FormalConjectures.OEIS.«105720»

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import FormalConjecturesUtil

Triangular matchstick numbers in the class of prime numbers

$a(n) = \sum_{k = n}^{2n} p_k$, where $p_k$ is the $k$-th prime.

References:

A105720

namespace OeisA105720open Nat Finset

The primary defining sequence a. $a(n)$ is the sum of primes from the $n$-th prime to the $2n$-th prime, with $p_1=2$. For $n=0$, we define $a(0) = 0$.

noncomputable def a (n : ℕ) : ℕ :=
  if n = 0 then 0
  else ∑ i ∈ Ico (n - 1) (2 * n), Nat.nth Nat.Prime i



@[category API, AMS 11]
lemma nth_prime_five : Nat.nth Nat.Prime 5 = 13 := by⊢ nth Nat.Prime 5 = 13
  have h1 : Nat.count Nat.Prime 13 = 5 := by decideh1:count Nat.Prime 13 = 5⊢ nth Nat.Prime 5 = 13h1:count Nat.Prime 13 = 5⊢ nth Nat.Prime 5 = 13
  have h2 : (13).Prime := by decideh1:count Nat.Prime 13 = 5h2:Nat.Prime 13⊢ nth Nat.Prime 5 = 13h1:count Nat.Prime 13 = 5h2:Nat.Prime 13⊢ nth Nat.Prime 5 = 13
  have h3 := Nat.nth_count h2h1:count Nat.Prime 13 = 5h2:Nat.Prime 13h3:nth Nat.Prime (count Nat.Prime 13) = 13⊢ nth Nat.Prime 5 = 13
  rwa [h1h1:count Nat.Prime 13 = 5h2:Nat.Prime 13h3:nth Nat.Prime 5 = 13⊢ nth Nat.Prime 5 = 13]h1:count Nat.Prime 13 = 5h2:Nat.Prime 13h3:nth Nat.Prime 5 = 13⊢ nth Nat.Prime 5 = 13 at h3



@[category API, AMS 11]
lemma nth_prime_six : Nat.nth Nat.Prime 6 = 17 := by⊢ nth Nat.Prime 6 = 17
  have h1 : Nat.count Nat.Prime 17 = 6 := by decideh1:count Nat.Prime 17 = 6⊢ nth Nat.Prime 6 = 17h1:count Nat.Prime 17 = 6⊢ nth Nat.Prime 6 = 17
  have h2 : (17).Prime := by decideh1:count Nat.Prime 17 = 6h2:Nat.Prime 17⊢ nth Nat.Prime 6 = 17h1:count Nat.Prime 17 = 6h2:Nat.Prime 17⊢ nth Nat.Prime 6 = 17
  have h3 := Nat.nth_count h2h1:count Nat.Prime 17 = 6h2:Nat.Prime 17h3:nth Nat.Prime (count Nat.Prime 17) = 17⊢ nth Nat.Prime 6 = 17
  rwa [h1h1:count Nat.Prime 17 = 6h2:Nat.Prime 17h3:nth Nat.Prime 6 = 17⊢ nth Nat.Prime 6 = 17]h1:count Nat.Prime 17 = 6h2:Nat.Prime 17h3:nth Nat.Prime 6 = 17⊢ nth Nat.Prime 6 = 17 at h3



@[category API, AMS 11]
lemma nth_prime_seven : Nat.nth Nat.Prime 7 = 19 := by⊢ nth Nat.Prime 7 = 19
  have h1 : Nat.count Nat.Prime 19 = 7 := by decideh1:count Nat.Prime 19 = 7⊢ nth Nat.Prime 7 = 19h1:count Nat.Prime 19 = 7⊢ nth Nat.Prime 7 = 19
  have h2 : (19).Prime := by decideh1:count Nat.Prime 19 = 7h2:Nat.Prime 19⊢ nth Nat.Prime 7 = 19h1:count Nat.Prime 19 = 7h2:Nat.Prime 19⊢ nth Nat.Prime 7 = 19
  have h3 := Nat.nth_count h2h1:count Nat.Prime 19 = 7h2:Nat.Prime 19h3:nth Nat.Prime (count Nat.Prime 19) = 19⊢ nth Nat.Prime 7 = 19
  rwa [h1h1:count Nat.Prime 19 = 7h2:Nat.Prime 19h3:nth Nat.Prime 7 = 19⊢ nth Nat.Prime 7 = 19]h1:count Nat.Prime 19 = 7h2:Nat.Prime 19h3:nth Nat.Prime 7 = 19⊢ nth Nat.Prime 7 = 19 at h3

macro "eval_a" : tactic => `(tactic| (
  repeat rw [Finset.sum_Ico_succ_top (by decide)]
  simp only [Finset.Ico_self, Finset.sum_empty, zero_add, add_zero]
  simp only [Nat.nth_prime_zero_eq_two, Nat.nth_prime_one_eq_three,
    Nat.nth_prime_two_eq_five, Nat.nth_prime_three_eq_seven,
    Nat.nth_prime_four_eq_eleven, nth_prime_five, nth_prime_six, nth_prime_seven]
  try rfl
))

Value of the sequence a at 0.


