967.lean`

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

6 declarations, 5 contain sorry, 0 declared as axioms

Erdős Problem 967

References:

erdosproblems.com/967
[ErIn64] Erdős, P. and Ingham, A. E., Arithmetical Tauberian theorems. Acta Arith. (1964), 341-356.
[Yi25] Yip, F., On a problem of Erdős and Ingham. arXiv:2512.16528 (2025).

Imports (1)

FormalConjecturesUtil

def Erdos967.summand The summand $\frac{1}{n^{1+it}}$, for a natural number $n$ and $t\in\mathbb{R}$ line 39 source
Source documentation
The summand $\frac{1}{n^{1+it}}$, for a natural number $n$ and $t\in\mathbb{R}$.

standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses: Real, Nat, Complex
Proof uses: HDiv.hDiv, OfNat.ofNat, HPow.hPow, HAdd.hAdd, HMul.hMul, Complex.I
14 instances, coercions and auxiliary constants left out.
```
noncomputable def summand (t : ℝ) (n : ℕ) : ℂ := 1 / (n : ℂ) ^ (1 + t * Complex.I : ℂ)
…
```
theorem Erdos967.erdos_967 sorry line 54 source
depends on sorry what this means
Statement uses: Iff, False, Nat, StrictMono, LT.lt, OfNat.ofNat, Summable, Real, HDiv.hDiv, SummationFilter.unconditional, Ne, Complex, HAdd.hAdd, tsum, Erdos967.summand
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
50 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem Erdos967.erdos_967.variants.yip Yip [Yi25] has proved that this is not always true - in fact, for any real sorry line 65 source
Source documentation
Yip [Yi25] has proved that this is not always true - in fact, for any real $t\neq 0$, there exists a sequence of integers $1<a_1<\cdots$ such that $\sum \frac{1}{a_i}<\infty$ and $1+\sum_{k}\frac{1}{a_k^{1+it}}=0$.

depends on sorry what this means
Statement uses: Real, Ne, OfNat.ofNat, Exists, Nat, And, StrictMono, LT.lt, Summable, HDiv.hDiv, SummationFilter.unconditional, Eq, Complex, HAdd.hAdd, tsum, Erdos967.summand
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
52 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem erdos_967.variants.yip (t : ℝ) (ht : t ≠ 0) :
    ∃ a : ℕ → ℕ, StrictMono a ∧ 1 < a 0 ∧ Summable (fun k : ℕ => 1 / (a k : ℝ)) ∧
      (1 + (∑' k, summand t (a k)) = 0) := by
…
```
Mentions: Erdos967.summand
theorem Erdos967.erdos_967.variants.finite It remains open whether this is true for every finite sequence of integers sorry line 74 source
Source documentation
It remains open whether this is true for every finite sequence of integers.

depends on sorry what this means
Statement uses: Iff, True, Finset, Nat, Membership.mem, LT.lt, OfNat.ofNat, Real, Ne, Complex, HAdd.hAdd, Finset.sum, Erdos967.summand
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
22 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem erdos_967.variants.finite : answer(sorry) ↔
    ∀ A : Finset ℕ, (∀ n ∈ A, 1 < n) → ∀ t : ℝ, 1 + (∑ n ∈ A, summand t n) ≠ 0 := by
…
```
Mentions: Erdos967.summand
theorem Erdos967.erdos_967.variants.two_three_five A question of Erdős and Ingham [ErIn64] sorry line 83 source
Source documentation
A question of Erdős and Ingham [ErIn64]. The simplest case they could not decide this question for was the finite sequence $\{2,3,5\}$.

depends on sorry what this means
Statement uses: Iff, True, Real, Ne, Complex, HAdd.hAdd, OfNat.ofNat, Finset.sum, Nat, Insert.insert, Finset, Singleton.singleton, Erdos967.summand
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
22 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem erdos_967.variants.two_three_five : answer(sorry) ↔
    ∀ t : ℝ, 1 + (∑ n ∈ ({2, 3, 5} : Finset ℕ), summand t n) ≠ 0 := by
…
```
Mentions: Erdos967.summand
theorem Erdos967.erdos_967.variants.tauberian Their interest in this problem arose from their proof that the statement that there are no such zeros is equivalent to the fact that, for any non-decreasing sorry line 95 source
Source documentation
Their interest in this problem arose from their proof that the statement that there are no such zeros is equivalent to the fact that, for any non-decreasing $f:\mathbb{R}\to \mathbb{R}_{\geq 0}$ which is bounded on every bounded interval and is $=0$ for $x<1$, the relationship $$f(x)+\sum_k f(x/a_k)=\left(1+\sum_k \frac{1}{a_k}+o(1)\right)x$$ implies $f(x)=(1+o(1))x$.

depends on sorry what this means
Statement uses: Nat, StrictMono, LT.lt, OfNat.ofNat, Summable, Real, HDiv.hDiv, SummationFilter.unconditional, Iff, Ne, Complex, HAdd.hAdd, tsum, Erdos967.summand, Monotone, LE.le, Eq, BddAbove, Set.image, Set.Icc and 3 more
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous
60 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem erdos_967.variants.tauberian (a : ℕ → ℕ) (ha : StrictMono a) (ha₀ : 1 < a 0)
    (hsum : Summable (fun k : ℕ => 1 / (a k : ℝ))) :
    (∀ t : ℝ, 1 + (∑' k, summand t (a k)) ≠ 0) ↔
      ∀ f : ℝ → ℝ, Monotone f → (∀ x : ℝ, 0 ≤ f x) → (∀ x < 1, f x = 0) →
        (∀ x y : ℝ, BddAbove (f '' Set.Icc x y)) →
        Tendsto (fun x : ℝ => (f x + ∑' k, f (x / (a k : ℝ))) / x) atTop
          (𝓝 (1 + ∑' k, 1 / (a k : ℝ))) →
        Tendsto (fun x : ℝ => f x / x) atTop (𝓝 1) := by
…
```
Mentions: Erdos967.summand