FormalConjectures.OEIS.«76495»

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import FormalConjecturesUtil

Smallest $x$ such that $\sigma(x) \bmod x = n$

The sequence $a(n)$ is the smallest positive integer $x$ such that $\sigma_1(x) \bmod x = n$, or $0$ if no such $x$ exists.

References:

A076495

namespace OeisA76495open ArithmeticFunctionopen Classical in

Smallest positive integer $x$ such that $\sigma_1(x) \bmod x = n$, or $0$ if no such $x$ exists.

noncomputable def a (n : ℕ) : ℕ :=
  if h : ∃ x, 0 < x ∧ (sigma 1 x : ℕ) % x = n then
    Nat.find h
  else
    0

Value of the sequence a at 1.


@[category test, AMS 11]
theorem a_1 : a 1 = 2 := by⊢ a 1 = 2
  classical
  dsimp [a]⊢ (if h : ∃ x, 0 < x ∧ (sigma 1) x % x = 1 then Nat.find h else 0) = 2
  split_ifs with hposh:∃ x, 0 < x ∧ (sigma 1) x % x = 1⊢ Nat.find h = 2negh:¬∃ x, 0 < x ∧ (sigma 1) x % x = 1⊢ False
  ·posh:∃ x, 0 < x ∧ (sigma 1) x % x = 1⊢ Nat.find h = 2 rw [Nat.find_eq_iffposh:∃ x, 0 < x ∧ (sigma 1) x % x = 1⊢ (0 < 2 ∧ (sigma 1) 2 % 2 = 1) ∧ ∀ n < 2, ¬(0 < n ∧ (sigma 1) n % n = 1)posh:∃ x, 0 < x ∧ (sigma 1) x % x = 1⊢ (0 < 2 ∧ (sigma 1) 2 % 2 = 1) ∧ ∀ n < 2, ¬(0 < n ∧ (sigma 1) n % n = 1)]posh:∃ x, 0 < x ∧ (sigma 1) x % x = 1⊢ (0 < 2 ∧ (sigma 1) 2 % 2 = 1) ∧ ∀ n < 2, ¬(0 < n ∧ (sigma 1) n % n = 1)
    decide +nativeAll goals completed! 🐙
  ·negh:¬∃ x, 0 < x ∧ (sigma 1) x % x = 1⊢ False exact (h ⟨2, byh:¬∃ x, 0 < x ∧ (sigma 1) x % x = 1⊢ 0 < 2 ∧ (sigma 1) 2 % 2 = 1 decide +nativeAll goals completed! 🐙⟩).elim

Value of the sequence a at 2.


@[category test, AMS 11]
theorem a_2 : a 2 = 20 := by⊢ a 2 = 20
  classical
  dsimp [a]⊢ (if h : ∃ x, 0 < x ∧ (sigma 1) x % x = 2 then Nat.find h else 0) = 20
  split_ifs with hposh:∃ x, 0 < x ∧ (sigma 1) x % x = 2⊢ Nat.find h = 20negh:¬∃ x, 0 < x ∧ (sigma 1) x % x = 2⊢ False
  ·posh:∃ x, 0 < x ∧ (sigma 1) x % x = 2⊢ Nat.find h = 20 rw [Nat.find_eq_iffposh:∃ x, 0 < x ∧ (sigma 1) x % x = 2⊢ (0 < 20 ∧ (sigma 1) 20 % 20 = 2) ∧ ∀ n < 20, ¬(0 < n ∧ (sigma 1) n % n = 2)posh:∃ x, 0 < x ∧ (sigma 1) x % x = 2⊢ (0 < 20 ∧ (sigma 1) 20 % 20 = 2) ∧ ∀ n < 20, ¬(0 < n ∧ (sigma 1) n % n = 2)]posh:∃ x, 0 < x ∧ (sigma 1) x % x = 2⊢ (0 < 20 ∧ (sigma 1) 20 % 20 = 2) ∧ ∀ n < 20, ¬(0 < n ∧ (sigma 1) n % n = 2)
    decide +nativeAll goals completed! 🐙
  ·negh:¬∃ x, 0 < x ∧ (sigma 1) x % x = 2⊢ False exact (h ⟨20, byh:¬∃ x, 0 < x ∧ (sigma 1) x % x = 2⊢ 0 < 20 ∧ (sigma 1) 20 % 20 = 2 decide +nativeAll goals completed! 🐙⟩).elim

Value of the sequence a at 3.


