That’s all that I can see, which doesn’t compare well with the nine particularly mathematical New Years Honours this year. Does anyone have any to add?

More information

UK Government Birthday Honours lists 2013.

New Zealand Birthday Honours lists 2013.

2013 Birthday Honours on Wikipedia.

In case you’re new to this: every now and then I encounter a paper or a book or an article that grabs my interest but isn’t directly useful for anything. It might be about some niche sub-sub-subtopic I’ve never heard of, or it might talk about something old from a new angle, or it might just have a funny title. I put these things in my Interesting Esoterica collection on Mendeley. And then when I’ve gathered up enough, I collect them here.

In this post the titles are links to the original sources, and I try to add some interpretation or explanation of why I think each thing is interesting below the abstract.

Some things might not be freely available, or even available for a reasonable price. Sorry.

Problems to sharpen the young

An annotated translation of Propositiones ad acuendos juvenes, the oldest mathematical problem collection in Latin, attributed to Alcuin of York.

This came up when we talked to David Singmaster. Alcuin was a cool dude! Closed access on JSTOR, $14.

Das 2:3-Ei – ein praktikables Eimodell

Or, “The 2:3 egg – a practical egg model”. Lengthy discussion (in German) about what makes a good egg shape, and how to construct one.

Constructing the Tits ovoid from an elliptic quadratic

This came up while I was searching for information on egg construction.

A smaller sleeping bag for a baby snake

By a sleeping bag for a baby snake in $d$ dimensions we mean a subset of $R^d$ which can cover, by rotation and translation, every curve of unit length. We construct sleeping bags which are smaller than any previously known in dimensions 3 and higher. In particular, we construct a three-dimensional sleeping bag of volume approximately 0.075803. For large $d$ we construct $d$-dimensional sleeping bags with volume less than $\frac{(c \sqrt{ \log d})^d}{d^{3d/2}}$ for some constant $c$.

To obtain the last result, we show that every curve of unit length in $R^d$ lies between two parallel hyperplanes at distance at most $c_1 d^{-3/2} \sqrt{\log d}$, for some constant $c_1$.

David Cushing told us about this puzzle at Newcastle MathsJam. This paper is also available in added-value form from Springer for £29.95.

The Urinal Problem

A man walks into a men’s room and observes n empty urinals. Which urinal should he pick so as to maximize his chances of maintaining privacy, i.e., minimize the chance that someone will occupy a urinal beside him? In this paper, we attempt to answer this question under a variety of models for standard men’s room behavior. Our results suggest that for the most part one should probably choose the urinal furthest from the door (with some interesting exceptions). We also suggest a number of variations on the problem that lead to many open problems.

Good knowledge to have, and I’m glad someone’s thinking about it.

Circuitry in 3d chess

This is the second of a projected three-part series of articles, which will ultimately prove the Turing-completeness of three-dimensional chess. In the first article, I described the basic rules of the game. In this article, I shall show how to construct basic logic gates using rooks, kings and pawns. The final article, which demonstrates Turing completeness, requires a fourth piece, namely the knight.

Familial Sinistrals Avoid Exact Numbers

We report data from an internet questionnaire of sixty number trivia. Participants were asked for the number of cups in their house, the number of cities they know and 58 other quantities. We compare the answers of familial sinistrals – individuals who are left-handed themselves or have a left-handed close blood-relative – with those of pure familial dextrals – right-handed individuals who reported only having right-handed close blood-relatives. We show that familial sinistrals use rounder numbers than pure familial dextrals in the survey responses. Round numbers in the decimal system are those that are multiples of powers of 10 or of half or a quarter of a power of 10. Roundness is a gradient concept, e.g. 100 is rounder than 50 or 200. We show that very round number like 100 and 1000 are used with 25% greater likelihood by familial sinistrals than by pure familial dextrals, while pure familial dextrals are more likely to use less round numbers such as 25, 60, and 200. We then use Sigurd’s (1988, Language in Society) index of the roundness of a number and report that familial sinistrals’ responses are significantly rounder on average than those of pure familial dextrals. To explain the difference, we propose that the cognitive effort of using exact numbers is greater for the familial sinistral group because their language and number systems tend to be more distributed over both hemispheres of the brain. Our data support the view that exact and approximate quantities are processed by two separate cognitive systems. Specifically, our behavioral data corroborates the view that the evolutionarily older, approximate number system is present in both hemispheres of the brain, while the exact number system tends to be localized in only one hemisphere.

