FormalConjectures/ErdosProblems/236.lean
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2 declarations, 1 contain sorry, 0 declared as axioms
Erdős Problem 236
Reference: erdosproblems.com/236
Imports (1)
- def
Erdos236.f$f(n)$ counts the number of solutions to $n=p+2^k$ for prime $p$ and $k\geq 0$ line 35 sourceSource documentation
$f(n)$ counts the number of solutions to $n=p+2^k$ for prime $p$ and $k\geq 0$.
standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses:
NatProof uses:
List.length,List.filter,Decidable.decide,Nat.Prime,HSub.hSub,HPow.hPow,OfNat.ofNat,List.range,HAdd.hAdd,Nat.log210 instances, coercions and auxiliary constants left out.
def f (n : ℕ) : ℕ := …
- theorem
Erdos236.erdos_236Let $f(n)$ count the number of solutions to $n=p+2^k$ for prime $p$ and $k\geq 0$ sorry line 42 sourceSource documentation
Let $f(n)$ count the number of solutions to $n=p+2^k$ for prime $p$ and $k\geq 0$. Show that $f(n)=o(\log n)$.
depends on sorry what this means
Statement uses:
Asymptotics.IsLittleO,Nat,Real,Filter.atTop, Erdos236.f,Real.logProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous,OfNat.ofNat9 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem erdos_236: (fun n => (f n : ℝ)) =o[atTop] (fun n => Real.log (n : ℝ)) := by …
Mentions: Erdos236.f