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Erdős problem 205

Is it true that all sufficiently large nn can be written as 2k+m2^k+m for some k≥0k\geq 0, where Ω(m)<log⁡log⁡m\Omega(m)<\log\log m? (Here Ω(m)\Omega(m) is the number of prime divisors of mm counted with multiplicity.)

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FormalConjectures/ErdosProblems/

205.lean

Retained formal statement1 of 5

Is it true that all sufficiently large nn can be written as 2k+m2^k+m for some k≥0k\geq 0, where Ω(m)<log⁡log⁡m\Omega(m)<\log\log m? (Here Ω(m)\Omega(m) is the number of prime divisors of mm counted with multiplicity.)

Barreto and Leeham, using ChatGPT and Aristotle, have proved a negative answer, which was quantified by Tao and Alexeev (see the comments): in fact there are infinitely many nn such that, for all kk with 2k<n2^k<n, n−2kn-2^k has at least ≫(log⁡nlog⁡log⁡n)1/2\gg \left(\frac{\log n}{\log\log n}\right)^{1/2} many prime factors.

FormalConjectures/ErdosProblems/205.leanErdos205.erdos_205.parts.i1 lineExact file
False ↔ ∀ᶠ (n : ℕ) in Filter.atTop, Erdos205.IsRepresentable (fun m => Real.log (Real.log ↑m)) n
SolvedStatement only, no proof