A Do-It-Yourself Paper Digital Computer, 1959.

I made one of these. It didn’t really work.

Division of labor in child care: A game-theoretic approach

This paper uses a game of repeated play to model parental child care in order to examine the gap between the expectations of egalitarian-minded couples before the transition to parenthood and the reality of parenthood, with its gendered roles. This is done first in a gender-free context in order to examine the mechanism by which the division of labor is established in a family – it is this same process through which gendered expectations have an impact. The analysis shows that it is difficult to achieve the equilibrium of equal sharing of child care, even when this is the preference of the parents. This leads to a discussion of alterations and meta-strategies for couples who want to share care equally. Gender differences between parents are also modeled, including the impact these have on outcomes and equilibria.

Cyclic twill-woven objects

Classical (or biaxial) twill is a textile weave in which the weft threads pass over and under two or more warp threads, with an offset between adjacent weft threads to give an appearance of diagonal lines. This paper introduces a theoretical framework for constructing twill-woven objects, i.e., cyclic twill-weavings on arbitrary surfaces, and it provides methods to convert polygonal meshes into twill-woven objects. It also develops a general technique to obtain exact triaxial-woven objects from an arbitrary polygonal mesh surface.

Richard Green posted about this paper on Google+. Closed access, Elsevier, $31.50.

Kindergarten Quantum Mechanics

These lecture notes survey some joint work with Samson Abramsky as it was presented by me at several conferences in the summer of 2005. It concerns `doing quantum mechanics using only pictures of lines, squares, triangles and diamonds’. This picture calculus can be seen as a very substantial extension of Dirac’s notation, and has a purely algebraic counterpart in terms of so-called Strongly Compact Closed Categories (introduced by Abramsky and I in quant-ph/0402130 and [4]) which subsumes my Logic of Entanglement quant-ph/0402014. For a survey on the `what’, the `why’ and the `hows’ I refer to a previous set of lecture notes quant-ph/0506132. In a last section we provide some pointers to the body of technical literature on the subject.

Using Monoidal Categories in the Transformational Study of Musical Time-Spans and Rhythms

Transformational musical theory has so far mainly focused on the study of groups acting on musical chords, one of the most famous example being the action of the dihedral group D24 on the set of major and minor chords. Comparatively less work has been devoted to the study of transformations of time-spans and rhythms. D. Lewin was the first to study group actions on time-spans by using a subgroup of the affine group in one dimension. In our previous work, the work of Lewin has been included in the more general framework of group extensions, and generalizations to time-spans on multiple timelines have been proposed. The goal of this paper is to show that such generalizations have a categorical background in free monodical categories generated by a group-as-category. In particular, symmetric monodical categories allow to deal with the possible interexchanges between timelines. We also show that more general time-spans can be considered, in which single time-spans are encapsulated in a “bracket” of time-spans, which allow for the description of complex rhythms.

I must look more closely into the combination of group theory and music theory at some point.

Six Ways to Sum a Series

A discussion of the sum of squares of the recipricals of the positive integers with a review of several proofs.

The Ubiquitous $\pi$

Some well-known and little-known appearances of $\pi$ in a variety of problems.

Closed access, JSTOR, $12.

Non-sexist solution of the ménage problem

The ménage problem asks for the number of ways of seating $n$ couples at a circular table, with men and women alternating, so that no one sits next to his or her partner. We present a straight-forward solution to this problem. What distinguishes our approach is that we do not seat the ladies first.

The greatest common divisor (gcd) of two or more integers is the greatest integer that evenly divides those integers. For example, the gcd of $8$ and $12$ is $4$ (usually written as $\gcd(8,12)=4$). Two integers are called coprime (or “relatively prime”) if their gcd is equal to $1$.

A reasonable question to ask is,

Given two randomly chosen integers $a$ and $b$, what is the probability that $\gcd(a,b)=1$?

“Randomly chosen integers” is a bit of a dodgy term but we’ll just ignore that and hope for the best.

