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Erdős Problem 18

Reference:

    erdosproblems.com/18

    [ErGr80] Erdős, P. and Graham, R. L. (1980). Old and New Problems and Results in Combinatorial Number Theory. Monographies de L'Enseignement Mathématique, 28. Université de Genève. (See the sections on Egyptian fractions or practical numbers).

    [Vo85] Vose, Michael D., Egyptian fractions. Bull. London Math. Soc. (1985), 21-24.

open Filter Asymptotics Realnamespace Erdos18

For a practical number $n$, $h(n)$ is the maximum over all $1 ≤ m ≤ n$ of the minimum number of divisors of $n$ needed to represent $m$ as a sum of distinct divisors.

noncomputable def practicalH (n : ℕ) : ℕ := Finset.sup (Finset.Icc 1 n) fun m => sInf {k | ∃ D : Finset ℕ, D ⊆ n.divisors ∧ D.card = k ∧ m ∈ subsetSums D}

$h(1) = 1$: we need the single divisor {1} to represent 1.

@[category test, AMS 11] theorem practicalH_one : practicalH 1 = 1 := ⊢ practicalH 1 = 1 All goals completed! 🐙

$h(2) = 1$: divisors are {1, 2}, each of m=1,2 needs only 1 divisor.

h1:sInf {k | ∃ D ⊆ {1, 2}, D.card = k ∧ 1 ∈ subsetSums ↑D} = 1h2:sInf {k | ∃ D ⊆ {1, 2}, D.card = k ∧ 2 ∈ subsetSums ↑D} = 1⊢ max (sInf {k | ∃ D ⊆ {1, 2}, D.card = k ∧ 1 ∈ subsetSums ↑D}) (sInf {k | ∃ D ⊆ {1, 2}, D.card = k ∧ 2 ∈ subsetSums ↑D}) = 1 All goals completed! 🐙

$h(6) = 2$: divisors are {1, 2, 3, 6}. The hardest m to represent is m=4 or m=5, each requiring 2 divisors: 4=1+3, 5=2+3.

