FormalConjectures/ErdosProblems/403.lean
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- Lean4 Domain Explanation Language
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Project sources: Pinned project documentation, Formal Conjectures paper, arXiv v1
1 declarations, 1 contain sorry, 0 declared as axioms
Erdős Problem 403
References:
- erdosproblems.com/403
- [ErGr80] Erdős, P. and Graham, R., Old and new problems and results in combinatorial number theory. Monographies de L'Enseignement Mathématique (1980).
- [Li76] Lin, S., On two problems of Erdős concerning sums of distinct factorials. Bell Laboratories internal memorandum (1960).
Imports (1)
- theorem
Erdos403.erdos_403Does the equation $$2^m=a_1!+\cdots+a_k!$$ with $a_1 0$ sorry line 58 sourceSource documentation
Does the equation $$2^m=a_1!+\cdots+a_k!$$ with $a_1<a_2<\cdots <a_k$ have only finitely many solutions?
Asked by Burr and Erdős. Frankl and Lin [Li76] independently showed that the answer is yes, and the largest solution is $$2^7=2!+3!+5!.$$ In fact Lin showed that the largest power of $2$ which can divide a sum of distinct factorials containing $2$ is $2^{254}$, and that there are only 5 solutions to $3^m=a_1!+\cdots+a_k!$ (when $m=0,1,2,3,6$).
See also [404].
A solution is encoded below as a pair $(m, s)$ where $s$ is the finite set $\{a_1 < a_2 < \cdots < a_k\}$ of positive integers, so the distinctness of the $a_i$ is given by set membership. The empty set contributes no solutions since $2^m \geq 1 > 0$.
The linked proof gives more than finiteness: it classifies the solutions outright, as $(0,\{1\})$, $(1,\{2\})$, $(3,\{2,3\})$, $(5,\{2,3,4\})$ and $(7,\{2,3,5\})$, so the set below has exactly five elements.
L4DEL explanation
Does the equation $$2^m=a_1!+\cdots+a_k!$$ with $a_1 0$. The linked proof gives more than finiteness: it classifies the solutions outright, as $(0,\{1\})$, $(1,\{2\})$, $(3,\{2,3\})$, $(5,\{2,3,4\})$ and $(7,\{2,3,5\})$, so the set below has exactly five elements.
Sources and attribution: Pinned Lean source (1646ca16afd6)
depends on sorry what this means
Statement uses:
Iff,True,Set.Finite,Prod,Nat,Finset,Set.ofPred,And,Membership.mem,Prod.snd,LT.lt,OfNat.ofNat,Eq,HPow.hPow,Prod.fst,Finset.sum,Nat.factorialProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous18 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem erdos_403 : answer(True) ↔ {p : ℕ × Finset ℕ | (∀ a ∈ p.2, 0 < a) ∧ 2 ^ p.1 = ∑ a ∈ p.2, a.factorial}.Finite := by …