If no prime divides $n$, then $n=1$, by the fundamental theorem of arithmetic. So we want to find the probability that no prime divides $\gcd(a,b)$. Let $a$ and $b$ be two randomly chosen integers and $p$ a fixed prime. If $p$ divides both $a$ and $b$ then $\gcd(a,b)\geq p$. We are interested in the probability that this does not occur for any $p$.

Now I need to say what I mean by “randomly chosen”. It’s very handwavey, but we will calculate the probability for an integer chosen uniformly at random between $1$ and $N$, and then let $N$ tend to infinity. We won’t actually do that, but that’s what we’re thinking of.

Every $p$^th integer is divisible by $p$, so the probability that $p$ divides $a$ and $b$ is $\frac{1}{p^{2}}$, and the probability that it divides at most one of $a$ or $b$ is $1-\frac{1}{p^{2}}$. That’s a good start, but what we want is this probability over all primes. The probabilities for each prime are independent of each other, so you can just multiply them all together. Therefore the probability that $\gcd(a,b)=1$ is

\[ \Pr \left( \gcd(a,b) = 1 \right) = \prod_{p \textrm{ prime}} \left( 1 - \frac{1}{p^2} \right). \]

We now have the problem of evaluating this number. For those in the know, our expression looks pretty darn close to the definition of the Riemann zeta function:

\[ \zeta(s) =\prod_{p \textrm{ prime}} \left( 1 - \frac{1}{p^s} \right)^{-1} = \sum_{n=1}^{\infty} n^{-s}. \]

(We used the Euler product formula to relate the product we had with the usual definition of the zeta function, which is a sum.)

By comparing the two equations, we can see that $\Pr( \gcd(a,b) = 1 ) = \frac{1}{\zeta(2)}$. Thankfully, $\zeta(2)$ is simply the sum of the reciprocals of the square numbers, which we know is $\frac{\pi^2}{6}$¹. Therefore the probability that two randomly chosen integers are coprime is the reciprocal of that: $\frac{6}{\pi^2}$. So we are done!

\[ \Pr( \gcd(a,b) = 1) = \frac{6}{\pi^2} \approx 0.6079 \]

Well, sort of. Now that we know the probability the gcd of two randomly chosen integers is $1$, why not ask a few more general questions? What is the probability that the gcd of two randomly chosen integers is a prime number $p$? More generally, what is the probability that the gcd of two randomly chosen integers is some $k \in\mathbb{N}$, and what’s the probability that it’s any prime?

First of all, what’s the probability that $\gcd(a,b)$ is a prime $p$? That means:

• $p$ divides both $a$ and $b$
• $p^2$, and any other prime, only divide at most one of $a$ and $b$.

As we said before, $\Pr(p \mid \gcd(a,b)) = \frac{1}{p^2}$. We want to exclude the cases when $p^2$ divides both $a$ and $b$, which happens $\frac{1}{p^4}$ of the time. For any other prime $q$, we want the same thing as before: $\Pr(q \nmid \gcd(a,b)) = 1 – \frac{1}{q^2}$. So, putting it all together, we have:

\begin{align} \Pr( \gcd(a,b) = p ) &= \left( \frac{1}{p^2} – \frac{1}{p^4} \right) \prod_{q \neq p \textrm{ prime}} \left( 1 – \frac{1}{q^2} \right) \\ &= \frac{1}{p^2} \left(1 – \frac{1}{p^2} \right) \prod_{q \neq p \textrm{ prime}} \left( 1 – \frac{1}{q^2} \right) \\ &= \frac{1}{p^2} \prod_{q \textrm{ prime}} \left( 1 – \frac{1}{q^2} \right) \\ &= \frac{6}{p^2\pi^2}.  \end{align}

Next, we can apply a similar argument to find the probability that $\gcd(a,b) = p^n$. $p^n$ divides both numbers $\frac{1}{\left( p^n \right)^2}$ of the time, and we need to exclude the case where $p^{n+1}$ divides both numbers, which happens $\frac{1}{\left( p^{n+1} \right)^2}$ of the time. Like last time, we can take a factor of $\frac{1}{p^{2n}}$ out of this, to get:

\begin{align} \Pr( \gcd(a,b) = p^n ) &= \frac{1}{p^{2n}} \prod_{q \textrm{ prime}} \left( 1 – \frac{1}{q^2} \right) \\ &= \frac{6}{\left( p^n \right)^2\pi^2} \end{align}