hdiv:Nat.divisors 6 = {1, 2, 3, 6}h4:4 ∈ Finset.Icc 1 6k:ℕB:Finset ℕhBsum:4 = ∑ i ∈ B, ihk:1 < 2d:ℕhDsub:{d} ⊆ Nat.divisors 6hBsub:B = ∅ ∨ B = {d}hd:d ∈ {1, 2, 3, 6}⊢ False hdiv:Nat.divisors 6 = {1, 2, 3, 6}h4:4 ∈ Finset.Icc 1 6k:ℕB:Finset ℕhBsum:4 = ∑ i ∈ B, ihk:1 < 2hDsub:{1} ⊆ Nat.divisors 6hBsub:B = ∅ ∨ B = {1}⊢ Falsehdiv:Nat.divisors 6 = {1, 2, 3, 6}h4:4 ∈ Finset.Icc 1 6k:ℕB:Finset ℕhBsum:4 = ∑ i ∈ B, ihk:1 < 2hDsub:{2} ⊆ Nat.divisors 6hBsub:B = ∅ ∨ B = {2}⊢ Falsehdiv:Nat.divisors 6 = {1, 2, 3, 6}h4:4 ∈ Finset.Icc 1 6k:ℕB:Finset ℕhBsum:4 = ∑ i ∈ B, ihk:1 < 2hDsub:{3} ⊆ Nat.divisors 6hBsub:B = ∅ ∨ B = {3}⊢ Falsehdiv:Nat.divisors 6 = {1, 2, 3, 6}h4:4 ∈ Finset.Icc 1 6k:ℕB:Finset ℕhBsum:4 = ∑ i ∈ B, ihk:1 < 2hDsub:{6} ⊆ Nat.divisors 6hBsub:B = ∅ ∨ B = {6}⊢ False hdiv:Nat.divisors 6 = {1, 2, 3, 6}h4:4 ∈ Finset.Icc 1 6k:ℕB:Finset ℕhBsum:4 = ∑ i ∈ B, ihk:1 < 2hDsub:{1} ⊆ Nat.divisors 6hBsub:B = ∅ ∨ B = {1}⊢ Falsehdiv:Nat.divisors 6 = {1, 2, 3, 6}h4:4 ∈ Finset.Icc 1 6k:ℕB:Finset ℕhBsum:4 = ∑ i ∈ B, ihk:1 < 2hDsub:{2} ⊆ Nat.divisors 6hBsub:B = ∅ ∨ B = {2}⊢ Falsehdiv:Nat.divisors 6 = {1, 2, 3, 6}h4:4 ∈ Finset.Icc 1 6k:ℕB:Finset ℕhBsum:4 = ∑ i ∈ B, ihk:1 < 2hDsub:{3} ⊆ Nat.divisors 6hBsub:B = ∅ ∨ B = {3}⊢ Falsehdiv:Nat.divisors 6 = {1, 2, 3, 6}h4:4 ∈ Finset.Icc 1 6k:ℕB:Finset ℕhBsum:4 = ∑ i ∈ B, ihk:1 < 2hDsub:{6} ⊆ Nat.divisors 6hBsub:B = ∅ ∨ B = {6}⊢ False hdiv:Nat.divisors 6 = {1, 2, 3, 6}h4:4 ∈ Finset.Icc 1 6k:ℕB:Finset ℕhBsum:4 = ∑ i ∈ B, ihk:1 < 2hDsub:{6} ⊆ Nat.divisors 6h:B = ∅⊢ Falsehdiv:Nat.divisors 6 = {1, 2, 3, 6}h4:4 ∈ Finset.Icc 1 6k:ℕB:Finset ℕhBsum:4 = ∑ i ∈ B, ihk:1 < 2hDsub:{6} ⊆ Nat.divisors 6h:B = {6}⊢ False hdiv:Nat.divisors 6 = {1, 2, 3, 6}h4:4 ∈ Finset.Icc 1 6k:ℕB:Finset ℕhBsum:4 = ∑ i ∈ B, ihk:1 < 2hDsub:{1} ⊆ Nat.divisors 6h:B = ∅⊢ Falsehdiv:Nat.divisors 6 = {1, 2, 3, 6}h4:4 ∈ Finset.Icc 1 6k:ℕB:Finset ℕhBsum:4 = ∑ i ∈ B, ihk:1 < 2hDsub:{1} ⊆ Nat.divisors 6h:B = {1}⊢ Falsehdiv:Nat.divisors 6 = {1, 2, 3, 6}h4:4 ∈ Finset.Icc 1 6k:ℕB:Finset ℕhBsum:4 = ∑ i ∈ B, ihk:1 < 2hDsub:{2} ⊆ Nat.divisors 6h:B = ∅⊢ Falsehdiv:Nat.divisors 6 = {1, 2, 3, 6}h4:4 ∈ Finset.Icc 1 6k:ℕB:Finset ℕhBsum:4 = ∑ i ∈ B, ihk:1 < 2hDsub:{2} ⊆ Nat.divisors 6h:B = {2}⊢ Falsehdiv:Nat.divisors 6 = {1, 2, 3, 6}h4:4 ∈ Finset.Icc 1 6k:ℕB:Finset ℕhBsum:4 = ∑ i ∈ B, ihk:1 < 2hDsub:{3} ⊆ Nat.divisors 6h:B = ∅⊢ Falsehdiv:Nat.divisors 6 = {1, 2, 3, 6}h4:4 ∈ Finset.Icc 1 6k:ℕB:Finset ℕhBsum:4 = ∑ i ∈ B, ihk:1 < 2hDsub:{3} ⊆ Nat.divisors 6h:B = {3}⊢ Falsehdiv:Nat.divisors 6 = {1, 2, 3, 6}h4:4 ∈ Finset.Icc 1 6k:ℕB:Finset ℕhBsum:4 = ∑ i ∈ B, ihk:1 < 2hDsub:{6} ⊆ Nat.divisors 6h:B = ∅⊢ Falsehdiv:Nat.divisors 6 = {1, 2, 3, 6}h4:4 ∈ Finset.Icc 1 6k:ℕB:Finset ℕhBsum:4 = ∑ i ∈ B, ihk:1 < 2hDsub:{6} ⊆ Nat.divisors 6h:B = {6}⊢ False hdiv:Nat.divisors 6 = {1, 2, 3, 6}h4:4 ∈ Finset.Icc 1 6k:ℕhk:1 < 2hDsub:{6} ⊆ Nat.divisors 6hBsum:4 = ∑ i ∈ {6}, i⊢ False hdiv:Nat.divisors 6 = {1, 2, 3, 6}h4:4 ∈ Finset.Icc 1 6k:ℕhk:1 < 2hDsub:{1} ⊆ Nat.divisors 6hBsum:4 = ∑ i ∈ ∅, i⊢ Falsehdiv:Nat.divisors 6 = {1, 2, 3, 6}h4:4 ∈ Finset.Icc 1 6k:ℕhk:1 < 2hDsub:{1} ⊆ Nat.divisors 6hBsum:4 = ∑ i ∈ {1}, i⊢ Falsehdiv:Nat.divisors 6 = {1, 2, 3, 6}h4:4 ∈ Finset.Icc 1 6k:ℕhk:1 < 2hDsub:{2} ⊆ Nat.divisors 6hBsum:4 = ∑ i ∈ ∅, i⊢ Falsehdiv:Nat.divisors 6 = {1, 2, 3, 6}h4:4 ∈ Finset.Icc 1 6k:ℕhk:1 < 2hDsub:{2} ⊆ Nat.divisors 6hBsum:4 = ∑ i ∈ {2}, i⊢ Falsehdiv:Nat.divisors 6 = {1, 2, 3, 6}h4:4 ∈ Finset.Icc 1 6k:ℕhk:1 < 2hDsub:{3} ⊆ Nat.divisors 6hBsum:4 = ∑ i ∈ ∅, i⊢ Falsehdiv:Nat.divisors 6 = {1, 2, 3, 6}h4:4 ∈ Finset.Icc 1 6k:ℕhk:1 < 2hDsub:{3} ⊆ Nat.divisors 6hBsum:4 = ∑ i ∈ {3}, i⊢ Falsehdiv:Nat.divisors 6 = {1, 2, 3, 6}h4:4 ∈ Finset.Icc 1 6k:ℕhk:1 < 2hDsub:{6} ⊆ Nat.divisors 6hBsum:4 = ∑ i ∈ ∅, i⊢ Falsehdiv:Nat.divisors 6 = {1, 2, 3, 6}h4:4 ∈ Finset.Icc 1 6k:ℕhk:1 < 2hDsub:{6} ⊆ Nat.divisors 6hBsum:4 = ∑ i ∈ {6}, i⊢ False All goals completed! 🐙

