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Erdős problem 377

Is there some absolute constant C>0C > 0 such that pn1p(2nn)1pC \sum_{p \leq n} 1_{p\nmid {2n \choose n}}\frac{1}{p} \leq C for all nn?

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FormalConjectures/ErdosProblems/

377.lean

Retained formal statement4 of 5

Erdos, Graham, Ruzsa, and Straus proved that if f(n)=pn1p(2nn)1p f(n) = \sum_{p \leq n} 1_{p\nmid {2n \choose n}}\frac{1}{p} and γ0=k=2logk2k \gamma_0 = \sum_{k = 2}^{\infty} \frac{\log k}{2^k} then limx1xnxf(n)2=γ02 \lim_{x\to\infty} \frac{1}{x}\sum_{n\leq x} f(n)^2 = \gamma_0^2

[EGRS75] Erdős, P. and Graham, R. L. and Ruzsa, I. Z. and Straus, E. G., _On the prime factors of (2nn)\binom{2n}{n}_. Math. Comp. (1975), 83-92.

FormalConjectures/ErdosProblems/377.leanErdos377.erdos_377.variants.limit.ii4 linesExact file
∀ (γ₀ : ℝ),  γ₀ = ∑' (k : ℕ), Real.log (↑k + 2) / 2 ^ (k + 2) →    Filter.Tendsto (fun x => 1 / ↑x * ∑ nFinset.Icc 1 x, Erdos377.sumInvPrimesNotDvdCentralBinom n ^ 2) Filter.atTop      (nhds (γ₀ ^ 2))
SolvedStatement only, no proof

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