It should be clear that you can take a couple of steps from here to derive that, for any $k \in \mathbb{N}$,

\[ \Pr( \gcd(a,b) = k ) = \frac{6}{k^2 \pi^2} \]

So what’s the probability that the gcd of two randomly chosen integers is any prime? Well, that’s simply

\begin{align} \sum_{p \textrm{ prime}} \Pr(\gcd(a,b)=p) &= \sum_{p \textrm{ prime}} \frac{6}{p^2 \pi^2} \\ &= \frac{6}{\pi^2} \cdot \sum_{p \textrm{ prime}} \frac{1}{p^2} \end{align}

To proceed any further we need to evaluate the sum of the reciprocals of the squares of the primes. When we did that with all the natural numbers, we used $\zeta(2)$. It turns out that there’s a related function which sums just over the primes, called the prime zeta function:

\[ P(s)=\sum_{p \textrm{ prime}} p^{-s}. \]

Replacing our sum with $P(2)$, we get:

\[  \Pr(\gcd(a,b) \textrm{ prime}) = \frac{6}{\pi^2} P(2) \approx \frac{6}{\pi^{2}} \times 0.45224 \approx 0.2749 \approx\frac{1}{4}. \]

So about a quarter.

Let’s end with one more generalisation. Let’s say we have 3 randomly chosen integers. No, scrap that. Let’s say we have 4 randomly chosen integers. No, that’s still not right… I know – let’s say we have $n$ randomly chosen integers!

Using similar arguments as above we can get that the probability of the gcd of $n$ integers chosen at random to be prime is \[ \frac{P(n)}{\zeta(n)}.\]

A nice simple(ish) answer and we are all done.

References

Riemann zeta function

Proof of the Euler product formula for the Riemann zeta function

Prime zeta function

There are three different proofs that $\sum_{n} \frac{1}{n^2} = \frac{\pi^2}{6}$ in Proofs from THE BOOK by Martin Aigner and Günter Ziegler. This was known as the Basel problem. Some more history is given in the comprehensive historical survey The Ubiquitous π by Dario Castellanos.

The Wolfram MathWorld page on “relatively prime” contains the excellent observation that $\frac{6}{\pi^2}$ is the fraction of points in the 2d lattice that you can see when standing at the origin. It also gives probabilities for pairs of numbers chosen from a few different principal ideal domains.

“On the distribution of the greatest common divisor” by Persi Diaconis and Paul Erdős does all this rigorously, with error terms and everything.

1. This infinite sum is the subject of the Basel problem, solved by Euler in 1735

Andy has a problem. He can’t solve it on his own – he needs your help. This problem vexed Andy so much that he spent four years trying to solve it on his own, to no avail. It really is a very difficult problem. Finally in 1997, out of what must have been sheer desperation, Andy reached out to his fellow man: maybe some kindly type out there could find a solution to his problem, which he would gladly reward with a small consideration.

Can you help a soul in need?

His name is D. Andrew “Andy” Beal, billionaire. He owns a mansion and a yacht¹. He’s a businessman from Dallas Texas, he’s worth 8.5 billion dollars, and he has previously spent that money on playing poker, almost launching rockets into space and avoiding prosecution for filing phony tax losses.

The news story is, “Andrew Beal, billionaire and mostly terrible person, will pay money to whoever solves this problem, which he thinks he thought of first but actually didn’t, all despite the complete lack of evidence, or even in spite of the abundance of evidence to the contrary, that mathematicians are motivated or even can be motivated by monetary reward.”

In 1997 an article written by Texas mathematician R. Daniel Mauldin was published in the Notices of the AMS, announcing both the problem and a prize of \$5,000 for solving it, offered by Andrew Beal. The problem is this:

Let $A$, $B$, $C$, $x$, $y$ and $z$ be positive integers, with $x,y,z \gt 2$. If \[ A^x + B^y = C^z,\] then $A$, $B$ and $C$ have a common prime factor.

It’s a fairly old problem and usually called the Tijdeman-Zagier conjecture, but Beal likes to call it “The Beal Conjecture”. Nobody has yet claimed the prize, and Beal has increased the prize fund several times over the years. If you’ve been sitting on a solution, maybe a million dollars can tempt you to reveal it? Last week, it was announced that Beal has increased the prize to \$1,000,000.

