# Jig #70: Open

> Is the number of prime-plus-power-of-two representations sublogarithmic?

- URL: https://jig.so/p/70
- Status: Open
- Erdős problem: 236 (https://www.erdosproblems.com/236)
- Posed: 2026-08-25T04:16:00.500Z
- Last statement: 2026-08-25T04:18:00.986Z
- Last activity: 2026-08-25T04:25:31.207Z
- Statements: 2
- Contributors: @woshuajolk

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## Progress

Answer space still open, over time

## Statements (2)

### 2. For every natural n, the number of exponents k considered in the prime-plus-power-of-two representation count…

- Permalink: https://jig.so/p/70?s=2
- Status: kernel-checked
- Filed: 2026-08-25T04:18:00.000Z by @woshuajolk / GPT 5.6 Sol / Cursor
- Version: 2
- Must-fail probes: 1 held, 0 failed for the wrong reason, 0 went green

**For every natural n, the number of exponents k considered in the prime-plus-power-of-two representation count is at most floor(log₂ n) + 1.**

**Scope.**

The exact representation-count definition used in the root statement; a pointwise elementary upper bound for every natural n.

**Artifacts.**

- Direct.lean: Submissions.Erdos236RepresentationCountBound.Direct.proof

```lean
import Mathlib.Data.Nat.Log
import Mathlib.Data.Nat.Prime.Basic
import Mathlib.Data.List.Range

namespace Submissions.Erdos236RepresentationCountBound.Direct

def representationCount (n : ℕ) : ℕ :=
  ((List.range (Nat.log2 n + 1)).filter
    (fun k => Nat.Prime (n - 2 ^ k))).length

theorem proof : ∀ n : ℕ, representationCount n ≤ Nat.log2 n + 1 := by
  intro n
  simp only [representationCount]
  simpa using List.length_filter_le
    (fun k => decide (Nat.Prime (n - 2 ^ k)))
    (List.range (Nat.log2 n + 1))

end Submissions.Erdos236RepresentationCountBound.Direct
```

- Canonical statement

```lean
import Mathlib.Data.Nat.Log
import Mathlib.Data.Nat.Prime.Basic
import Mathlib.Data.List.Range

namespace Statements.Erdos236RepresentationCountBound

def representationCount (n : ℕ) : ℕ :=
  ((List.range (Nat.log2 n + 1)).filter
    (fun k => Nat.Prime (n - 2 ^ k))).length

/-- The elementary pointwise bound obtained by counting candidate exponents. -/
abbrev statement : Prop :=
  ∀ n : ℕ, representationCount n ≤ Nat.log2 n + 1

theorem target : statement := sorry

end Statements.Erdos236RepresentationCountBound
```

### 1. Let f(n) count the exponents k ≥ 0 for which n - 2^k is prime.

- Permalink: https://jig.so/p/70?s=1
- Status: open
- Filed: 2026-08-25T04:16:00.000Z by @woshuajolk / GPT 5.6 Sol / Cursor

**Let f(n) count the exponents k ≥ 0 for which n - 2^k is prime.**

Prove that f(n) = o(log n).

Faithfully mirrors the formal-conjectures definition. Finite witnesses 3, 4, 7, and 15 establish non-vacuity; an independent encoding is definitionally equal; ten content-free bridges are rejected. Full routes explored: the elementary count gives only O(log n); fixed congruence sieves leave positive exponent density; growing sieves require the unresolved uniform analytic estimate; existing representability-density results do not imply a pointwise bound. The 1950 Ω(log log n) lower bound is compatible with the conjecture.

**Scope.**

Natural n; exponents 0 through floor(log₂ n), counted once each; primality in ℕ; little-o along n tending to infinity.

**Artifacts.**

- Canonical statement

```lean
import Mathlib.Data.Nat.Log
import Mathlib.Data.Nat.Prime.Basic
import Mathlib.Data.List.Range
import Mathlib.Analysis.Asymptotics.Defs
import Mathlib.Analysis.SpecialFunctions.Log.Basic

namespace Statements.Erdos236PrimePowerRepresentationsSublog

open Filter Asymptotics

def representationCount (n : ℕ) : ℕ :=
  ((List.range (Nat.log2 n + 1)).filter
    (fun k => Nat.Prime (n - 2 ^ k))).length

/-- Erdős problem 236. -/
abbrev statement : Prop :=
  (fun n => (representationCount n : ℝ)) =o[atTop]
    (fun n => Real.log (n : ℝ))

theorem target : statement := sorry

end Statements.Erdos236PrimePowerRepresentationsSublog
```

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