FormalConjectures/ErdosProblems/69.lean
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2 declarations, 2 contain sorry, 0 declared as axioms
Erdős Problem 69
Reference: erdosproblems.com/69
Imports (1)
- theorem
Erdos69.erdos_69Is $$ \sum_{n\geq 2}\frac{\omega(n)}{2^n} $$ irrational? sorry line 40 sourceSource documentation
Is $$ \sum_{n\geq 2}\frac{\omega(n)}{2^n} $$ irrational? (Here $\omega(n)$ counts the number of distinct prime divisors of $n$.)
depends on sorry what this means
Statement uses:
Irrational,tsum,Real,Nat,HDiv.hDiv,ArithmeticFunction,ArithmeticFunction.cardDistinctFactors,HAdd.hAdd,OfNat.ofNat,HPow.hPow,SummationFilter.unconditionalProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous46 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem erdos_69 : Irrational <| ∑' n, ω (n + 2) / 2 ^ (n + 2) := by …
- theorem
Erdos69.erdos_69.variants.specialisation_of_erdos_257Tao observed thaterdos_69is a special case oferdos_257, since $$ \sum_{n\geq 2}\frac{\omega(n)}{2^n} = \sum_p \frac{1}{2^p - 1}. $$ sorry line 50 sourceSource documentation
Tao observed that
erdos_69is a special case oferdos_257, since $$ \sum_{n\geq 2}\frac{\omega(n)}{2^n} = \sum_p \frac{1}{2^p - 1}. $$depends on sorry what this means
Statement uses:
Set,Nat,Set.ofPred,Nat.Prime,Eq,Real,tsum,HDiv.hDiv,ArithmeticFunction,ArithmeticFunction.cardDistinctFactors,HAdd.hAdd,OfNat.ofNat,HPow.hPow,SummationFilter.unconditional,HSub.hSub,Membership.memProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous60 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem erdos_69.variants.specialisation_of_erdos_257 : let A := { n : ℕ | n.Prime } …