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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FormalConjectures/ErdosProblems/

448.lean

Retained formal statement3 of 8

Erdős and Graham asked whether there is a good inequality for $\sum_{n \le x} \tau^+(n)$ . This was answered by Ford [Fo08], who proved $\sum_{n \le x} \tau^+(n) \asymp x \cdot \frac{(\log x)^{1 - \alpha}}{(\log\log x)^{3/2}},$ where $\alpha = 1 - \frac{1 + \log\log 2}{\log 2} = 0.08607\ldots$ .

FormalConjectures/ErdosProblems/448.leanErdos448.erdos_448.variants.ford3 linesExact file

1(fun x => ∑ n ∈ Finset.Icc 1 x, ↑(Erdos448.tauPlus n)) =Θ[Filter.atTop] fun x =>2  have α := 1 - (1 + Real.log (Real.log 2)) / Real.log 2;3  ↑x * Real.log ↑x ^ (1 - α) / Real.log (Real.log ↑x) ^ (3 / 2)

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