If you don’t have a solution but the prospect of a million dollars has bought your attention, first of all what’s wrong with you, and secondly, an overview paper by Noam Elkies on $ABC$-type problems is a good place to start your investigations.

An unattributed website on the conjecture and prize, bealconjecture.com, is registered to “Beal Aerospace Technologies, Inc.”, a company which was going to do private satellite launches. bealconjecture.com contains a digression complaining about Granville et al’s comments re the attribution of the conjecture, and bealaerospace.com contains two letters whinging about NASA and that hard-working schmoes like Dennis Tito had to pay Russia to have a go on the ISS.

It isn’t unheard of for prizes to be offered to solve a mathematical conjecture – in the late 19th century the King of Sweden offered a prize for a solution to the $n$-body problem; there are the Clay Millennium prizes, offered by another American businessman and famously turned down by Perelman; and Erdős very often handed out small amounts of money for solutions to problems. What those all have in common is that the bounty offered didn’t increase, and what I object to with Beal is this idea he apparently has that he just needs to find the market price of a solution and it will appear.

Take it away, Pink Floyd!

More information

The Beal Prize information page at the American Mathematical Society.

Beal’s own page about the conjecture.

A question titled “Status of Beal, Granville, Tijdeman-Zagier Conjecture” at MathOverflow.

A Generalization of Fermat’s Last Theorem: The Beal Conjecture and Prize Problem by R. Daniel Mauldin in Notices of the AMS.

1. probably

When Yitang Zhang unexpectedly announced a proof that that there are infinitely many pairs of primes less than 70 million apart from each other – a step on the way to the twin primes conjecture – certain internet wags amused themselves and a minority of others with the question, “is it a bigger jump from infinity to 70 million, or from 70 million to 2?”.

Of course the answer is that it’s a really short distance from 70 million to 2, and here’s my evidence: the bound of 70 million has in just over three weeks been reduced to just a shade over 100,000.

There were two nice things about Yitang Zhang’s paper: he explains the techniques he used very well; and the bound of 70,000,000 that he announced could be whittled down fairly easily – the big achievement was showing that there is a bound at all, never mind how big it is. In fact, Zhang rounded his actual bound of 63,374,611 up to 70 million, because it doesn’t really matter.

But if there is wiggle room, people will inevitably wiggle. Straight away, mathematicians around the world set to work  fine-tuning the bounds on the various quantities involved in Zhang’s proof. The core result is that if a set $S$ of $3.5 \times 10^6$ numbers is admissible, then there are infinitely many $n$ such that $n+S$ contains at least two primes. If the difference between the largest and the smallest elements of the admissible set is $k$, then you know that there are infinitely many prime numbers at most $k$ apart. So constructing a narrower admissible set is the name of the game.

As far as I can tell, Tim Trudgian was first to bid, with a bound of 59,874,594, which he called “a poor man’s improvement” in a note submitted to the arXiv. Soon after that, Scott Morrison of the Secret Blogging Seminar found a way to reduce it ever so slightly to 59,470,640.

The bound quickly tumbled much further than that, and Terry Tao initiated a new Polymath project to coordinate efforts. Participants are trying to optimise three constants:

• $H$ is a quantity such that there are infinitely many pairs of consecutive primes of distance at most $H$ apart. Would like to be as small as possible.
  
• $k_0$ is a quantity such that every admissible $k_0$-tuple has infinitely many translates which each contain at least two primes. Would like to be as small as possible.
• $\varpi$ is a technical parameter related to a specialized form of the Elliott-Halberstam conjecture. Would like to be as large as possible. Improvements in $\varpi$ lead to improvements in $k_0$.

Records are being kept on the Polymath wiki, and progress is swift: just yesterday eighteen different values for $H$ were recorded, along with four values of $k_0$. As I write, they currently think that $H \leq 108,540$. Of course, in the rough and tumble of mathematicians trading arguments through comments pages it’s entirely possible that mistakes will go unnoticed, but it’s certainly fun to watch!

Interested followers should take a look at an online reading seminar conducted by Terry Tao which is going through the original paper.

