FormalConjectures.OEIS.«111114»

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import FormalConjecturesUtil

Integer part of $\mathrm{prime}(n)/\pi(n)$

Here $\mathrm{prime}(n)$ is the $n$-th prime number, and $\pi(n)$ is the prime-counting function.

References:

A111114

namespace OeisA111114open Nat

a n is the integer part of $\mathrm{prime}(n)/\pi(n)$. Here $\mathrm{prime}(n)$ is the $n$-th prime number, and $\pi(n)$ is the prime-counting function. The sequence is defined for $n \ge 2$.

noncomputable def a (n : ℕ) : ℕ :=
  (Nat.nth Nat.Prime (n - 1)) / (Nat.primeCounting n)



@[category test, AMS 11]
theorem a_2 : a 2 = 3 := by⊢ a 2 = 3
  have : Nat.nth Nat.Prime 1 = 3 := Nat.nth_prime_one_eq_threethis:nth Nat.Prime 1 = 3⊢ a 2 = 3
  unfold athis:nth Nat.Prime 1 = 3⊢ nth Nat.Prime (2 - 1) / primeCounting 2 = 3
  rw [thisthis:nth Nat.Prime 1 = 3⊢ 3 / primeCounting 2 = 3this:nth Nat.Prime 1 = 3⊢ 3 / primeCounting 2 = 3]this:nth Nat.Prime 1 = 3⊢ 3 / primeCounting 2 = 3
  decideAll goals completed! 🐙



@[category test, AMS 11]
theorem a_3 : a 3 = 2 := by⊢ a 3 = 2
  have : Nat.nth Nat.Prime 2 = 5 := Nat.nth_prime_two_eq_fivethis:nth Nat.Prime 2 = 5⊢ a 3 = 2
  unfold athis:nth Nat.Prime 2 = 5⊢ nth Nat.Prime (3 - 1) / primeCounting 3 = 2
  rw [thisthis:nth Nat.Prime 2 = 5⊢ 5 / primeCounting 3 = 2this:nth Nat.Prime 2 = 5⊢ 5 / primeCounting 3 = 2]this:nth Nat.Prime 2 = 5⊢ 5 / primeCounting 3 = 2
  decideAll goals completed! 🐙



@[category test, AMS 11]
theorem a_4 : a 4 = 3 := by⊢ a 4 = 3
  have : Nat.nth Nat.Prime 3 = 7 := Nat.nth_prime_three_eq_seventhis:nth Nat.Prime 3 = 7⊢ a 4 = 3
  unfold athis:nth Nat.Prime 3 = 7⊢ nth Nat.Prime (4 - 1) / primeCounting 4 = 3
  rw [thisthis:nth Nat.Prime 3 = 7⊢ 7 / primeCounting 4 = 3this:nth Nat.Prime 3 = 7⊢ 7 / primeCounting 4 = 3]this:nth Nat.Prime 3 = 7⊢ 7 / primeCounting 4 = 3
  decideAll goals completed! 🐙



@[category test, AMS 11]
theorem a_5 : a 5 = 3 := by⊢ a 5 = 3
  have : Nat.nth Nat.Prime 4 = 11 := Nat.nth_prime_four_eq_eleventhis:nth Nat.Prime 4 = 11⊢ a 5 = 3
  unfold athis:nth Nat.Prime 4 = 11⊢ nth Nat.Prime (5 - 1) / primeCounting 5 = 3
  rw [thisthis:nth Nat.Prime 4 = 11⊢ 11 / primeCounting 5 = 3this:nth Nat.Prime 4 = 11⊢ 11 / primeCounting 5 = 3]this:nth Nat.Prime 4 = 11⊢ 11 / primeCounting 5 = 3
  decideAll goals completed! 🐙

open Filter

Conjecture: As $n \rightarrow \infty$, there are infinitely many n's such that $a(n)$ is greater than $a(n+1)$.

@[category research open, AMS 11]
theorem conjecture : ∃ᶠ n in atTop, a n > a (n + 1) := by⊢ ∃ᶠ (n : ℕ) in atTop, a n > a (n + 1)
  sorryAll goals completed! 🐙

end OeisA111114