FormalConjectures.OEIS.«70823»

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import FormalConjecturesUtil

Recurrence $a(n+2) = |a(n+1)a(n) - a(n)a(n+1)|$ via concatenation

The sequence starts with $a(1) = 0, a(2) = 1$. For $n \ge 1$, $$a(n+2) = |\text{concat}(a(n+1), a(n)) - \text{concat}(a(n), a(n+1))|$$ where concatenation is in decimal representation.

References:

A070823

namespace OeisA70823

Number of digits of a natural number $n$ in base 10 (with 1 digit for 0).

def numDigitsBase10 (n : ℕ) : ℕ :=
  if n = 0 then 1 else Nat.log 10 n + 1

Concatenate $x$ followed by $y$ in base 10.

def concatenate (x y : ℕ) : ℕ :=
  x * 10 ^ (numDigitsBase10 y) + y

The sequence $a(1)=0, a(2)=1, a(n+2)=|\text{concat}(a(n+1),a(n))-\text{concat}(a(n),a(n+1))|$.

def a : ℕ → ℕ
  | 0 => 0
  | 1 => 0
  | 2 => 1
  | n + 3 =>
    let cat1 := concatenate (a (n + 2)) (a (n + 1))
    let cat2 := concatenate (a (n + 1)) (a (n + 2))
    ((cat1 : ℤ) - (cat2 : ℤ)).natAbs

Value of the sequence a at 1.

@[category test, AMS 11]
theorem a_1 : a 1 = 0 := by⊢ a 1 = 0
  decideAll goals completed! 🐙

Value of the sequence a at 2.

@[category test, AMS 11]
theorem a_2 : a 2 = 1 := by⊢ a 2 = 1
  decideAll goals completed! 🐙

Value of the sequence a at 3.

@[category test, AMS 11]
theorem a_3 : a 3 = 9 := by⊢ a 3 = 9
  decideAll goals completed! 🐙

Value of the sequence a at 4.

@[category test, AMS 11]
theorem a_4 : a 4 = 72 := by⊢ a 4 = 72
  decideAll goals completed! 🐙

Value of the sequence a at 5.

@[category test, AMS 11]
theorem a_5 : a 5 = 243 := by⊢ a 5 = 243
  decideAll goals completed! 🐙

$a(n) \equiv 0 \pmod 3$ if $n > 2$. Is $a(n)$ always of the form $2^j \cdot 3^k \cdot s$ where $s$ is a squarefree number?

Answer: False, $a(20)$ is divisible by $13^2$ but not by $13^3$.

@[category research solved, AMS 11,
  formal_proof using lean4 at
    "https://github.com/KitaKen1/oeis-a070823-counterexample/blob/51399770e734616c6463be034e41f7469991d752/lean/OeisA70823CounterexampleFC.lean#L72-L81"]
theorem conjecture :
    answer(False) ↔ ∀ n : ℕ, 2 < n →
      a n ≡ 0 [MOD 3] ∧
        ∃ j k s : ℕ, a n = 2 ^ j * 3 ^ k * s ∧ Squarefree s := by⊢ False ↔ ∀ (n : ℕ), 2 < n → OeisA70823.a n ≡ 0 [MOD 3] ∧ ∃ j k s, OeisA70823.a n = 2 ^ j * 3 ^ k * s ∧ Squarefree s
  sorryAll goals completed! 🐙

end OeisA70823