More information

Bounded gaps between primes by Yitang Zhang

A poor man’s improvement on Zhang’s result: there are infinitely many prime gaps less than 60 million by T.S. Trudgian

I just can’t resist: there are infinitely many pairs of primes at most 59470640 apart by Scott Morrison

Polymath proposal: bounded gaps between primes by Terence Tao

Online reading seminar for Zhang’s “bounded gaps between primes”  at Terry Tao’s blog

Bounded gaps between primes at the Polymath project wiki

Game of proofs boosts prime pair result by millions at New Scientist

Readers of The Aperiodical may recall three excellent posts on the Maths of Star Trek by Jim ‘But Not As We Know It’ Grime. At the same time, Jim discussed the topic in glorious audio with Andy Holding and Will Thompson, hosts of the Science of Fiction podcast (worth listening to, but at least visit the page to see a picture of Jim nursing a tribble). As part of this, the hosts asked Jim what uniform colour mathematicians on the Enterprise would wear.

JIM: Science and medics, those are the blue shirts.

HOST: Where do mathematicians go? Scientists?

JIM: That’s right, yes, science.

HOST: You’re safe?

JIM: Yes, I am, I’m in the blue shirt category.

Jim is pleased to say that mathematicians wear blue because, as he explains, gold and red uniformed crew were much more likely to be killed during the famous five-year mission than those in blue. I’ve written in the past about maths and mathematicians being everywhere, for example when asserting that most of the Nobel prizes are for mathematics. Was Jim right about those blue-shirted mathematicians?

Jim apparently said blue because blue is the colour worn by scientists (and medics). Gold uniforms are worn by command crew, which included command officers, pilots and some others. Finally, red, the famous uniform colour that spawned an archetype, is worn by operations, which includes engineers, security and tactical, communications and others. Somehow, I doubt the Enterprise carries many pure mathematicians or, say, mathematical artists, so I think we’re at the dirty end of the spectrum here: applications. Let’s consider the three divisions.

• Operations: Are mathematicians currently employed in engineering? Of course mathematicians are employed in engineering! How about communications? Certainly this is a very mathematical topic; for example Brigadier Edward (Ted) M. Flint, former head of the Royal Corps of Signals, studied mathematics and engineering at university. Security and tactical is harder to pin down. Certainly, we use mathematics and mathematicians in crime-fighting, defence and tactics.
• Command: in the Royal Navy, it seems shiphandling, seamanship and bridge watchkeeping comes under the Warfare division. According to a case study of a ‘Lieutenant Ben‘, to be eligible requires a first degree in “an engineering, maths or physics-based subject”. In civilian life, mathematicians certainly make good managers.
• Science and medicine: Of course, mathematicians certainly work across science and in medicine, at least in medicical physics and medical statistics.

So it is not clear to me that a mathematician working on Captain Kirk’s Enterprise would necessarily be given a blue uniform. As mathematicians do everything, so might they also wear every colour. Ah, and have every chance of being the ill-fated ‘red shirt’.

Now, I’m off to find a podcast called ‘The Pedantry of Science of Fiction Podcasts’, on which I can explain my findings.

For a lark, David and I have decided to review some of the Encyclopedia’s sequences. We’re rating sequences on four axes: Novelty, Aesthetics, Explicability and Completeness.

Following last week’s palaver, we’re going to do our best to be serious this time. Game faces on.

A052486
Achilles numbers – powerful but imperfect: writing n=product(p_i^e_i) then none of the e_i=1 (i.e. powerful(1)) but the highest common factor of the e_i>1 is 1 (so not perfect powers).

72, 108, 200, 288, 392, 432, 500, 648, 675, 800, 864, 968, 972, 1125, 1152, 1323, 1352, 1372, 1568, 1800, 1944, 2000, 2312, 2592, 2700, 2888, 3087, 3200, 3267, 3456, 3528, 3872, 3888, 4000, 4232, 4500, 4563, 4608, 5000, 5292, 5324, 5400, 5408, 5488, ...

Explicability

Christian: “Powerful but imperfect”… a bit like you, then?

David: ?!

Christian: Actually, we can draw a Venn diagram. Hang on…

…

Christian: Oh no I got it wrong. Continue hanging on…

Christian: Good work on the stamp.

