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6
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Tenenbaum [Sur un probleme de crible et ses applications (1986)] showed that
$$ F(n)/n=\max_{1\le i < \tau(n)} \frac{d_{i+1}(n)}{d_i(n)},$$
where $1=d_1(n)< \ldots < d_{\tau(n)}(n)=n$ is the increasing sequence of divisors, and
$F$ is given by $F(1)=1$ and $F(n)=\max \{ d P^{-}(d) : d|n,\, d>1 \}$ for $n\ge 2$, where $P^{-}(n)$ is the smallest prime divisor of $n$.
For $n=p_1^{\alpha_1}\cdots p_k^{\alpha_k}$, with $p_1<\ldots < p_k$, this leads to the formula
$$ \max_{1\le i < \tau(n)} \frac{d_{i+1}(n)}{d_i(n)}
= \max_{1 \le j \le k} \{ p_j p_j^{\alpha_j}\cdots p_k^{\alpha_k} \}/n$$
Let $D(x,t)=|\{n\le x: F(n)/n \le t\}|$. I showed [Integers with dense divisors, 2 (2004)] that $D(x,t)=x\, d(w) \{1+O(1/\log t) \}$, for $x\ge 3$, $x\ge t \ge \exp\{(\log(\log(x)))^{5/3+\varepsilon}\}$, where $w=\log x / \log t$ and $d(w)$ is a continuous function that can be expressed in terms of Dickman's function and which satisfies $d(w) \asymp 1/w$.
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5
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Tenenbaum [Sur un probleme de crible et ses applications (1986)] showed that
$$ F(n)/n=\max_{1\le i < \tau(n)} \frac{d_{i+1}(n)}{d_i(n)},$$
where $1=d_1(n)< \ldots < d_{\tau(n)}(n)=n$ is the increasing sequence of divisors, and
$F$ is given by $F(1)=1$ and $F(n)=\max \{ d P^{-}(d) : d|nd|n,\, d>1 \}$ for $n\ge 2$, where $P^{-}(n)$ is the smallest prime divisor of $n$.
For $n=p_1^{\alpha_1}\cdots p_k^{\alpha_k}$ this leads to the formula
$$ \max_{1\le i < \tau(n)} \frac{d_{i+1}(n)}{d_i(n)}
= \max_{1 \le j \le k} \{ p_j p_j^{\alpha_j}\cdots p_k^{\alpha_k} \}/n$$
Let $D(x,t)=|\{n\le x: F(n)/n \le t\}|$. I showed [Integers with dense divisors, 2 (2004)] that $D(x,t)=x D(x,t)=x\, d(w) \{1+O(1/\log t) \}$, for $x\ge 3$, $x\ge t \ge \exp\{(\log(\log(x)))^{5/3+\varepsilon}\}$, where $w=\log x / \log t$ and $d(w)$ is a continuous function that can be expressed in terms of Dickman's function and which satisfies $d(w) \asymp 1/w$.
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4
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Tenenbaum [Sur un probleme de crible et ses applications (1986)] showed that
$$ F(n)/n=\max_{1\le i \le < \tau(n)} \frac{d_{i+1}(n)}{d_i(n)},$$
where $1=d_1(n)< \ldots < d_{\tau(n)}(n)=n$ is the increasing sequence of divisors, and
$F$ is given by $F(1)=1$ and $F(n)=\max \{ d P^{-}(d) : d|n, d>1 \}$ for $n\ge 2$, where $P^{-}(n)$ is the smallest prime divisor of $n$.
