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import FormalConjecturesUtilFibonacci-Collatz sequence
Fibonacci-Collatz sequence: $a(1)=1, a(2)=2$; for $n > 2$, let $\mathrm{fib} = a(n-1) + a(n-2)$; if $\mathrm{fib}$ is odd then $a(n) = 3 \cdot \mathrm{fib} + 1$ else $a(n) = \mathrm{fib}/2$.
References:
namespace OeisA105801
The primary defining sequence a.
$a(n)$ is the Fibonacci-Collatz sequence: $a(1)=1, a(2)=2$; for $n > 2$,
let $\mathrm{fib} = a(n-1) + a(n-2)$;
if $\mathrm{fib}$ is odd then $a(n) = 3 \cdot \mathrm{fib} + 1$ else $a(n) = \mathrm{fib}/2$.
def a : ℕ → ℕ
| 0 => 0 -- The sequence is 1-indexed, a(0) is a conventional filler.
| 1 => 1
| 2 => 2
| n + 3 =>
let fib := a (n + 2) + a (n + 1)
if fib % 2 = 1 then 3 * fib + 1 else fib / 2Term theorems verifying the first few values of the sequence against the official OEIS b-file
@[category test, AMS 11]
theorem a_1 : a 1 = 1 := ⊢ a 1 = 1 All goals completed! 🐙@[category test, AMS 11]
theorem a_2 : a 2 = 2 := ⊢ a 2 = 2 All goals completed! 🐙@[category test, AMS 11]
theorem a_3 : a 3 = 10 := ⊢ a 3 = 10 All goals completed! 🐙@[category test, AMS 11]
theorem a_4 : a 4 = 6 := ⊢ a 4 = 6 All goals completed! 🐙@[category test, AMS 11]
theorem a_5 : a 5 = 8 := ⊢ a 5 = 8 All goals completed! 🐙Conjecture: for every $k > 0$ there is an index $m$ such that all the $a(n)$ with $n > m$ have the same residue $\bmod 3^k$.
@[category research solved, AMS 11,
formal_proof using lean4 at
"https://github.com/KitaKen1/oeis-a105801-lean/blob/68642a80db062bee7061437b60d31c3e7d626595/lean/OeisA105801FC.lean#L323-L351"]
theorem conjecture :
∀ k : ℕ, 0 < k → ∃ m : ℕ, ∀ n : ℕ, m < n → a n ≡ a (m + 1) [MOD (3^k)] := ⊢ ∀ (k : ℕ), 0 < k → ∃ m, ∀ (n : ℕ), m < n → OeisA105801.a n ≡ OeisA105801.a (m + 1) [MOD 3 ^ k]
All goals completed! 🐙end OeisA105801