David: Anyway, back to the explanation. A number is POWERFUL if all the powers in its prime decomposition are bigger than 1 – so if a prime divides it, its square does as well.

Christian: Ah, we are punning. Very droll.

David: I didn’t make a pun!

Christian: Not you, whoever invented powerful numbers.

David: You mean Achilles?

Christian: Have we explained it? Can we give a score?

David: Hold your horses. An Achilles number is a POWERFUL number which is not a perfect power.

Christian: OK! Four out of five. Yes?

\[ \frac{4}{5} \]

Novelty

Christian: I don’t like this category any more.

David: Why?

Christian: All the interesting sequences were thought up ages ago. Actually, maybe this one wasn’t. I might have been misled by Achilles. You could call my tendency to jump to conclusions my… Achilles heel!

Christian: It was added to the OEIS in 2000 with no references. So it’s fairly new.

David: I was going to say it should lose 1 for being a subsequence of another sequence, but every sequence is a subsequence of another sequence.

Christian: I put it to you, Cushing, that it’s the very act of taking a subsequence that is novel. Anyway, what fresh new ideas has this one got going for it?

David: I don’t care any more. Three.

Christian: No, it’s a new idea to me. More: four.

David: We found it when I thought I had a new sequence. So it’s not that novel.

\[ \frac{3}{5} \]

Aesthetics

David: All these numbers are sexy. They’re so full and yet so empty.

Christian: Like Quavers?

David: EXACTLY like Quavers. I am worried about all the zeroes.

Christian: I don’t think they turn up particularly often. You need to have both 2 and 5 as factors, which asymptotically almost no Achilles numbers do.

(David points out all the zeroes in the first 80 entries)

Christian: Have faith! Once we get to the bigger primes there will be fewer zeroes. Because of combinatorics, you’ll get lots of fives and twos at the start.

David: What about 5,000,000,000? That’s a lot of zeroes. And it’s Achilles.

Christian: I’m not denying that there are zeroes, I’m just saying there aren’t too many.

David: You pick the score. I’m sad now I’ve run out of chocolate raisins.

\[ \frac{4}{5} \]

Christian: Actually, we still need to discuss this. I want to give it a high score, but it hasn’t moved me to tears of joy.

David: 72 is such a beautiful number! I think all the numbers here are beautiful by themselves. But you can have too much beauty. Three out of five.

Christian: We’ll stick with four.

\[ \frac{4}{5} \]

Completeness

David: Is there a way to generate them all?

Christian: Like a generating function? I’m sure there is.

David: To the OEIS!

<batman noises>

Christian: The OEIS doesn’t have one. Why not? We just want vectors in… what do you call the space with infinitely many dimensions?

David: $\mathbb{Z}^{\infty}$. Is that uncountable?

Christian: Yeah. There’s a bijection with the reals.

David: So there are uncountably many Achilles numbers. Now this is a palaver. Let’s quickly move on and give it a 4.

Christian: Let’s play it safe. Five.

\[ \frac{5}{5} \]

Total score

David: STOP THE PRESSES! We only want vectors from $\mathbb{Z}^{\infty}$ with finitely many non-zero terms. Maths is saved!

Christian: Can you write down a generating function yet?

David: I’ll do it later.

Christian: Well, pending that, we’ll add up a provisional score.

\[ \frac{4+3+4+4+5}{25} = \frac{20}{25} = \frac{4}{5} \]

Christian: Something seems wrong. Oh well. We’ve got bigger troubles at the moment.

David: We should review it again next week but give it a Novelty score of 1 because we’ve seen it before. And I’ll work on a generating function.

Will our intrepid heroes find a generating function? Will they ever review a sequence sensibly? Will David get credited next time this series appears in the Carnival of Maths? Find out next week, same Bat-time, same Bat-place!

The Carnival rounds up maths blog posts from all over the internet, including some from our own Aperiodical. See our Carnival of Mathematics page for more information.

• Dara O Briain’s School of Hard Sums (S025)
  Festival Guest Director and comedian Dara O Briain presents a live version of the TV maths game show.
• The Hazards of Life, with David Spiegelhalter (S102)
  Winton Professor of the Public Understanding of Risk and Winter Wipeout contestant David Spiegelhalter presents a show about risk and statistics.
• Stand-Up Mathematics 2013, with Matt Parker (S117)
  Aperiodi-pal and comedian Matt Parker performs his new stand-up show, covering all his favourite bits of maths.