$h(12) = 3$: divisors are {1, 2, 3, 4, 6, 12}. The hardest m is m=11, requiring 3 divisors: 11=1+4+6.

hdiv:Nat.divisors 12 = {1, 2, 3, 4, 6, 12}h11:11 ∈ Finset.Icc 1 12k:ℕB:Finset ℕhBsum:11 = ∑ i ∈ B, ihk:2 < 3a:ℕb:ℕhab:a ≠ bhDsub:{a, b} ⊆ Nat.divisors 12hBsub:↑B ⊆ ↑{a, b}ha:a ∈ {1, 2, 3, 4, 6, 12}hb:b ∈ {1, 2, 3, 4, 6, 12}hBp:B ⊆ {a, b}key:∀ S ∈ {a, b}.powerset, ∑ x ∈ S, x ≠ 11⊢ False All goals completed! 🐙

For any practical number $n$, $h(n)$ ≤ number of divisors of $n$.

@[category test, AMS 11] theorem practicalH_le_divisors (n : ℕ) (hn : Nat.IsPractical n) : practicalH n ≤ n.divisors.card := n:ℕhn:n.IsPractical⊢ practicalH n ≤ n.divisors.card n:ℕhn:n.IsPractical⊢ ∀ (b : ℕ), 1 ≤ b ∧ b ≤ n → sInf {k | ∃ D ⊆ n.divisors, D.card = k ∧ b ∈ subsetSums ↑D} ≤ n.divisors.card All goals completed! 🐙

$h(n!)$ is well-defined since $n!$ is practical for $n ≥ 1$.

