FormalConjectures.OEIS.«100475»

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import FormalConjecturesUtil

Prime-th recurrence with reversal at each step

$$a(n) = \operatorname{reversal}(p_{a(n-1)})$$ with $a(0)=1$, where $p_k$ is the $k$-th prime number.

References:

A100475

namespace OeisA100475open Nat List

Reverses the base 10 digits of a natural number.

def reverseDigits (k : ℕ) : ℕ :=
  Nat.ofDigits 10 (Nat.digits 10 k |>.reverse)

The primary defining sequence a. a n is the Prime-th recurrence with reversal at each step. $a(n) = \operatorname{reversal}(p_{a(n-1)})$ with $a(0)=1$, where $p_k$ is the $k$-th prime number (i.e., $p_1=2, p_2=3, \dots$).

noncomputable def a : ℕ → ℕ
  | 0 => 1
  | n + 1 =>
    let k := a n
    if k = 0 then 0
    else reverseDigits (Nat.nth Nat.Prime (k - 1))

@[category API, AMS 11]
lemma a_succ (n : ℕ) :
    a (n + 1) = if a n = 0 then 0 else reverseDigits (Nat.nth Nat.Prime (a n - 1)) := byn:ℕ⊢ a (n + 1) = if a n = 0 then 0 else reverseDigits (nth Nat.Prime (a n - 1))
  rflAll goals completed! 🐙

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

@[category test, AMS 11]
theorem a_0 : a 0 = 1 := by⊢ a 0 = 1 rflAll goals completed! 🐙



@[category test, AMS 11]
theorem a_1 : a 1 = 2 := by⊢ a 1 = 2
  rw [a_succ⊢ (if a 0 = 0 then 0 else reverseDigits (nth Nat.Prime (a 0 - 1))) = 2⊢ (if a 0 = 0 then 0 else reverseDigits (nth Nat.Prime (a 0 - 1))) = 2]⊢ (if a 0 = 0 then 0 else reverseDigits (nth Nat.Prime (a 0 - 1))) = 2
  rw [a_0⊢ (if 1 = 0 then 0 else reverseDigits (nth Nat.Prime (1 - 1))) = 2⊢ (if 1 = 0 then 0 else reverseDigits (nth Nat.Prime (1 - 1))) = 2]⊢ (if 1 = 0 then 0 else reverseDigits (nth Nat.Prime (1 - 1))) = 2
  simp [Nat.nth_prime_zero_eq_two, reverseDigits]All goals completed! 🐙



@[category test, AMS 11]
theorem a_2 : a 2 = 3 := by⊢ a 2 = 3
  rw [a_succ⊢ (if a 1 = 0 then 0 else reverseDigits (nth Nat.Prime (a 1 - 1))) = 3⊢ (if a 1 = 0 then 0 else reverseDigits (nth Nat.Prime (a 1 - 1))) = 3]⊢ (if a 1 = 0 then 0 else reverseDigits (nth Nat.Prime (a 1 - 1))) = 3
  rw [a_1⊢ (if 2 = 0 then 0 else reverseDigits (nth Nat.Prime (2 - 1))) = 3⊢ (if 2 = 0 then 0 else reverseDigits (nth Nat.Prime (2 - 1))) = 3]⊢ (if 2 = 0 then 0 else reverseDigits (nth Nat.Prime (2 - 1))) = 3
  simp [Nat.nth_prime_one_eq_three, reverseDigits]All goals completed! 🐙



@[category test, AMS 11]
theorem a_3 : a 3 = 5 := by⊢ a 3 = 5
  rw [a_succ⊢ (if a 2 = 0 then 0 else reverseDigits (nth Nat.Prime (a 2 - 1))) = 5⊢ (if a 2 = 0 then 0 else reverseDigits (nth Nat.Prime (a 2 - 1))) = 5]⊢ (if a 2 = 0 then 0 else reverseDigits (nth Nat.Prime (a 2 - 1))) = 5
  rw [a_2⊢ (if 3 = 0 then 0 else reverseDigits (nth Nat.Prime (3 - 1))) = 5⊢ (if 3 = 0 then 0 else reverseDigits (nth Nat.Prime (3 - 1))) = 5]⊢ (if 3 = 0 then 0 else reverseDigits (nth Nat.Prime (3 - 1))) = 5
  simp [Nat.nth_prime_two_eq_five, reverseDigits]All goals completed! 🐙



@[category test, AMS 11]
theorem a_4 : a 4 = 11 := by⊢ a 4 = 11
  rw [a_succ⊢ (if a 3 = 0 then 0 else reverseDigits (nth Nat.Prime (a 3 - 1))) = 11⊢ (if a 3 = 0 then 0 else reverseDigits (nth Nat.Prime (a 3 - 1))) = 11]⊢ (if a 3 = 0 then 0 else reverseDigits (nth Nat.Prime (a 3 - 1))) = 11
  rw [a_3⊢ (if 5 = 0 then 0 else reverseDigits (nth Nat.Prime (5 - 1))) = 11⊢ (if 5 = 0 then 0 else reverseDigits (nth Nat.Prime (5 - 1))) = 11]⊢ (if 5 = 0 then 0 else reverseDigits (nth Nat.Prime (5 - 1))) = 11
  simp [Nat.nth_prime_four_eq_eleven, reverseDigits]⊢ ofDigits 10 [1, 1] = 11
  rflAll goals completed! 🐙

Definition of the generalized sequence starting at x.

noncomputable def aStartAt (x : ℕ) : ℕ → ℕ
  | 0 => x
  | n + 1 =>
    let k := aStartAt x n
    if k = 0 then 0
    else reverseDigits (Nat.nth Nat.Prime (k - 1))

A sequence $f : \mathbb{N} \to \mathbb{N}$ is ultimately periodic if there exist $N, P \in \mathbb{N}$, with $P>0$, such that for all $n \ge N$, $f(n+P) = f(n)$.

def IsUltimatelyPeriodic (f : ℕ → ℕ) : Prop :=
  ∃ N P, P > 0 ∧ ∀ n, n ≥ N → f (n + P) = f n

Starting at other than $a(n) = 1$, does this sequence ever go into a loop?

@[category research open, AMS 11]
theorem conjecture (x : ℕ) (h : x ≠ 1) :
    answer(sorry) = IsUltimatelyPeriodic (aStartAt x) := byx:ℕh:x ≠ 1⊢ sorry = IsUltimatelyPeriodic (aStartAt x)
  sorryAll goals completed! 🐙

end OeisA100475