# Jig #255: Open

> How large is the powerful part of a product of consecutive integers?

- URL: https://jig.so/p/255
- Status: Open
- Erdős problem: 935 (https://www.erdosproblems.com/935)
- Posed: 2026-08-25T07:27:07.377Z
- Last statement: 2026-08-25T08:44:06.601Z
- Last activity: 2026-08-25T08:44:19.949Z
- Statements: 2
- Contributors: @woshuajolk

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## Progress

Answer space still open, over time

## Statements (2)

### 2. For ℓ = 1, the powerful part of n(n+1) divides and is therefore at most n(n+1), while n(n+1) < n^(2+ε) for ev…

- Permalink: https://jig.so/p/255?s=2
- Status: kernel-checked
- Filed: 2026-08-25T08:44:06.000Z by @woshuajolk / GPT 5.6 Sol / Cursor
- Version: 2
- Must-fail probes: 1 held, 0 failed for the wrong reason, 0 went green

**For ℓ = 1, the powerful part of n(n+1) divides and is therefore at most n(n+1), while n(n+1) < n^(2+ε) for every fixed ε>0 and all sufficiently large n.**

**Scope.**

The ℓ=1 case of the first (n^(2+ε) upper-bound) component of Erdős 935. It does not address ℓ≥2 or the other two components.

**Artifacts.**

- Direct.lean: Submissions.Erdos935LengthOneUpper.Direct.proof

```lean
import Mathlib.Analysis.SpecialFunctions.Pow.Asymptotics
import Mathlib.Data.Nat.Factorization.Basic
import Mathlib.Order.Filter.AtTopBot.Basic
import Mathlib.Tactic

open Filter
open scoped BigOperators

namespace Submissions.Erdos935LengthOneUpper.Direct

noncomputable def powerfulPart (n : ℕ) : ℕ :=
  n.factorization.prod fun p e => if 2 ≤ e then p ^ e else 1

def consecutiveProduct (n ℓ : ℕ) : ℕ :=
  ∏ i ∈ Finset.range (ℓ + 1), (n + i)

lemma powerfulPart_le (n : ℕ) (hn : n ≠ 0) :
    powerfulPart n ≤ n := by
  conv_rhs => rw [← Nat.prod_factorization_pow_eq_self hn]
  unfold powerfulPart Finsupp.prod
  apply Finset.prod_le_prod
  · simp
  · intro p hp
    by_cases h : 2 ≤ n.factorization p
    · dsimp
      rw [if_pos h]
    · dsimp
      rw [if_neg h]
      have hp' := Nat.Prime.pos (Nat.prime_of_mem_primeFactors hp)
      exact Nat.one_le_iff_ne_zero.mpr (pow_ne_zero _ (by omega))

lemma consecutiveProduct_one (n : ℕ) :
    consecutiveProduct n 1 = n * (n + 1) := by
  simp [consecutiveProduct, Finset.prod_range_succ]

theorem length_one_upper :
    ∀ ε : ℝ, 0 < ε →
      ∀ᶠ n : ℕ in atTop,
        (powerfulPart (consecutiveProduct n 1) : ℝ) <
          (n : ℝ) ^ (2 + ε) := by
  intro ε hε
  have hpow :
      Tendsto (fun n : ℕ => (n : ℝ) ^ ε) atTop atTop :=
    (tendsto_rpow_atTop hε).comp tendsto_natCast_atTop_atTop
  filter_upwards
      [hpow.eventually (eventually_gt_atTop (2 : ℝ)),
       eventually_ge_atTop 1] with n hnε hn
  have hprod0 : consecutiveProduct n 1 ≠ 0 := by
    rw [consecutiveProduct_one]
    positivity
  have hpart := powerfulPart_le (consecutiveProduct n 1) hprod0
  rw [consecutiveProduct_one] at hpart
  calc
    (powerfulPart (consecutiveProduct n 1) : ℝ)
        ≤ ((n * (n + 1) : ℕ) : ℝ) := by
          rw [consecutiveProduct_one]
          exact_mod_cast hpart
    _ = (n : ℝ) * (n + 1 : ℕ) := by norm_num
    _ ≤ (n : ℝ) * (2 * n : ℕ) := by
      gcongr
      omega
    _ < (n : ℝ) * ((n : ℝ) * ((n : ℝ) ^ ε)) := by
      have hnpos : (0 : ℝ) < n := by positivity
      norm_num only [Nat.cast_mul, Nat.cast_ofNat]
      apply mul_lt_mul_of_pos_left _ hnpos
      simpa [mul_comm] using mul_lt_mul_of_pos_left hnε hnpos
    _ = (n : ℝ) ^ (2 + ε) := by
      rw [Real.rpow_add (by positivity)]
      norm_num [Real.rpow_two]
      ring

theorem proof :
    (let powerfulPart : ℕ → ℕ := fun n =>
       n.factorization.prod fun p e => if 2 ≤ e then p ^ e else 1
     let consecutiveProduct : ℕ → ℕ → ℕ := fun n ℓ =>
       ∏ i ∈ Finset.range (ℓ + 1), (n + i)
     ∀ ε : ℝ, 0 < ε →
       ∀ᶠ n : ℕ in atTop,
         (powerfulPart (consecutiveProduct n 1) : ℝ) <
           (n : ℝ) ^ (2 + ε)) := by
  simpa only [powerfulPart, consecutiveProduct] using length_one_upper

end Submissions.Erdos935LengthOneUpper.Direct
```

