FormalConjectures/ErdosProblems/1003.lean
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3 declarations, 3 contain sorry, 0 declared as axioms
Erdős Problem 1003
Reference: erdosproblems.com/1003
Imports (1)
- theorem
Erdos1003.erdos_1003Are there infinitely many solutions to $\phi(n) = \phi(n+1)$, where $\phi$ is the Euler totient function? sorry line 38 sourceSource documentation
Are there infinitely many solutions to $\phi(n) = \phi(n+1)$, where $\phi$ is the Euler totient function?
depends on sorry what this means
Statement uses:
Iff,True,Set.Infinite,Nat,Set.ofPred,Eq,Nat.totient,HAdd.hAdd,OfNat.ofNatProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous6 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem erdos_1003 : answer(sorry) ↔ Set.Infinite {n | φ n = φ (n + 1)} := by … - theorem
Erdos1003.erdos_1003.variants.IccErdős [Er85e] says that, presumably, for every $k \geq 1$ the equation $$\phi(n) = \phi(n+1) = \cdots = \phi (n+k)$$ has infinitely many solutions sorry line 48 sourceSource documentation
Erdős [Er85e] says that, presumably, for every $k \geq 1$ the equation $$\phi(n) = \phi(n+1) = \cdots = \phi (n+k)$$ has infinitely many solutions.
[Er85e] Erdős, P., Some problems and results in number theory. Number theory and combinatorics. Japan 1984 (Tokyo, Okayama and Kyoto, 1984) (1985), 65-87.
depends on sorry what this means
Statement uses:
Iff,True,Nat,GE.ge,OfNat.ofNat,Set.Infinite,Set.ofPred,Membership.mem,Set,Set.Icc,Eq,Nat.totient,HAdd.hAddProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous12 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem erdos_1003.variants.Icc : answer(sorry) ↔ ∀ k ≥ 1, {n | ∀ i ∈ Set.Icc 1 k, φ n = φ (n + i)}.Infinite := by … - theorem
Erdos1003.erdos_1003.variants.eps87Erdős, Pomerance, and Sárközy [EPS87] proved that for all large $x$, the number of $n \leq x$ with $\phi(n) = \phi(n+1)$ is at most $$\frac{x}{\exp((\log x)^{1/3})}$$ sorry line 59 sourceSource documentation
Erdős, Pomerance, and Sárközy [EPS87] proved that for all large $x$, the number of $n \leq x$ with $\phi(n) = \phi(n+1)$ is at most $$\frac{x}{\exp((\log x)^{1/3})}$$.
[EPS87] Erd\H os, Paul and Pomerance, Carl and S'ark"ozy, Andr'as, On locally repeated values of certain arithmetic functions. {II}. Proc. Amer. Math. Soc. (1987), 1--7.
depends on sorry what this means
Statement uses:
Filter.Eventually,Real,LE.le,Set.ncard,Nat,Set.ofPred,Erdos1003.erdos_1003.variants.eps87.match_1,And,Eq,Nat.totient,HAdd.hAdd,OfNat.ofNat,HDiv.hDiv,Real.exp,HPow.hPow,Real.log,Filter.atTopProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous34 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem erdos_1003.variants.eps87 : ∀ᶠ x in atTop, {(n : ℕ) | (n ≤ x) ∧ φ n = φ (n + 1)}.ncard ≤ x / Real.exp ((x.log) ^ ((1 : ℝ) / 3)) := by …