FormalConjectures/ErdosProblems/1106.lean
Explanation legend Lean4 Domain Explanation Language (L4DEL)
L4DEL explanations are a separate layer. They do not change the Lean source. Intuition is explanatory prose; numbered items are a reading path; sources link the wording back to evidence.
- L4DEL
- Lean4 Domain Explanation Language
Credit: The upstream authors and contributors retain credit for the project and mathematics. lean4-explainer only renders their pinned source and adds the separately disclosed explanation layer.
Project sources: Pinned project documentation, Formal Conjectures paper, arXiv v1
3 declarations, 2 contain sorry, 0 declared as axioms
Erdős Problem 1106
References:
- erdosproblems.com/1106
- [ScWi87] Schinzel, A. and Wirsing, E., Multiplicative properties of the partition function. Proc. Indian Acad. Sci. Math. Sci. (1987), 297--303.
Imports (1)
- def
Erdos1106.pThe partition function p(n) is the number of ways to write n as a sum of positive integers (where the order of the summands does not matter) line 37 sourceSource documentation
The partition function p(n) is the number of ways to write n as a sum of positive integers (where the order of the summands does not matter).
standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses:
NatProof uses:
Fintype.card,Nat.Partition1 instances, coercions and auxiliary constants left out.
def p : ℕ → ℕ := fun n => Fintype.card (Nat.Partition n) …
- theorem
Erdos1106.erdos_1106.parts.iLet $p(n)$ be the partition number of $n$ and $F(n)$ be the number of distinct prime factors of $∏_{i= 1} ^ {n} p(n)$, then $F(n)$ tends to infinity when $n$ tends to infinity sorry line 46 sourceSource documentation
Let $p(n)$ be the partition number of $n$ and $F(n)$ be the number of distinct prime factors of $∏_{i= 1} ^ {n} p(n)$, then $F(n)$ tends to infinity when $n$ tends to infinity.
The answer is yes: Schinzel and Wirsing [ScWi87] proved $F(n) \gg \log n$.
depends on sorry what this means
Statement uses:
Iff,True,Filter.Tendsto,Nat,Finset.card,Nat.primeFactors,Finset.prod,Finset.Icc,OfNat.ofNat, Erdos1106.p,Filter.atTopProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous8 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem erdos_1106.parts.i : answer(True) ↔ Tendsto (fun n => #(∏ i ∈ Icc 1 n, p i).primeFactors) atTop atTop := by …Mentions: Erdos1106.p
- theorem
Erdos1106.erdos_1106.parts.iiLet $p(n)$ be the partition number of $n$ and $F(n)$ be the number of distinct prime factors of $∏_{i= 1} ^ {n} p(n)$, $F(n)>n$ for sufficiently large $n$ sorry line 55 sourceSource documentation
Let $p(n)$ be the partition number of $n$ and $F(n)$ be the number of distinct prime factors of $∏_{i= 1} ^ {n} p(n)$, $F(n)>n$ for sufficiently large $n$.
depends on sorry what this means
Statement uses:
Iff,True,Filter.Eventually,Nat,GT.gt,Finset.card,Nat.primeFactors,Finset.prod,Finset.Icc,OfNat.ofNat, Erdos1106.p,Filter.atTopProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous10 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem erdos_1106.parts.ii : answer(sorry) ↔ ∀ᶠ n in atTop, #(∏ i ∈ Icc 1 n, p i).primeFactors > n := by …Mentions: Erdos1106.erdos_1106.parts.i, Erdos1106.p