## Archive for November, 2010

### Random update

November 23, 2010

The reason for the fact that is has been more than 2 weeks since my last update is simple; contrary to popular belief, I have been busy. With a few things actually. I have been a bit busy studying (who would’ve thought). I have been busy writing my first article (thank you very much), which will be finished end of the week, hopefully. I have been busy playing guitar (obviously, I suck). And, last but not least, I have been juggling like there was no tomorrow (again, I suck). When I’ll improve, I might put some videos of me juggling here. That ought to be fun 🙂 My personal bests are, as of this moment:

3 balls: 12 minutes (close to 2500 catches)
4 balls: 57 seconds (over 200 catches)
5 balls: 15 catches
6 balls: 6 catches
7+ balls: err, no

I’ll update you on all of these things as soon as I’ve got some news.

EDIT 25 november: Woohoo! I flashed 6 balls 😀
EDIT 30 november: I updated a few personal bests. As of tomorrow I won’t do that anymore to be able to track my progress. But right now I’m too happy with my clean run of 15 😀

### PGOM LD AD LD CD

November 5, 2010

A variant on a well-known poem;

The title (basically) said it and I again will say
I’m in my prime; today it’s my 21st birthday!

### Erdösproblem 3, part III

November 4, 2010

So far I found about $10$ more, what I like to call, NP-primitive counterexamples to the question asked here. When do I call a counterexample NP-primitive? When it’s easy to show (by the reasoning from the previous post) that it’s indeed a counterexample and if no (strict) subset of it is a NP-primitive counterexample. And given a NP-primitive counterexample, it’s easy to construct other easy to check counterexamples. The question now is of course: are there infinitely many NP-primitive counterexamples? Conceivably, this is not too hard. The ones I found so far:

$\{2,3,4,5,6,8,9,10,12,15,16\}$
$\{2,3,4,5,6,7,8,9,15,16,24\}$
$\{2,3,4,5,6,7,8,9,10,15,230\}$
$\{2,3,4,5,6,8,9,10,12,15,18,180\}$
$\{2,3,4,5,6,7,8,9,14,15,31,2474\}$
$\{2,3,4,5,6,7,8,9,11,15,75,8153\}$
$\{2,3,4,5,6,7,8,9,12,13,93,44155\}$
$\{2,3,4,5,6,7,8,10,12,14,60,105\}$
$\{2,3,4,5,6,7,8,10,12,15,42,210,420\}$
$\{3,4,5,6,7,8,9,10,11,12,13,14,15,16,18,21,72,453\}$
$\{3,4,5,6,7,8,9,10,11,12,14,15,16,18,20,21,30,154,315\}$
$\{3,4,5,6,7,8,9,10,11,12,13,14,15,18,20,24,30,840,450450\}$

And you can even make more by e.g. replacing $60$ in the fifth to last set with $120$, $140$ and $840$. A more general, and to me more sensible, conjecture of Erdös, Graham and Spencer is the following:

If $\displaystyle \sum_{i=1}^k \dfrac{1}{x_i} < N - \dfrac{1}{m}$, then the $x_i$ can be split into $N$ subsets
$A_1$, $A_2$, .., $A_n$ such that for all $j$ with $1 \le j \le N$ we have: $\displaystyle \sum_{a_i \in A_j} \dfrac{1}{a_i} \le 1$, as long as $m$ $\le 30$.

This is known for $m = \dfrac{7}{2}$ (see here) and $\{2,3,3,5,5,5,5\}$ shows that $m = 30$ is tight. Very probably this conjecture is cute enough to deserve a closer look at sometime.

EDIT: A counterexample to this last conjecture was found by Song Guo (Electronic J. of Combinatorics 15 (2008), #N43). You can find it online, if you’re interested. That paper also refers to other articles about these conjectures. The main question of finding the maximal $m$ is still open, in any case

### Erdösproblem 3, part II

November 3, 2010

I checked my dictionary, but apparently there is no english word for the dutch word ‘tangconstructie’. Anyway, remember how I, rhetorically, asked ‘how else could it be?’ yesterday? Well, I am more than delighted to say that I am able to give a counterexample to the metaconjecture that every proof of a counterexample to the question of Erdös and Graham I talked about in the previous post, works via a case-by-case-argument.

