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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} then there is some constant c<1c < 1 such that for all large nn f(n)cloglogn. f(n) \leq c \log\log n.

[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.ub1 lineExact file
c < 1, ∀ᶠ (n : ℕ) in Filter.atTop, Erdos377.sumInvPrimesNotDvdCentralBinom nc * Real.log (Real.logn)
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