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FormalConjectures.ErdosProblems.«316»

Erdős Problem 316 #

References:

theorem Erdos316.erdos_316 :
False ∀ (A : Finset ), 0A1AnA, 1 / n < 2∃ (A₁ : Finset ) (A₂ : Finset ), Disjoint A₁ A₂ A = A₁ A₂ nA₁, 1 / n < 1 nA₂, 1 / n < 1

Is it true that if $A \subseteq \mathbb{N}\setminus\{1\}$ is a finite set with $\sum_{n \in A} \frac{1}{n} < 2$ then there is a partition $A=A_1 \sqcup A_2$ such that $\sum_{n \in A_i} \frac{1}{n} < 1$ for $i=1,2$?

This is not true in general, as shown by Sándor [Sa97].

The minimal counterexample is $\{2,3,4,5,6,7,10,11,13,14,15\}$, found by Tom Stobart.

This was formalized in Lean by Mehta.

theorem Erdos316.erdos_316.variants.multiset :
∃ (A : Multiset ), 0A 1A (Multiset.map (fun (x : ) => 1 / x) do let aA pure a).sum < 2 ∀ (A₁ A₂ : Multiset ), A = A₁ + A₂1 (Multiset.map (fun (x : ) => 1 / x) do let aA₁ pure a).sum 1 (Multiset.map (fun (x : ) => 1 / x) do let aA₂ pure a).sum

This is not true if $A$ is a multiset, for example $2,3,3,5,5,5,5$.

theorem Erdos316.erdos_316.variants.generalized (n : ) (hn : 2 n) :
∃ (A : Finset ), A.Nonempty 0A 1A kA, 1 / k < n ∀ (P : Finpartition A), P.parts.card = npP.parts, 1 np, 1 / n

This is not true in general, as shown by Sándor [Sa97], who observed that the proper divisors of $120$ form a counterexample. More generally, Sándor shows that for any $n\geq 2$ there exists a finite set $A\subseteq \mathbb{N}\backslash\{1\}$ with $\sum_{k\in A}\frac{1}{k} < n$ and no partition into $n$ parts each of which has $\sum_{k\in A_i}\frac{1}{k}<1$.