1147.lean`

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Project sources: Pinned project documentation, Formal Conjectures paper, arXiv v1

4 declarations, 3 contain sorry, 0 declared as axioms

Erdős Problem 1147

References:

erdosproblems.com/1147
[Va99] Various, Some of Paul's favorite problems. Booklet produced for the conference "Paul Erdős and his mathematics", Budapest, July 1999 (1999).
[Ko16b] Konieczny, Jakub, Sets of recurrence as bases for the positive integers. Acta Arith. (2016), 309-338.

Imports (1)

FormalConjecturesUtil

def Erdos1147.recurrenceSet The set $\{ n\geq 1: \| \alpha n^2\| < \epsilon(n)\}$ line 38 source
Source documentation
The set $\{ n\geq 1: \| \alpha n^2\| < \epsilon(n)\}$.

standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses: Real, Nat, Set
Proof uses: Set.ofPred, And, LE.le, OfNat.ofNat, LT.lt, distToNearestInt, HMul.hMul, HPow.hPow
11 instances, coercions and auxiliary constants left out.
```
def recurrenceSet (α : ℝ) (ε : ℕ → ℝ) : Set ℕ :=
…
```
theorem Erdos1147.erdos_1147 sorry line 55 source
depends on sorry what this means
Statement uses: Iff, False, Real, LT.lt, OfNat.ofNat, Irrational, Set.IsAsymptoticAddBasisOfOrder, Nat, Erdos1147.recurrenceSet, HDiv.hDiv, Real.log
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
24 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem Erdos1147.erdos_1147.variants.sqrt_two sorry line 63 source
depends on sorry what this means
Statement uses: Not, Set.IsAsymptoticAddBasisOfOrder, Nat, Erdos1147.recurrenceSet, Real.sqrt, OfNat.ofNat, Real, HDiv.hDiv, Real.log
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
24 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem Erdos1147.erdos_1147.variants.almost_every Documented declaration: almost_every sorry line 73 source
Source documentation
Konieczny [Ko16b] showed that, given any $\epsilon(n)\to 0$, the set $\{ n\geq 1: \| \alpha n^2\| < \epsilon(n)\}$ is not an additive basis of order $2$ for almost every $\alpha>0$.

depends on sorry what this means
Statement uses: Nat, Real, Filter.Tendsto, Filter.atTop, nhds, OfNat.ofNat, Filter.Eventually, LT.lt, Not, Set.IsAsymptoticAddBasisOfOrder, Erdos1147.recurrenceSet, MeasureTheory.ae, MeasureTheory.Measure, MeasureTheory.MeasureSpace.volume
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
26 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem erdos_1147.variants.almost_every : ∀ ε : ℕ → ℝ, Tendsto ε atTop (nhds 0) →
    ∀ᵐ α : ℝ, 0 < α → ¬ (recurrenceSet α ε).IsAsymptoticAddBasisOfOrder 2 := by
…
```
Mentions: Erdos1147.recurrenceSet