User andreas weingartner - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T02:11:08Z http://mathoverflow.net/feeds/user/12947 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/115711/ratio-of-consecutive-divisors-and-average/115766#115766 Answer by Andreas Weingartner for Ratio of consecutive divisors and average Andreas Weingartner 2012-12-08T01:50:06Z 2012-12-08T03:53:01Z <p>Tenenbaum [Sur un probleme de crible et ses applications (1986)] showed that $$F(n)/n=\max_{1\le i &lt; \tau(n)} \frac{d_{i+1}(n)}{d_i(n)},$$ where $1=d_1(n)&lt; \ldots &lt; d_{\tau(n)}(n)=n$ 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 $p_1&lt;\ldots &lt; p_k$, this leads to the formula $$\max_{1\le i &lt; \tau(n)} \frac{d_{i+1}(n)}{d_i(n)} = \max_{1 \le j \le k} \{ p_j p_j^{\alpha_j}\cdots p_k^{\alpha_k} \}/n$$</p> <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 $d(w) \asymp 1/w$. </p> http://mathoverflow.net/questions/55381/partitioning-the-integers-1-through-n-so-that-the-product-of-the-elements-in/55424#55424 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 Andreas Weingartner 2011-02-14T17:11:09Z 2011-02-15T20:52:17Z <p><strong>First (failed) attempt:</strong></p> <p>For $n \equiv 0 \mod{3}$ or $n\equiv 2 \mod{3}$, there is a second solution: $P=\{ 1, 2, \frac{n(n+1)}{6}-1 \}$.</p> <p>Edit: As the comments point out, the last term is too large.</p> <p><strong>Second attempt:</strong></p> <p>When $n=12m+3$, a second solution is $P=\{ 1, 8m+1, 9m+2 \}$.</p> <p>When $n=30m+24$, a second solution is $P=\{ 1, 18m+14, 25m+19\}$.</p> <p>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. </p> <p>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.</p> http://mathoverflow.net/questions/33755/erdos-kac-for-sum-of-divisors/55273#55273 Answer by Andreas Weingartner for Erdos-Kac for sum of divisors Andreas Weingartner 2011-02-13T02:15:33Z 2011-02-13T02:15:33Z <p>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 $0.2474 &lt; F(2) &lt; 0.2480$. His method can be adapted to get bounds for other fixed values of $x$. </p> <p>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.</p> http://mathoverflow.net/questions/115711/ratio-of-consecutive-divisors-and-average/115766#115766 Comment by Andreas Weingartner Andreas Weingartner 2012-12-08T19:17:23Z 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. http://mathoverflow.net/questions/115711/ratio-of-consecutive-divisors-and-average/115766#115766 Comment by Andreas Weingartner Andreas Weingartner 2012-12-08T04:40:42Z 2012-12-08T04:40:42Z I don't think there is a typo. Because you are looking for a maximum, you can take $\beta_i = \alpha_i$. http://mathoverflow.net/questions/115711/ratio-of-consecutive-divisors-and-average/115766#115766 Comment by Andreas Weingartner Andreas Weingartner 2012-12-08T02:17:26Z 2012-12-08T02:17:26Z Yes, $d|n$ implies $d\le n$ in the definition of $F$. http://mathoverflow.net/questions/113959/a-function-whose-fixed-points-are-the-primes Comment by Andreas Weingartner Andreas Weingartner 2012-11-21T03:06:58Z 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 $n$.