n:ℕih:n.factorial.IsPracticalm:ℕhm:m ≤ (n + 1) * n.factorialhle:n.factorial < mq:ℕ := m / (n + 1)r:ℕ := m % (n + 1)h_div:m = (n + 1) * q + rh_r_lt:r < n + 1B:Finset ℕhB_sub:↑B ⊆ ↑n.factorial.divisorshB_sum:q = ∑ i ∈ B, ihdvd:∀ d ∈ B, d * (n + 1) ∈ (n + 1).factorial.divisorshB'_sum:(Finset.image (fun x ↦ x * (n + 1)) B).sum id = (n + 1) * qhr:¬r = 0h_disj:Disjoint (Finset.image (fun x ↦ x * (n + 1)) B) {r}x:ℕhx:x ∈ ↑(Finset.image (fun x ↦ x * (n + 1)) B ∪ {r})h:x ∈ {r}⊢ r ∈ ↑(n + 1).factorial.divisors; exact Nat.mem_divisors.mpr ⟨(Nat.dvd_factorial (n:ℕih:n.factorial.IsPracticalm:ℕhm:m ≤ (n + 1) * n.factorialhle:n.factorial < mq:ℕ := m / (n + 1)r:ℕ := m % (n + 1)h_div:m = (n + 1) * q + rh_r_lt:r < n + 1B:Finset ℕhB_sub:↑B ⊆ ↑n.factorial.divisorshB_sum:q = ∑ i ∈ B, ihdvd:∀ d ∈ B, d * (n + 1) ∈ (n + 1).factorial.divisorshB'_sum:(Finset.image (fun x ↦ x * (n + 1)) B).sum id = (n + 1) * qhr:¬r = 0h_disj:Disjoint (Finset.image (fun x ↦ x * (n + 1)) B) {r}x:ℕhx:x ∈ ↑(Finset.image (fun x ↦ x * (n + 1)) B ∪ {r})h:x ∈ {r}⊢ 0 < r All goals completed! 🐙) (n:ℕih:n.factorial.IsPracticalm:ℕhm:m ≤ (n + 1) * n.factorialhle:n.factorial < mq:ℕ := m / (n + 1)r:ℕ := m % (n + 1)h_div:m = (n + 1) * q + rh_r_lt:r < n + 1B:Finset ℕhB_sub:↑B ⊆ ↑n.factorial.divisorshB_sum:q = ∑ i ∈ B, ihdvd:∀ d ∈ B, d * (n + 1) ∈ (n + 1).factorial.divisorshB'_sum:(Finset.image (fun x ↦ x * (n + 1)) B).sum id = (n + 1) * qhr:¬r = 0h_disj:Disjoint (Finset.image (fun x ↦ x * (n + 1)) B) {r}x:ℕhx:x ∈ ↑(Finset.image (fun x ↦ x * (n + 1)) B ∪ {r})h:x ∈ {r}⊢ r ≤ n All goals completed! 🐙)).trans (Nat.factorial_dvd_factorial n.le_succ), Nat.factorial_ne_zero _⟩, rfl⟩

Conjecture 1. Are there infinitely many practical numbers $m$ such that $h(m) < (\log \log m)^{O(1)}$?

More precisely: does there exist a constant $C > 0$ such that for infinitely many practical numbers $m$, we have $h(m) < (\log \log m)^C$?

@[category research open, AMS 11] theorem erdos_18a : answer(sorry) ↔ ∃ C : ℝ, 0 < C ∧ ∃ᶠ m in atTop, Nat.IsPractical m ∧ (practicalH m : ℝ) < (log (log m)) ^ C := ⊢ True ↔ ∃ C, 0 < C ∧ ∃ᶠ (m : ℕ) in atTop, m.IsPractical ∧ ↑(practicalH m) < log (log ↑m) ^ C All goals completed! 🐙

Conjecture 2. Is it true that $h(n!) < n^{o(1)}$? That is, for all $\varepsilon > 0$, is $h(n!) < n^\varepsilon$ for sufficiently large $n$?

@[category research open, AMS 11] theorem erdos_18b : answer(sorry) ↔ ∀ ε : ℝ, 0 < ε → ∀ᶠ n : ℕ in atTop, (practicalH n.factorial : ℝ) < (n : ℝ) ^ ε := ⊢ True ↔ ∀ (ε : ℝ), 0 < ε → ∀ᶠ (n : ℕ) in atTop, ↑(practicalH n.factorial) < ↑n ^ ε All goals completed! 🐙

Conjecture 3. Or perhaps even $h(n!) < (\log n)^{O(1)}$?

Erdős offered $250 for a proof or disproof.

@[category research open, AMS 11] theorem erdos_18c : answer(sorry) ↔ ∃ C : ℝ, 0 < C ∧ ∀ᶠ n : ℕ in atTop, (practicalH n.factorial : ℝ) < (log n) ^ C := ⊢ True ↔ ∃ C, 0 < C ∧ ∀ᶠ (n : ℕ) in atTop, ↑(practicalH n.factorial) < log ↑n ^ C All goals completed! 🐙

Erdős's Theorem. Erdős proved that $h(n!) < n$ for all $n \ge 1$.

@[category research solved, AMS 11] theorem erdos_18_upper_bound : ∀ᶠ n : ℕ in atTop, practicalH (Nat.factorial n) < n := ⊢ ∀ᶠ (n : ℕ) in atTop, practicalH n.factorial < n All goals completed! 🐙

Vose's Theorem. Vose proved the existence of infinitely many practical numbers $m$ such that $h(m) \ll (\log m)^{1/2}$. This gives a positive answer to a weaker form of Conjecture 1.

@[category research solved, AMS 11] theorem erdos_18_vose : ∃ C : ℝ, 0 < C ∧ ∃ᶠ m in atTop, Nat.IsPractical m ∧ (practicalH m : ℝ) < C * (log m) ^ (1 / 2 : ℝ) := ⊢ ∃ C, 0 < C ∧ ∃ᶠ (m : ℕ) in atTop, m.IsPractical ∧ ↑(practicalH m) < C * log ↑m ^ (1 / 2) All goals completed! 🐙end Erdos18