User andreas weingartner - MathOverflow

Answer by Andreas Weingartner for Ratio of consecutive divisors and average

2012-12-08T01:50:06Z

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 $d(w) \asymp 1/w$.

Answer by Andreas Weingartner for Partitioning the integers $1$ through $n$ so that the product of the elements in one set is equal to the sum of the elements in the other

2011-02-14T17:11:09Z

First (failed) attempt:

For $n \equiv 0 \mod{3}$ or $n\equiv 2 \mod{3}$, there is a second solution: $P=\{ 1, 2, \frac{n(n+1)}{6}-1 \} $.

Edit: As the comments point out, the last term is too large.

Second attempt:

When $n=12m+3$, a second solution is $P=\{ 1, 8m+1, 9m+2 \}$.

When $n=30m+24$, a second solution is $P=\{ 1, 18m+14, 25m+19\}$.

We can generate an infinite family of such solutions as follows. We look for solutions of the form $P=\{1, a, b\}$, which leads to the equation $$ \frac{n(n+1)}{2}= 1ab+a+b+1=(a+1)(b+1) .$$ One solution of this is the original solution given in the question. We can get other solutions by exchanging divisors of (a+1) and (b+1) while keeping both factors $\le n+1$. For example, assuming $n$ is odd and $3|n$ and $2|\frac{n+1}{2}$, we exchange the divisors $2$ and $3$ to get the new solution $a+1=2n/3$ and $b+1=3(n+1)/4$. This leads to the solution for $n=12m+3$ given above. Assuming $n$ is even, and exchanging the factors $3$ and $5$ leads to the other solution given above.

This way we can generate an infinite number of linear congruences for $n$ with corresponding solutions $P$. The question is what proportion of integers is covered by all these congruences. When $n$ and $\frac{n+1}{2}$ are both prime, which is probably true for infinitely many $n$, this method does not generate a second solution.

Answer by Andreas Weingartner for Erdos-Kac for sum of divisors

2011-02-13T02:15:33Z

Let $F(x)$ be the proportion of integers $n$ such that $\frac{\sigma(n)}{n} \ge x$. Deleglise [Experiment. Math. 7 (1998), no. 2, 137-143] showed that $F(2)$, the density of abundant numbers, satisfies $0.2474 < F(2) < 0.2480$. His method can be adapted to get bounds for other fixed values of $x$.

As $x\to \infty$, $F(x)\to 0$ very rapidly. In [Proc. Amer. Math. Soc. 135 (9) (2007) 2677–2681] I showed that $$F(x) = \exp(-e^{x e^{-\gamma}} (1+O(x^{-2}))),$$ where $\gamma$ is Euler's constant. This result also holds for the distribution function of $\frac{n}{\varphi(n)}$, where $\varphi(n)$ is Euler's totient function.

Comment by Andreas Weingartner

2012-12-08T19:17:23Z

We have $D(x,t) \asymp x \log t / \log x$ for $x \ge t \ge 2$, a result due to Saias. So for fixed $t$, the integers with $F(n)/n \le t$ have zero density.

Comment by Andreas Weingartner

2012-12-08T04:40:42Z

I don't think there is a typo. Because you are looking for a maximum, you can take $\beta_i = \alpha_i$.

Comment by Andreas Weingartner

2012-12-08T02:17:26Z

Yes, $d|n$ implies $d\le n$ in the definition of $F$.

Comment by Andreas Weingartner

2012-11-21T03:06:58Z

Another problem of the same flavor (and just as difficult) is $f(n)=\sigma(n)-1$, where $\sigma(n)$ is the sum of the divisors of $n$.