FormalConjectures/ErdosProblems/695.lean
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2 declarations, 2 contain sorry, 0 declared as axioms
Erdős Problem 695
Reference: erdosproblems.com/695
Imports (1)
Imported by 1 file
- theorem
Erdos695.erdos_695Let $q_1 < q_2 < \cdots$ be a sequence of primes such that $q_{i + 1} \equiv 1 \pmod{q_i}$ sorry line 39 sourceSource documentation
Let $q_1 < q_2 < \cdots$ be a sequence of primes such that $q_{i + 1} \equiv 1 \pmod{q_i}$. Is it true that $$ \lim_{k \to \infty} q_k^{1/k} = \infty? $$
depends on sorry what this means
Statement uses:
Iff,True,Nat,StrictMono,Nat.Prime,Eq,HMod.hMod,HAdd.hAdd,OfNat.ofNat,Filter.Tendsto,Real,HPow.hPow,HDiv.hDiv,Filter.atTopProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous32 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem erdos_695 : answer(sorry) ↔ ∀ {q : ℕ → ℕ}, StrictMono q → (∀ i, (q i).Prime) → (∀ i, q (i + 1) % q i = 1) → Tendsto (fun k => (q k : ℝ) ^ (1 / k : ℝ)) atTop atTop := by … - theorem
Erdos695.erdos_695.variants.upperBoundIs there a sequence of primes $q_1 < q_2 < \cdots$ such that $q_{i + 1} \equiv 1 \pmod{q_i}$ and $$ q(k) \leq \exp(k (\log k)^{1 + o(1)})? $$ sorry line 54 sourceSource documentation
Is there a sequence of primes $q_1 < q_2 < \cdots$ such that $q_{i + 1} \equiv 1 \pmod{q_i}$ and $$ q(k) \leq \exp(k (\log k)^{1 + o(1)})? $$
depends on sorry what this means
Statement uses:
Iff,True,Exists,Nat,And,StrictMono,Nat.Prime,Eq,HMod.hMod,HAdd.hAdd,OfNat.ofNat,Real,Asymptotics.IsLittleO,Filter.atTop,LE.le,Real.exp,HMul.hMul,HPow.hPow,Real.logProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous36 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem erdos_695.variants.upperBound : answer(sorry) ↔ ∃ q : ℕ → ℕ, StrictMono q ∧ (∀ i, (q i).Prime) ∧ (∀ i, q (i + 1) % q i = 1) ∧ ∃ o : ℕ → ℝ, (o =o[atTop] (1 : ℕ → ℝ)) ∧ -- We use `(k + 1)` here as the informal statement is 1-indexed. ∀ k, q k ≤ exp ((k + 1) * log (k + 1) ^ (1 + o k)) := by …