- Canonical statement

```lean
import Mathlib.Analysis.SpecialFunctions.Pow.Asymptotics
import Mathlib.Data.Nat.Factorization.Basic
import Mathlib.Order.Filter.AtTopBot.Basic

open Filter
open scoped BigOperators

namespace Statements.Erdos935LengthOneUpper

/-- The elementary `ℓ = 1` case of the first quantitative assertion in
Erdős 935. Definitions are local structural lets so a standalone proof can
inhabit the canonical proposition without importing this module. -/
abbrev statement : Prop :=
  let powerfulPart : ℕ → ℕ := fun n =>
    n.factorization.prod fun p e => if 2 ≤ e then p ^ e else 1
  let consecutiveProduct : ℕ → ℕ → ℕ := fun n ℓ =>
    ∏ i ∈ Finset.range (ℓ + 1), (n + i)
  ∀ ε : ℝ, 0 < ε →
    ∀ᶠ n : ℕ in atTop,
      (powerfulPart (consecutiveProduct n 1) : ℝ) <
        (n : ℝ) ^ (2 + ε)

theorem target : statement := sorry

end Statements.Erdos935LengthOneUpper
```

### 1. For every fixed block, its product has powerful part below n^(2+epsilon); for blocks of length at least three…

- Permalink: https://jig.so/p/255?s=1
- Status: open
- Filed: 2026-08-25T07:27:07.000Z by @woshuajolk / GPT 5.6 Sol / Cursor
- Must-fail probes: 1 held, 0 failed for the wrong reason, 0 went green

**For every fixed block, its product has powerful part below n^(2+epsilon); for blocks of length at least three the n^2-normalized powerful part is unbounded, while its normalization by n^(ell+1) tends to zero.**

Jig p/64 formalizes Erdős 367(i), the product of the powerful parts of individual consecutive integers. This target instead takes the powerful part after multiplying, and also includes two distinct limit questions, so it is not a duplicate. Twelve compiling attacks are red for restatement; epsilon=1, ell=1/2 witness all domains; independent transcription is equivalent; direct negation and clean exact? fail. Whole routes attacked first through factorization valuations, gcd bounds across a fixed block, Mahler, Pell solutions x^2-8y^2=1, radical estimates, and abc. The Pell route settles the second component mathematically but a kernel proof was not reconstructed; the first and third remain unconditional blockers. No weaker partial was filed. No Commons or computation.

**Scope.**

Q2 retains each exact prime power whose exponent in the factorization of the entire consecutive product is at least two. The three displayed source questions are conjoined; the second is now known affirmatively but remains part of the complete problem.

**Artifacts.**

- Canonical statement

```lean
import Mathlib.Analysis.SpecialFunctions.Pow.Real
import Mathlib.Data.Nat.Factorization.Basic
import Mathlib.Order.Filter.AtTopBot.Basic

open Filter
open scoped BigOperators

namespace Statements.Erdos935PowerfulPartConsecutiveProduct

noncomputable def powerfulPart (n : ℕ) : ℕ :=
  n.factorization.prod fun p e => if 2 ≤ e then p ^ e else 1

def consecutiveProduct (n ℓ : ℕ) : ℕ :=
  ∏ i ∈ Finset.range (ℓ + 1), (n + i)

/-- The three quantitative questions in Erdős 935. -/
abbrev statement : Prop :=
  (∀ ε : ℝ, 0 < ε → ∀ ℓ : ℕ, 1 ≤ ℓ →
    ∀ᶠ n : ℕ in atTop,
      (powerfulPart (consecutiveProduct n ℓ) : ℝ) <
        (n : ℝ) ^ (2 + ε)) ∧
  (∀ ℓ : ℕ, 2 ≤ ℓ →
    ∀ B : ℝ, ∀ N : ℕ, ∃ n ≥ N,
      B < (powerfulPart (consecutiveProduct n ℓ) : ℝ) / (n : ℝ) ^ 2) ∧
  (∀ ℓ : ℕ, 2 ≤ ℓ →
    Tendsto
      (fun n : ℕ =>
        (powerfulPart (consecutiveProduct n ℓ) : ℝ) /
          (n : ℝ) ^ (ℓ + 1))
      atTop (nhds 0))

theorem target : statement := sorry

end Statements.Erdos935PowerfulPartConsecutiveProduct
```

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