FormalConjectures/OEIS/103425.lean

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Explanation legend Lean4 Domain Explanation Language (L4DEL)

L4DEL explanations are a separate layer. They do not change the Lean source. Intuition is explanatory prose; numbered items are a reading path; sources link the wording back to evidence.

L4DEL
Lean4 Domain Explanation Language

Credit: The upstream authors and contributors retain credit for the project and mathematics. lean4-explainer only renders their pinned source and adds the separately disclosed explanation layer.

Project sources: Pinned project documentation, Formal Conjectures paper, arXiv v1

10 declarations, 1 contain sorry, 0 declared as axioms

$a(n) = 3a(n-1) + a(n-2) - 3a(n-3)$

References:

Imports (1)
  1. def OeisA103425.a The primary defining sequence a line 36 source

    Source documentation

    The primary defining sequence a. $a(n)$ is defined by the recurrence relation $a(n) = 3 a(n-1) + a(n-2) - 3 a(n-3)$ with initial terms $a(0)=1, a(1)=3, a(2)=5$.

    no axioms what this means

    Statement uses: Nat

    2 instances, coercions and auxiliary constants left out.

    def a : ℕ → ℕ
      | 0 => 1
      | 1 => 3
      | 2 => 5
      | n + 3 => 3 * a (n + 2) + a (n + 1) - 3 * a n
    
    /-- Term theorems verifying the first few values of the sequence against the official OEIS b-file -/
    @[category test, AMS 11]

    Mentions: test, AMS

  2. theorem OeisA103425.a_0 Term theorems verifying the first few values of the sequence against the official OEIS b-file line 44 source

    Source documentation

    Term theorems verifying the first few values of the sequence against the official OEIS b-file

    no axioms what this means

    Statement uses: Eq, Nat, OeisA103425.a, OfNat.ofNat

    Proof uses: of_decide_eq_true, id, Bool, Decidable.decide, Bool.true, Eq.refl

    3 instances, coercions and auxiliary constants left out.

    Tactic trace: 1 steps

    Open to load the checked before-and-after goal states.

    theorem a_0 : a 0 = 1 := by decide
    …

    Mentions: OeisA103425.a

  3. theorem OeisA103425.a_1 line 47 source

    no axioms what this means

    Statement uses: Eq, Nat, OeisA103425.a, OfNat.ofNat

    Proof uses: of_decide_eq_true, id, Bool, Decidable.decide, Bool.true, Eq.refl

    3 instances, coercions and auxiliary constants left out.

    Tactic trace: 1 steps

    Open to load the checked before-and-after goal states.

  4. theorem OeisA103425.a_2 line 50 source

    no axioms what this means

    Statement uses: Eq, Nat, OeisA103425.a, OfNat.ofNat

    Proof uses: of_decide_eq_true, id, Bool, Decidable.decide, Bool.true, Eq.refl

    3 instances, coercions and auxiliary constants left out.

    Tactic trace: 1 steps

    Open to load the checked before-and-after goal states.

  5. theorem OeisA103425.a_3 line 53 source

    no axioms what this means

    Statement uses: Eq, Nat, OeisA103425.a, OfNat.ofNat

    Proof uses: of_decide_eq_true, id, Bool, Decidable.decide, Bool.true, Eq.refl

    3 instances, coercions and auxiliary constants left out.

    Tactic trace: 1 steps

    Open to load the checked before-and-after goal states.

  6. theorem OeisA103425.a_4 line 56 source

    no axioms what this means

    Statement uses: Eq, Nat, OeisA103425.a, OfNat.ofNat

    Proof uses: of_decide_eq_true, id, Bool, Decidable.decide, Bool.true, Eq.refl

    3 instances, coercions and auxiliary constants left out.

    Tactic trace: 1 steps

    Open to load the checked before-and-after goal states.

  7. def OeisA103425.IsWeightedTribonacci line 58 source

    no axioms what this means

    Statement uses: Int, Nat

    Proof uses: Eq, HAdd.hAdd, OfNat.ofNat, HMul.hMul

    6 instances, coercions and auxiliary constants left out.

  8. theorem OeisA103425.conjecture sorry line 73 source

    depends on sorry what this means

    Statement uses: Iff, True, Exists, Int, Nat, And, Eq, Nat.gcd, Int.gcd, Int.natAbs, OfNat.ofNat, OeisA103425.IsWeightedTribonacci, Not, Nat.Prime

    Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous

    2 instances, coercions and auxiliary constants left out.

    Tactic trace: 1 steps

    Open to load the checked before-and-after goal states.