This is easier to compute because it requires finding the maximum of only $\omega(n)$ quantities (if you fix For $p=P^{-}(d)$, you only need n=p_1^{\alpha_1}\cdots p_k^{\alpha_k}$ this leads to consider divisors $d$ that have all the prime divisors of formula
$n$ that are $ \ge p$).\max_{1\le i < \tau(n)} \frac{d_{i+1}(n)}{d_i(n)}
= \max_{1 \le j \le k} \{ p_j p_j^{\alpha_j}\cdots p_k^{\alpha_k} \}/n$$
Let $D(x,t)=|\{n\le x: F(n)/n \le t\}|$. I showed [Integers with dense divisors, 2 (2004)] that $D(x,t)=x d(w) \{1+O(1/\log t) \}$, for $x\ge 3$, $x\ge t \ge \exp\{(\log(\log(x)))^{5/3+\varepsilon}\}$, where $w=\log x / \log t$ and $d(w)$ is a continuous function that can be expressed in terms of Dickman's function and which satisfies $d(w) \asymp 1/w$.
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3
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Tenenbaum [Sur un probleme de crible et ses applications (1986)] showed that
$$ F(n)/n=\max_{1\le i \le \tau(n)} \frac{d_{i+1}(n)}{d_i(n)},$$
where $1=d_1(n)< \ldots < d_{\tau(n)}(n)=n$ is the increasing sequence of divisors, and
$F$ is given by $F(1)=1$ and $F(n)=\max \{ d P^{-}(d) : d|n, d>1 \}$ for $n\ge 2$, where $P^{-}(n)$ is the smallest prime divisor of $n$. This is easier to compute because it requires finding the maximum of only $\omega(n)$ quantities (if you fix $p=P^{-}(d)$, you only need to consider divisors $d$ that have all the prime divisors of $n$ that are $\ge p$).
Let $D(x,t)=|\{n\le x: F(n)/n \le t\}|$. I showed [Integers with dense divisors, 2 (2004)] that $D(x,t)=x d(w) \{1+O(1/\log t) \}$, for $x\ge 3$, $x\ge t \ge \exp\{(\log(\log(x)))^{5/3+\varepsilon}\}$, where $w=\log x / \log t$ and $d(w)$ is a continuous function that can be expressed in terms of Dickman's function and which satisfies $d(w) \asymp 1/w$.
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2
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Tenenbaum [Sur un probleme de crible et ses applications (1986)] showed that
the maximum you are looking for is $F(n)/n$, $ F(n)/n=\max_{1\le i \le \tau(n)} \frac{d_{i+1}(n)}{d_i(n)},$$
where $1=d_1(n)< \ldots < d_{\tau(n)}(n)=n$ is the increasing sequence of divisors, and
$F$ is given by $F(1)=1$ and $F(n)=\max \{ d P^{-}(d) : d|n, d>1 \}$ for $n\ge 2$, where $P^{-}(n)$ is the smallest prime divisor of $n$. This is easier to compute because it requires finding the maximum of only $\omega(n)$ quantities (if you fix $p=P^{-}(d)$, you only need to consider divisors $d$ that have all the prime divisors of $n$ that are $\ge p$).
Let $D(x,t)=|\{n\le x: F(n)/n \le t\}|$. I showed [Integers with dense divisors, 2 (2004)] that $D(x,t)=x d(w) \{1+O(1/\log t) \}$, where $d(w)$ is a continuous function that can be expressed in terms of Dickman's function and which satisfies $d(w) \asymp 1/w$.
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1
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Tenenbaum [Sur un probleme de crible et ses applications (1986)] showed that the maximum you are looking for is $F(n)/n$, where $F(1)=1$ and $F(n)=\max \{ d P^{-}(d) : d|n, d>1 \}$ for $n\ge 2$, where $P^{-}(n)$ is the smallest prime divisor of $n$. This is easier to compute because it requires finding the maximum of only $\omega(n)$ quantities (if you fix $p=P^{-}(d)$, you only need to consider divisors $d$ that have all the prime divisors of $n$ that are $\ge p$).
Let $D(x,t)=|\{n\le x: F(n)/n \le t\}|$. I showed [Integers with dense divisors, 2 (2004)] that $D(x,t)=x d(w) \{1+O(1/\log t) \}$, where $d(w)$ is a continuous function that can be expressed in terms of Dickman's function and which satisfies $d(w) \asymp 1/w$.
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