FormalConjectures.OEIS.«78680»

/-
Copyright 2026 The Formal Conjectures Authors.

Licensed under the Apache License, Version 2.0 (the "License");
you may not use this file except in compliance with the License.
You may obtain a copy of the License at

    https://www.apache.org/licenses/LICENSE-2.0

Unless required by applicable law or agreed to in writing, software
distributed under the License is distributed on an "AS IS" BASIS,
WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
See the License for the specific language governing permissions and
limitations under the License.
-/

import FormalConjecturesUtil

Smallest $m > 0$ such that $n \cdot 2^m + 1$ is prime

The sequence $a(n)$ is the smallest positive integer $m$ such that $n \cdot 2^m + 1$ is prime, or $0$ if no such $m$ exists.

References:

A078680

namespace OeisA78680open Classical in

Smallest $m > 0$ such that $n \cdot 2^m + 1$ is prime, or 0 if no such $m$ exists.

noncomputable def a (n : ℕ) : ℕ :=
  if h : ∃ m, 0 < m ∧ (n * 2 ^ m + 1).Prime then
    Nat.find h
  else
    0

Value of the sequence a at 1.


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

Value of the sequence a at 2.


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

Value of the sequence a at 3.


@[category test, AMS 11]
theorem a_3 : a 3 = 1 := by⊢ a 3 = 1
  classical
  dsimp [a]⊢ (if h : ∃ m, 0 < m ∧ Nat.Prime (3 * 2 ^ m + 1) then Nat.find h else 0) = 1
  split_ifs with hposh:∃ m, 0 < m ∧ Nat.Prime (3 * 2 ^ m + 1)⊢ Nat.find h = 1negh:¬∃ m, 0 < m ∧ Nat.Prime (3 * 2 ^ m + 1)⊢ False
  ·posh:∃ m, 0 < m ∧ Nat.Prime (3 * 2 ^ m + 1)⊢ Nat.find h = 1 rw [Nat.find_eq_iffposh:∃ m, 0 < m ∧ Nat.Prime (3 * 2 ^ m + 1)⊢ (0 < 1 ∧ Nat.Prime (3 * 2 ^ 1 + 1)) ∧ ∀ n < 1, ¬(0 < n ∧ Nat.Prime (3 * 2 ^ n + 1))posh:∃ m, 0 < m ∧ Nat.Prime (3 * 2 ^ m + 1)⊢ (0 < 1 ∧ Nat.Prime (3 * 2 ^ 1 + 1)) ∧ ∀ n < 1, ¬(0 < n ∧ Nat.Prime (3 * 2 ^ n + 1))]posh:∃ m, 0 < m ∧ Nat.Prime (3 * 2 ^ m + 1)⊢ (0 < 1 ∧ Nat.Prime (3 * 2 ^ 1 + 1)) ∧ ∀ n < 1, ¬(0 < n ∧ Nat.Prime (3 * 2 ^ n + 1))
    decide +nativeAll goals completed! 🐙
  ·negh:¬∃ m, 0 < m ∧ Nat.Prime (3 * 2 ^ m + 1)⊢ False exact (h ⟨1, byh:¬∃ m, 0 < m ∧ Nat.Prime (3 * 2 ^ m + 1)⊢ 0 < 1 ∧ Nat.Prime (3 * 2 ^ 1 + 1) decide +nativeAll goals completed! 🐙⟩).elim

Value of the sequence a at 4.


