Erdős problem 448 · problems.science

Let $\tau(n)$ count the divisors of $n$ and $\tau^+(n)$ count the number of $k$ such that $n$ has a divisor in $[2^k, 2^{k+1})$ . Is it true that, for all $\epsilon > 0$ , $\tau^+(n) < \epsilon \cdot \tau(n)$ for almost all $n$ ?

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8 retained statements2415f78e850a

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FormalConjectures/ErdosProblems/

448.lean

Retained formal statement2 of 8

Quantitative form of the (negative) answer to erdos_448. Erdős and Tenenbaum [ErTe81] showed that the upper density of $\{n : \tau^+(n) < \epsilon \cdot \tau(n)\}$ is in fact $\asymp \epsilon^{1 - o(1)}$ , where the $o(1)$ in the exponent tends to $0$ as $\epsilon \to 0$ . Equivalently, $\log(\text{upper density}) / \log \epsilon \to 1$ as $\epsilon \to 0^+$ .

FormalConjectures/ErdosProblems/448.leanErdos448.erdos_448.variants.erdos_tenenbaum2 linesExact file

1Filter.Tendsto (fun ε => Real.log {n | ↑(Erdos448.tauPlus n) < ε * ↑n.divisors.card}.upperDensity / Real.log ε)2  (nhdsWithin 0 (Set.Ioi 0)) (nhds 1)

SolvedStatement only, no proof

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