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Erdős problem 69

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Is ∑n≥2ω(n)2n \sum_{n\geq 2}\frac{\omega(n)}{2^n} irrational? (Here ω(n)\omega(n) counts the number of distinct prime divisors of nn.)

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FormalConjectures/ErdosProblems/

69.lean

Retained formal statement2 of 2

Tao observed that erdos_69 is a special case of erdos_257, since ∑n≥2ω(n)2n=∑p12p−1. \sum_{n\geq 2}\frac{\omega(n)}{2^n} = \sum_p \frac{1}{2^p - 1}.

FormalConjectures/ErdosProblems/69.leanErdos69.erdos_69.variants.specialisation_of_erdos_2572 linesExact file
have A := {n | Nat.Prime n};∑' (n : ℕ), ↑(ArithmeticFunction.cardDistinctFactors (n + 2)) / 2 ^ (n + 2) = ∑' (p : ↑A), 1 / (2 ^ ↑p - 1)
SolvedStatement only, no proof