  9. theorem OeisA103425.conjecture.variants.nonconstant Is there a non-constant $(a, b, c)$ weighted tribonacci sequence with $a, b, c$ relatively prime which is prime-free? line 87 source

    Source documentation

    Is there a non-constant $(a, b, c)$ weighted tribonacci sequence with $a, b, c$ relatively prime which is prime-free?

    Yes: take $(a, b, c) = (1, 1, -1)$ and $x(n) = 4(n + 1)$.

    standard axioms: Classical.choice, Quot.sound, propext what this means

    Statement uses: Iff, True, Exists, Int, Nat, And, Eq, Nat.gcd, Int.gcd, Int.natAbs, OfNat.ofNat, OeisA103425.IsWeightedTribonacci, Ne, Not, Nat.Prime

    Proof uses: Iff.intro, Exists.intro, Neg.neg, HMul.hMul, HAdd.hAdd, And.intro, Mathlib.Meta.NormNum.isNat_eq_true, Mathlib.Meta.NormNum.isNat_gcd, Mathlib.Meta.NormNum.isInt_gcd, Int.ofNat, Mathlib.Meta.NormNum.IsNat.to_isInt, Mathlib.Meta.NormNum.isNat_ofNat, Eq.refl, Mathlib.Meta.NormNum.int_gcd_helper, Int.natAbs_natCast, Nat.gcd_one_left, Mathlib.Meta.NormNum.isNat_natAbs_neg, Mathlib.Meta.NormNum.isInt_neg, Int.negOfNat, Eq.mpr and 63 more

    56 instances, coercions and auxiliary constants left out.

    Tactic trace: 14 steps

    Open to load the checked before-and-after goal states.

    theorem conjecture.variants.nonconstant : answer(True) ↔
        ∃ (a b c : ℤ) (x : ℕ → ℤ),
          Nat.gcd (Int.gcd a b) c.natAbs = 1 ∧
          IsWeightedTribonacci a b c x ∧
          (∃ m n, x m ≠ x n) ∧
          ∀ n, ¬ (x n).natAbs.Prime := by
    …

    Mentions: OeisA103425.a, c, OeisA103425.IsWeightedTribonacci

  10. theorem OeisA103425.conjecture.variants.nonconstant_distinct_coeffs Is there a non-constant $(a, b, c)$ weighted tribonacci sequence with $a, b, c$ relatively prime and pairwise distinct which is prime-free? line 109 source

    Source documentation

    Is there a non-constant $(a, b, c)$ weighted tribonacci sequence with $a, b, c$ relatively prime and pairwise distinct which is prime-free?

    Yes: take $(a, b, c) = (3, -3, 1)$ and $x(n) = (n + 2)^2$.

    standard axioms: Quot.sound, propext what this means

    Statement uses: Iff, True, Exists, Int, Nat, And, Eq, Nat.gcd, Int.gcd, Int.natAbs, OfNat.ofNat, Ne, OeisA103425.IsWeightedTribonacci, Not, Nat.Prime

    Proof uses: Iff.intro, Exists.intro, Neg.neg, HPow.hPow, HAdd.hAdd, And.intro, Mathlib.Meta.NormNum.isNat_eq_true, Mathlib.Meta.NormNum.isNat_gcd, Mathlib.Meta.NormNum.isInt_gcd, Int.ofNat, Int.negOfNat, Mathlib.Meta.NormNum.IsNat.to_isInt, Mathlib.Meta.NormNum.isNat_ofNat, Eq.refl, Mathlib.Meta.NormNum.isInt_neg, Mathlib.Meta.NormNum.int_gcd_helper, Int.natAbs_natCast, Int.natAbs_neg, Mathlib.Meta.NormNum.nat_gcd_helper_dvd_left, Mathlib.Meta.NormNum.isNat_natAbs_pos and 70 more

    56 instances, coercions and auxiliary constants left out.

    Tactic trace: 16 steps

    Open to load the checked before-and-after goal states.

    theorem conjecture.variants.nonconstant_distinct_coeffs : answer(True) ↔
        ∃ (a b c : ℤ) (x : ℕ → ℤ),
          Nat.gcd (Int.gcd a b) c.natAbs = 1 ∧
          a ≠ b ∧ a ≠ c ∧ b ≠ c ∧
          IsWeightedTribonacci a b c x ∧
          (∃ m n, x m ≠ x n) ∧
          ∀ n, ¬ (x n).natAbs.Prime := by
    …

    Mentions: OeisA103425.a, c, OeisA103425.IsWeightedTribonacci