Consider the set $X = \{2,3,4,5,6,8,9,10,12,15,16\}$ and let $lcm\{S\}$ be the least common multiple of all members of $S$, for a set $S$ of positive integers. Then $\displaystyle \sum_{x_i \in X} \dfrac{1}{x_i} = 2 - \dfrac{1}{lcm\{X\}}$. Now, assume we have a partition of $X$ in two sets $A$ and $B$ with $\displaystyle \sum_{a_i \in A} \dfrac{1}{a_i} < 1$ and $\displaystyle \sum_{b_i \in B} \dfrac{1}{b_i} < 1$. Then in particular we have:

$2 - \dfrac{1}{lcm\{X\}}$
$= \displaystyle \sum_{x_i \in X} \dfrac{1}{x_i}$
$= \displaystyle \sum_{a_i \in A} \dfrac{1}{a_i}$ $+$ $\displaystyle \sum_{b_i \in B} \dfrac{1}{b_i}$
$\le 1 - \dfrac{1}{lcm\{A\}} + 1 - \dfrac{1}{lcm\{B\}}$
$\le 2 - \dfrac{2}{lcm\{X\}}$

And boom goes the dynamite.

### Solving a third Erdösproblem (?)

November 2, 2010

In their monograph on problems in combinatorial number theory, Erdös and Graham ask the following question:

Is it true that if $1 < x_1 < x_2 < .. < x_k$ and $\displaystyle \sum_{i=1}^k \dfrac{1}{x_i} < 2$ then there exist $\epsilon_i = 0$ or $1$ so that $\displaystyle \sum_{i=1}^k \dfrac{\epsilon_i}{x_i} < 1$ and $\displaystyle \sum_{i=1}^k \dfrac{1-\epsilon_i}{x_i} < 1$?

And I can’t find a way to do this where the $x_i$‘s are $\{2,3,4,5,6,8,9,11,12,13,16\}$. And, if this indeed is a counterexample, I’m bold enough to conjecture that it’s the easiest. $\{2,3,4,5,6,7,8,11,12,13,33\}$ could very well be a candidate too, but I haven’t checked that one thoroughly.

Edit: I now proved that the first one of these is indeed a counterexample, but of course, the proof is a case-by-case-argument (how else could it be?), so I won’t put it here, because of uglyness.

### Weighed cointossing

November 2, 2010

Assume $S = \{s_1, s_2, .. \}$ is a set with $1 \le$ $s_1$ $\le s_2 \le ..$, such that $\displaystyle \sum \dfrac{1}{s_i}$ diverges.  Let $a_i = \pm 1$, each with probability $\dfrac{1}{2}$. What can we say about $\displaystyle \sum_{i=1}^n \dfrac{a_i}{s_i}$ when $n$ goes to infinity? Must there, for example, be infinitely many sign changes with probability $1$? It can’t be hard to show this for bounded $s_i$, but what about $S = \aleph$ for example? I honestly have no clue whatsoever.

EDIT 22 Februari 2011
I can show that there is a positive probability that $\left|\displaystyle \sum_{i=1}^n \dfrac{a_i}{s_i}\right| > c$ for any $c$, for $n$ large enough; Let $k = k(c)$ be the smallest integer such that $\displaystyle \sum_{i=1}^k \dfrac{1}{s_i} > c$ (note that this $k$ exists since the sum of $\dfrac{1}{s_i}$ is assumed to go to infinity).
Then either the probability that $\displaystyle \sum_{i=k+1}^n \dfrac{a_i}{s_i}$ $\ge 0$ is at least $\dfrac{1}{2}$ or the probability that $\displaystyle \sum_{i=k+1}^n \dfrac{a_i}{s_i}$ $\le 0$ is at least $\dfrac{1}{2}$ (*). Assume the first case (second case goes analogous), then we have: $P\left(\displaystyle \sum_{i=1}^n \dfrac{a_i}{s_i} > c\right) \ge P\left(\displaystyle \sum_{i=k+1}^n \dfrac{a_i}{s_i} \ge 0\right) \displaystyle \prod_{j=1}^k P(a_j = 1) \ge \dfrac{1}{2}\dfrac{1}{2^k} = \dfrac{1}{2^{k+1}} > 0$

(*) It is fun to note that both are the case, independent of $S$. This can be used to show that $P\left(\left|\displaystyle \sum_{i=1}^n \dfrac{a_i}{s_i}\right| > c\right) \ge \dfrac{1}{2^k}$ for all $n$ $\ge$ $k$