@[category test, AMS 11]
theorem a_3 : a 3 = 4 := by⊢ a 3 = 4
  classical
  dsimp [a]⊢ (if h : ∃ x, 0 < x ∧ (sigma 1) x % x = 3 then Nat.find h else 0) = 4
  split_ifs with hposh:∃ x, 0 < x ∧ (sigma 1) x % x = 3⊢ Nat.find h = 4negh:¬∃ x, 0 < x ∧ (sigma 1) x % x = 3⊢ False
  ·posh:∃ x, 0 < x ∧ (sigma 1) x % x = 3⊢ Nat.find h = 4 rw [Nat.find_eq_iffposh:∃ x, 0 < x ∧ (sigma 1) x % x = 3⊢ (0 < 4 ∧ (sigma 1) 4 % 4 = 3) ∧ ∀ n < 4, ¬(0 < n ∧ (sigma 1) n % n = 3)posh:∃ x, 0 < x ∧ (sigma 1) x % x = 3⊢ (0 < 4 ∧ (sigma 1) 4 % 4 = 3) ∧ ∀ n < 4, ¬(0 < n ∧ (sigma 1) n % n = 3)]posh:∃ x, 0 < x ∧ (sigma 1) x % x = 3⊢ (0 < 4 ∧ (sigma 1) 4 % 4 = 3) ∧ ∀ n < 4, ¬(0 < n ∧ (sigma 1) n % n = 3)
    decide +nativeAll goals completed! 🐙
  ·negh:¬∃ x, 0 < x ∧ (sigma 1) x % x = 3⊢ False exact (h ⟨4, byh:¬∃ x, 0 < x ∧ (sigma 1) x % x = 3⊢ 0 < 4 ∧ (sigma 1) 4 % 4 = 3 decide +nativeAll goals completed! 🐙⟩).elim

Value of the sequence a at 4.


@[category test, AMS 11]
theorem a_4 : a 4 = 9 := by⊢ a 4 = 9
  classical
  dsimp [a]⊢ (if h : ∃ x, 0 < x ∧ (sigma 1) x % x = 4 then Nat.find h else 0) = 9
  split_ifs with hposh:∃ x, 0 < x ∧ (sigma 1) x % x = 4⊢ Nat.find h = 9negh:¬∃ x, 0 < x ∧ (sigma 1) x % x = 4⊢ False
  ·posh:∃ x, 0 < x ∧ (sigma 1) x % x = 4⊢ Nat.find h = 9 rw [Nat.find_eq_iffposh:∃ x, 0 < x ∧ (sigma 1) x % x = 4⊢ (0 < 9 ∧ (sigma 1) 9 % 9 = 4) ∧ ∀ n < 9, ¬(0 < n ∧ (sigma 1) n % n = 4)posh:∃ x, 0 < x ∧ (sigma 1) x % x = 4⊢ (0 < 9 ∧ (sigma 1) 9 % 9 = 4) ∧ ∀ n < 9, ¬(0 < n ∧ (sigma 1) n % n = 4)]posh:∃ x, 0 < x ∧ (sigma 1) x % x = 4⊢ (0 < 9 ∧ (sigma 1) 9 % 9 = 4) ∧ ∀ n < 9, ¬(0 < n ∧ (sigma 1) n % n = 4)
    decide +nativeAll goals completed! 🐙
  ·negh:¬∃ x, 0 < x ∧ (sigma 1) x % x = 4⊢ False exact (h ⟨9, byh:¬∃ x, 0 < x ∧ (sigma 1) x % x = 4⊢ 0 < 9 ∧ (sigma 1) 9 % 9 = 4 decide +nativeAll goals completed! 🐙⟩).elim

Value of the sequence a at 6.


@[category test, AMS 11]
theorem a_6 : a 6 = 25 := by⊢ a 6 = 25
  classical
  dsimp [a]⊢ (if h : ∃ x, 0 < x ∧ (sigma 1) x % x = 6 then Nat.find h else 0) = 25
  split_ifs with hposh:∃ x, 0 < x ∧ (sigma 1) x % x = 6⊢ Nat.find h = 25negh:¬∃ x, 0 < x ∧ (sigma 1) x % x = 6⊢ False
  ·posh:∃ x, 0 < x ∧ (sigma 1) x % x = 6⊢ Nat.find h = 25 rw [Nat.find_eq_iffposh:∃ x, 0 < x ∧ (sigma 1) x % x = 6⊢ (0 < 25 ∧ (sigma 1) 25 % 25 = 6) ∧ ∀ n < 25, ¬(0 < n ∧ (sigma 1) n % n = 6)posh:∃ x, 0 < x ∧ (sigma 1) x % x = 6⊢ (0 < 25 ∧ (sigma 1) 25 % 25 = 6) ∧ ∀ n < 25, ¬(0 < n ∧ (sigma 1) n % n = 6)]posh:∃ x, 0 < x ∧ (sigma 1) x % x = 6⊢ (0 < 25 ∧ (sigma 1) 25 % 25 = 6) ∧ ∀ n < 25, ¬(0 < n ∧ (sigma 1) n % n = 6)
    decide +nativeAll goals completed! 🐙
  ·negh:¬∃ x, 0 < x ∧ (sigma 1) x % x = 6⊢ False exact (h ⟨25, byh:¬∃ x, 0 < x ∧ (sigma 1) x % x = 6⊢ 0 < 25 ∧ (sigma 1) 25 % 25 = 6 decide +nativeAll goals completed! 🐙⟩).elim

At present, the 0 entry for $n = 5$ is only a conjecture. That is, it is conjectured that there is no positive integer $x$ such that $\sigma_1(x) \bmod x = 5$.

@[category research open, AMS 11]
theorem conjecture : a 5 = 0 := by⊢ a 5 = 0
  sorryAll goals completed! 🐙

end OeisA76495