@[category test, AMS 11]
theorem a_4 : a 4 = 2 := by⊢ a 4 = 2
  classical
  dsimp [a]⊢ (if h : ∃ m, 0 < m ∧ Nat.Prime (4 * 2 ^ m + 1) then Nat.find h else 0) = 2
  split_ifs with hposh:∃ m, 0 < m ∧ Nat.Prime (4 * 2 ^ m + 1)⊢ Nat.find h = 2negh:¬∃ m, 0 < m ∧ Nat.Prime (4 * 2 ^ m + 1)⊢ False
  ·posh:∃ m, 0 < m ∧ Nat.Prime (4 * 2 ^ m + 1)⊢ Nat.find h = 2 rw [Nat.find_eq_iffposh:∃ m, 0 < m ∧ Nat.Prime (4 * 2 ^ m + 1)⊢ (0 < 2 ∧ Nat.Prime (4 * 2 ^ 2 + 1)) ∧ ∀ n < 2, ¬(0 < n ∧ Nat.Prime (4 * 2 ^ n + 1))posh:∃ m, 0 < m ∧ Nat.Prime (4 * 2 ^ m + 1)⊢ (0 < 2 ∧ Nat.Prime (4 * 2 ^ 2 + 1)) ∧ ∀ n < 2, ¬(0 < n ∧ Nat.Prime (4 * 2 ^ n + 1))]posh:∃ m, 0 < m ∧ Nat.Prime (4 * 2 ^ m + 1)⊢ (0 < 2 ∧ Nat.Prime (4 * 2 ^ 2 + 1)) ∧ ∀ n < 2, ¬(0 < n ∧ Nat.Prime (4 * 2 ^ n + 1))
    decide +nativeAll goals completed! 🐙
  ·negh:¬∃ m, 0 < m ∧ Nat.Prime (4 * 2 ^ m + 1)⊢ False exact (h ⟨2, byh:¬∃ m, 0 < m ∧ Nat.Prime (4 * 2 ^ m + 1)⊢ 0 < 2 ∧ Nat.Prime (4 * 2 ^ 2 + 1) decide +nativeAll goals completed! 🐙⟩).elim

Value of the sequence a at 5.


@[category test, AMS 11]
theorem a_5 : a 5 = 1 := by⊢ a 5 = 1
  classical
  dsimp [a]⊢ (if h : ∃ m, 0 < m ∧ Nat.Prime (5 * 2 ^ m + 1) then Nat.find h else 0) = 1
  split_ifs with hposh:∃ m, 0 < m ∧ Nat.Prime (5 * 2 ^ m + 1)⊢ Nat.find h = 1negh:¬∃ m, 0 < m ∧ Nat.Prime (5 * 2 ^ m + 1)⊢ False
  ·posh:∃ m, 0 < m ∧ Nat.Prime (5 * 2 ^ m + 1)⊢ Nat.find h = 1 rw [Nat.find_eq_iffposh:∃ m, 0 < m ∧ Nat.Prime (5 * 2 ^ m + 1)⊢ (0 < 1 ∧ Nat.Prime (5 * 2 ^ 1 + 1)) ∧ ∀ n < 1, ¬(0 < n ∧ Nat.Prime (5 * 2 ^ n + 1))posh:∃ m, 0 < m ∧ Nat.Prime (5 * 2 ^ m + 1)⊢ (0 < 1 ∧ Nat.Prime (5 * 2 ^ 1 + 1)) ∧ ∀ n < 1, ¬(0 < n ∧ Nat.Prime (5 * 2 ^ n + 1))]posh:∃ m, 0 < m ∧ Nat.Prime (5 * 2 ^ m + 1)⊢ (0 < 1 ∧ Nat.Prime (5 * 2 ^ 1 + 1)) ∧ ∀ n < 1, ¬(0 < n ∧ Nat.Prime (5 * 2 ^ n + 1))
    decide +nativeAll goals completed! 🐙
  ·negh:¬∃ m, 0 < m ∧ Nat.Prime (5 * 2 ^ m + 1)⊢ False exact (h ⟨1, byh:¬∃ m, 0 < m ∧ Nat.Prime (5 * 2 ^ m + 1)⊢ 0 < 1 ∧ Nat.Prime (5 * 2 ^ 1 + 1) decide +nativeAll goals completed! 🐙⟩).elim

There is a conjecture that the first zero is $n = 65536 = 2^{16}$ (which is equivalent to the statement that $2^{2^k} + 1$ is composite for $k > 4$). - T. D. Noe, Feb 25 2011

@[category research open, AMS 11]
theorem conjecture :
    a (2 ^ 16) = 0 ∧ ∀ n : ℕ, 1 ≤ n ∧ n < 2 ^ 16 → a n ≠ 0 := by⊢ a (2 ^ 16) = 0 ∧ ∀ (n : ℕ), 1 ≤ n ∧ n < 2 ^ 16 → OeisA78680.a n ≠ 0
  sorryAll goals completed! 🐙

end OeisA78680