# Ratio of consecutive divisors and average

Let $2\leq d_1 < d_2,...,d_l < n$ be all the proper nontrivial divisors of $n$. I like to understand how much these divisors deviates from each other. Here are two questions in this regard:

(1) What is the maximum of the set $\{d_i/d_{i-1}: 1\leq i \leq l\}$. Say it $M$.

Assume that you know the prime factorization of $n (= p_1^{\alpha_1}..p_r^{\alpha_r})$. Can I have a formula in terms of $p_i$'s and $\alpha_i$'s. One of the crude upper bound can be $p_r$ but this is really bad if $n$ has many distinct prime factors.

(2) Instead of maximum, the average may be more interesting and useful. So can we estimate the mean and variance,

$\mu = (1/l)\sum_{i=1}^{l}{d_i/d_{i-1}}$ and $\sigma^2 = (1/l)\sum_{i=1}^{l}{(d_i/d_{i-1}-\mu)^2}$

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I would ask Kevin Ford at the University of Illinois. A few years he published a long paper in the Annals about divisors in an interval. –  M Khan Dec 7 '12 at 13:55
But $H(x, y, z) - H(x-\Delta, y, z)$ can be computed only if $\Delta \geq x/\log^{10}{z}$, (as in Introduction, Theorem 2, in above mentioned paper). Then recovering the integer whose divisor we have to look at is not possible by this formula. –  Kamalakshya Dec 7 '12 at 16:37
This may be worth studying if n has many divisors. What significance do you see it having for n having fewer than 5 divisors? Or 10? Gerhard "Ask Me About System Design" Paseman, 2012.12.07 –  Gerhard Paseman Dec 7 '12 at 17:11
Also, assuming d_0 is 1, it is clear that M is at least p_1, where I assume the prime factors are indexed in order of increasing magnitude, and that for n with many factors, log of the mean will be close to if not equal to log of n divided by number of divisors of n. There may be some bizarre exceptions like a twice a large ppprime taken to a large power, but they should be easy to characterize. Gerhard "Ask Me About Gut Feeling" Paseman, 2012.12.07 –  Gerhard Paseman Dec 7 '12 at 17:21

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 $d(w) \asymp 1/w$.
In definition of $F(n)$ is $d\leq n$ ? –  Kamalakshya Dec 8 '12 at 2:11
Yes, $d|n$ implies $d\le n$ in the definition of $F$. –  Andreas Weingartner Dec 8 '12 at 2:17
small typo $\max_{1\le i < \tau(n)} \frac{d_{i+1}(n)}{d_i(n)} = \max_{\substack{0 < \beta_j \le \alpha_j \\ 0 \le \beta_{j+l} \le \alpha_{j+l}\\ 1 \le l}} \{ p_j p_j^{\beta_j}\cdots p_k^{\beta_k} \}/n$ –  Kamalakshya Dec 8 '12 at 4:30
I don't think there is a typo. Because you are looking for a maximum, you can take $\beta_i = \alpha_i$. –  Andreas Weingartner Dec 8 '12 at 4:40