by
    unfold a; split <;> [omega; rfl]
  rw [ha3, h_least.csInf_eq]

/-- Value of the sequence `a` at 4. -/
@[category test, AMS 11]
theorem a_4 : a 4 = 7 := by
  have h_least : IsLeast { k : ℕ | k > 4 ∧ S 4 ∣ S k } 7 := by
    constructor
    · simp only [Set.mem_ofPred_eq]
      refine ⟨by decide, ?_⟩
      rw [S_4, S_7]
      decide
    · intro k hk
      simp only [Set.mem_ofPred_eq] at hk
      by_contra! h
      have hk_gt := hk.1
      interval_cases k
      · have hdiv := hk.2
        rw [S_4, S_5] at hdiv
        revert hdiv; decide
      · have hdiv := hk.2
        rw [S_4, S_6] at hdiv
        revert hdiv; decide
  have ha4 : a 4 = sInf { k : ℕ | k > 4 ∧ S 4 ∣ S k } := by
    unfold a; split <;> [omega; rfl]
  rw [ha4, h_least.csInf_eq]

/--
Conjecture: $f(n, k) = 2$ for infinitely many cases, where $k = a(n)$.

We assume $a(n) \ne 0$ (i.e., that a suitable $k > n$ always exists), as `sInf` evaluates to $0$
on an empty set.
-/
@[category research open, AMS 11]
theorem conjecture1 (M : ℕ) (ha : ∀ n > 0, a n ≠ 0) :
    ∃ n : ℕ, n ≥ M ∧ n > 0 ∧ S (a n) = 2 * S n := by
  sorry

/--
Open problem: Whether the ratio $f(n, k)$ is bounded, where $k = a(n)$.

We assume $a(n) \ne 0$ (i.e., that a suitable $k > n$ always exists), as `sInf` evaluates to $0$
on an empty set.
-/
@[category research open, AMS 11]
theorem conjecture2 (ha : ∀ n > 0, a n ≠ 0) :
    ∃ C : ℕ, ∀ n : ℕ, n > 0 → S (a n) / S n ≤ C := by
  sorry


end OeisA167918

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