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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

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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 for almost all integers f(m)=γ0+o(1)f(m) = \gamma_0 + o(1).

[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.ae5 linesExact file
∀ (γ₀ : ℝ),  γ₀ = ∑' (k : ℕ), Real.log (↑k + 2) / 2 ^ (k + 2) →o,      ∃ (_ : Filter.Tendsto o Filter.atTop (nhds 0)),        ∀ᶠ (n : ℕ) in Filter.cofinite, Erdos377.sumInvPrimesNotDvdCentralBinom n = γ₀ + o n
SolvedStatement only, no proof

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