FormalConjectures.OEIS.«114831»

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import FormalConjecturesUtil

Each term is previous term plus floor of harmonic mean of two previous terms.

$a(1) = 1, a(2) = 2$ and $a(n) = a(n-1) + \lfloor \frac{2 a(n-1) a(n-2)}{a(n-1) + a(n-2)} \rfloor$ for $n \ge 3$.

References:

A114831

namespace OeisA114831open Filter Real Topology

The primary defining sequence a. Each term is previous term plus floor of harmonic mean of two previous terms.

noncomputable def a : ℕ → ℕ
| 0 => 0 -- Dummy value, sequence starts at index 1
| 1 => 1
| 2 => 2
| n + 3 =>
  let an1 : ℕ := a (n + 2)
  let an2 : ℕ := a (n + 1)
  let num : ℚ := (2 * an1 * an2 : ℕ).cast
  let den : ℚ := (an1 + an2 : ℕ).cast
  let harmonicTermFloor : ℕ := Int.toNat (num / den).floor
  an1 + harmonicTermFloor

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



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

@[category test, AMS 11]
theorem a_3 : a 3 = 3 := by⊢ a 3 = 3
  norm_num [a, Rat.floor, Int.toNat]All goals completed! 🐙

@[category test, AMS 11]
theorem a_4 : a 4 = 5 := by⊢ a 4 = 5
  norm_num [a, Rat.floor, Int.toNat]All goals completed! 🐙

Conjecture based on OEIS A114831: What is this sequence, asymptotically? If the limit exists, the ratio of consecutive terms must tend to $\sqrt{3}$: $$ \lim_{n \to \infty} \frac{a(n+1)}{a(n)} = \sqrt{3}. $$ That's because $a(n)$ is positive, monotonically increasing ($a(n) > a(n-1)$) and $a(n+2) \geq a(n+1) + a(n)$. So $a(n)$ grows exponentially, at least as fast as the Fibonnaci numbers. Assuming $\frac{a(n+1)}{a(n)}$ tend to a limit L, solving for L in the definition of $a(n)$ gives $L=\sqrt{3}$.

@[category research solved, AMS 11,
  formal_proof using lean4 at
    "https://github.com/KitaKen1/oeis-a114831-asymptotic/blob/54cdeeed2ef5838e3aa61a3a228e6867802d20df/lean/OeisA114831FC.lean#L252-L285"]
theorem conjecture3 :
    Tendsto (fun n ↦ (a (n + 1) : ℝ) / (a n : ℝ)) atTop (nhds (Real.sqrt 3)) := by⊢ Tendsto (fun n ↦ ↑(a (n + 1)) / ↑(a n)) atTop (𝓝 √3)
  sorryAll goals completed! 🐙

end OeisA114831