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← problemsCB006openMathematics/ Combinatorics

Singmaster's conjecture

Maintainer — open

Is there a bound on how often a number appears in Pascal's triangle?

Motivation

David Singmaster, 1971. Every number greater than one appears finitely often; 3003 appears eight times, and no number is known to appear more.

Singmaster conjectured a universal constant — possibly as small as eight or ten. Even proving some absolute bound exists is open; the best known is O(log n / log log n) occurrences for the value n.

David Singmaster, 1971.

Lean API
import Mathlib.Data.Finset.Card
import Mathlib.Data.Nat.Choose.Basic

namespace Conjectura.CB006

/-- Is there an absolute bound on how many times a number greater than one can appear
in Pascal's triangle? Only 3003 is known to appear eight times; Singmaster conjectured
a universal constant, and even a bound is unproved. -/
def goal : Prop :=
  ∃ C : ℕ, ∀ a : ℕ, 1 < a →
    ∀ S : Finset (ℕ × ℕ), (∀ p ∈ S, Nat.choose p.1 p.2 = a) → S.card ≤ C

end Conjectura.CB006
Definition3
FinsetFinset

structure

chooseNat.choose

def

cardS.card

theorem

Related work0
    For your AI
    Download as .md
    I am proving a theorem in Lean 4 and submitting it to Conjectura.
    
    ## Problem CB006 — Singmaster's conjecture
    
    Is there a bound on how often a number appears in Pascal's triangle?
    
    ## Environment (fixed — do not assume anything newer)
    
    - Lean toolchain: `leanprover/lean4:v4.33.0-rc1`
    - Mathlib: `v4.33.0-rc1`
    
    If a lemma you want does not exist in that Mathlib, prove it inline instead of
    importing something newer.
    
    ## The exact statement I must prove
    
    ```lean
    theorem solution : ∃ C : ℕ, ∀ a : ℕ, 1 < a → ∀ S : Finset (ℕ × ℕ),
        (∀ p ∈ S, Nat.choose p.1 p.2 = a) → S.card ≤ C := by
      sorry
    ```
    
    ## The file I submit
    
    ```lean
    import Conjectura.Problems.CB006.Statement
    
    namespace Submission
    
    theorem solution : Conjectura.CB006.goal := by
      sorry
    
    end Submission
    
    ```
    
    ## The Lean definitions of every term in this problem
    
    These are the actual definitions your proof will be checked against. Do not
    substitute your own version of any of them.
    
    ### Finset
    
    structure
    
    ```lean
    Finset
    ```
    Defined in Mathlib.
    
    ### choose
    
    def
    
    ```lean
    Nat.choose
    ```
    Defined in Mathlib.
    
    ### card
    
    theorem
    
    ```lean
    S.card
    ```
    Defined in Mathlib.
    
    ## Rules — submissions violating these are rejected automatically
    
    1. **Do not change the name or type of `solution`.** It must satisfy the
       statement above exactly.
    2. **Do not redefine or shadow anything from the problem's Statement module.**
       Declaring your own `goal`, or redefining a name it depends on, produces a
       proof of a *different* statement and is rejected. This is the single most
       common rejection.
    3. **No `sorry`** anywhere, including in helper lemmas. It surfaces as the
       axiom `sorryAx` and is detected transitively through imports.
    4. **No `native_decide`** — it surfaces as `Lean.ofReduceBool` and is not
       accepted, because it trusts compiled code rather than the kernel.
    5. Only these axioms are permitted: `propext`, `Classical.choice`,
       `Quot.sound`.
    6. Follow Mathlib style: hypotheses left of the colon, explicit types,
       `snake_case` theorem names, `UpperCamelCase` types.
    
    ## What I want from you
    
    Here is my argument in informal mathematics:
    
    > [PASTE YOUR PROOF SKETCH HERE]
    
    Turn it into Lean 4 that compiles under the environment above and satisfies the
    statement exactly. Where you are unsure a lemma exists in this Mathlib version,
    say so explicitly rather than guessing a name.

    Submissions are not open yet

    Conjectura is in beta. You can read every statement, every definition and the Lean behind them, and download the exact files the checker uses — but proofs are not being accepted yet.

    The reason is a deliberate order of operations. Accepting a proof means running a stranger’s code and standing behind a verdict, and no statement here yet carries a researcher’s name. A machine-checked answer to a question nobody has vouched for is worth very little, so the vouching comes first.

    The English write-ups are also switched off during the beta. Nothing on this page is generated by a model.

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