@[category test, AMS 11]
theorem a_0 : a 0 = 0 := by⊢ a 0 = 0
  rw [a,⊢ (if 0 = 0 then 0 else ∑ i ∈ Ico (0 - 1) (2 * 0), nth Nat.Prime i) = 0All goals completed! 🐙 if_pos rfl⊢ 0 = 0All goals completed! 🐙]All goals completed! 🐙

Value of the sequence a at 1.


@[category test, AMS 11]
theorem a_1 : a 1 = 5 := by⊢ a 1 = 5
  rw [a,⊢ (if 1 = 0 then 0 else ∑ i ∈ Ico (1 - 1) (2 * 1), nth Nat.Prime i) = 5⊢ ∑ i ∈ Ico (1 - 1) (2 * 1), nth Nat.Prime i = 5 if_neg (by⊢ ¬1 = 0⊢ ∑ i ∈ Ico (1 - 1) (2 * 1), nth Nat.Prime i = 5 decideAll goals completed! 🐙⊢ ∑ i ∈ Ico (1 - 1) (2 * 1), nth Nat.Prime i = 5)]⊢ ∑ i ∈ Ico (1 - 1) (2 * 1), nth Nat.Prime i = 5
  eval_aAll goals completed! 🐙

Value of the sequence a at 2.


@[category test, AMS 11]
theorem a_2 : a 2 = 15 := by⊢ a 2 = 15
  rw [a,⊢ (if 2 = 0 then 0 else ∑ i ∈ Ico (2 - 1) (2 * 2), nth Nat.Prime i) = 15⊢ ∑ i ∈ Ico (2 - 1) (2 * 2), nth Nat.Prime i = 15 if_neg (by⊢ ¬2 = 0⊢ ∑ i ∈ Ico (2 - 1) (2 * 2), nth Nat.Prime i = 15 decideAll goals completed! 🐙⊢ ∑ i ∈ Ico (2 - 1) (2 * 2), nth Nat.Prime i = 15)]⊢ ∑ i ∈ Ico (2 - 1) (2 * 2), nth Nat.Prime i = 15
  eval_aAll goals completed! 🐙

Value of the sequence a at 3.


@[category test, AMS 11]
theorem a_3 : a 3 = 36 := by⊢ a 3 = 36
  rw [a,⊢ (if 3 = 0 then 0 else ∑ i ∈ Ico (3 - 1) (2 * 3), nth Nat.Prime i) = 36⊢ ∑ i ∈ Ico (3 - 1) (2 * 3), nth Nat.Prime i = 36 if_neg (by⊢ ¬3 = 0⊢ ∑ i ∈ Ico (3 - 1) (2 * 3), nth Nat.Prime i = 36 decideAll goals completed! 🐙⊢ ∑ i ∈ Ico (3 - 1) (2 * 3), nth Nat.Prime i = 36)]⊢ ∑ i ∈ Ico (3 - 1) (2 * 3), nth Nat.Prime i = 36
  eval_aAll goals completed! 🐙

Value of the sequence a at 4.


@[category test, AMS 11]
theorem a_4 : a 4 = 67 := by⊢ a 4 = 67
  rw [a,⊢ (if 4 = 0 then 0 else ∑ i ∈ Ico (4 - 1) (2 * 4), nth Nat.Prime i) = 67⊢ ∑ i ∈ Ico (4 - 1) (2 * 4), nth Nat.Prime i = 67 if_neg (by⊢ ¬4 = 0⊢ ∑ i ∈ Ico (4 - 1) (2 * 4), nth Nat.Prime i = 67 decideAll goals completed! 🐙⊢ ∑ i ∈ Ico (4 - 1) (2 * 4), nth Nat.Prime i = 67)]⊢ ∑ i ∈ Ico (4 - 1) (2 * 4), nth Nat.Prime i = 67
  eval_aAll goals completed! 🐙

Terms are squares at only(?) three values of $n = 3, 6, 4072$: corresponding terms are 6^2, 13^2, and 15735^2.

@[category research open, AMS 11]
theorem conjecture :
    ∀ n : ℕ, 0 < n → (IsSquare (a n) ↔ (n = 3 ∨ n = 6 ∨ n = 4072)) := by⊢ ∀ (n : ℕ), 0 < n → (IsSquare (OeisA105720.a n) ↔ n = 3 ∨ n = 6 ∨ n = 4072)
  sorryAll goals completed! 🐙

end OeisA105720