Aside from these events, there are some other events which will certainly contain mathematical content (ignoring, of course, the fact that all science is maths). My picks include:

• Famelab is a competition to find scientists and researchers who can present their topic well. The contest requires them to give a three-minute talk without slides and with minimal props, and communicate something engaging in a short time. The International Famelab competition holds its two semi-finals (part 1 and part 2) and the final at Cheltenham, and two of this year’s semi-finalists - Galya (Bulgaria) and Eduardo (Spain) – are mathematicians. Watch the acts, and choose whether to support your country, or your subject, or the one with the best hair.
• Call My Genetically Engineered Bluff (S133), a hilarious science quiz, includes as panellists mathematician Matt Parker, and maths fans Timandra Harkness and Helen Arney, who will presumably all be trying to inject as much maths as possible.
• The festival’s education strand, which has its own separate programme, includes one maths show – Magical Maths, presented by Dr Matt Pritchard.

Aside from these, there doesn’t appear to be much maths in the festival – which is a shame: they’ve picked a real maths fan as a festival director, in the form of Dara (some of his other events include a live version of his BBC show Science Club and an interview with Peter Higgs) but somehow that hasn’t injected much maths into the programme. The festival has historically been really good for this and I hope it’s something they have their eye on.

1 Miss Germany who studies maths

“”Miss Germany”: Mathe-Studentin aus Hannover ist die Schönste” in Der Spiegel

There can only be one Miss Germany 2013. After what was surely a hard-fought battle and in no way an anachronism which demeans us all, Caroline Noeding emerged victorious as the fairest of them all.
Now here’s the bit that’s crazy: she’s a maths student! See if you can get your head round that, Mr/Mrs PRECONCEPTIONS.

Actually, it’s not even like there isn’t precedent for maths students doing well in beauty contests: Aoibhínn ní Shúilleabhaín won the Rose of Tralee contest before continuing on to an academic career in maths education.

Noeding made it through to the final after winning the Miss Lower Saxony contest in January. Her other hobbies include dancing tango, singing, playing the piano, languages, philosophy and French. But there’s only one question we’re interested in: is she pure or applied?

2 thirds of people are confused by supermarket pricing

“Supermarket promotions confuse two thirds of shoppers, survey reveals” in The Grocer

Two thirds of a survey of 1,010 shoppers failed to identify which of a selection of special offers represented the best value. 42% failed to identify the cheapest deal on milk, when faced with four options. It’s not clear whether this is a result of the general public’s level of numeracy, or the fact that with multi-buy deals and differing pack sizes it’s getting increasingly confusing to actually work out the per-unit cost of a product. Following the supermarkets’ logic, if I were in charge of a supermarket all the prices would be in binary or in the form of alphametic puzzles, and if you couldn’t work out the best deal, it’d be your own fault.

3 years is not an eternity

“Man behind ‘Why I Don’t Have a Girlfriend’ theory to marry” at TODAY.com

Maths PhD student Peter Backus received mild media attention for a spectacularly low level of confidence in his marriageability a few years ago. Having been single for three years, Peter released a tongue-in-cheek “paper”, wherein he applied the Drake Equation (more commonly used to calculate the probability of making contact with extra-terrestrial life, and mentioned in our recent series of Star Trek articles by James Grime) to the problem of whether he’d be able to find a girlfriend.

The results were mildly disappointing: he reckoned he had a 1 in 285,000 chance of meeting his future wife on a given night out in London. However, in a highly unlikely/result-disproving twist, he’s met a girl and they’re getting married, which is nice. Assuming he worked hard at it and went out every Saturday night in the intervening two years, he won a 1 in 2,741 lottery¹.

If this heartwarming tale made you uncomfortably optimistic about the human condition, try this article about the story on CNET in which the journalist speculates about whether he’s managed to find the happy couple’s private Pinterest page, and includes more confusing metaphors and possibly defamatory non sequiturs than I’ve ever seen in one place.

4 is banned in a Toronto suburb

“Richmond Hill, Toronto Suburb, Bans Number 4 For New Addresses” at The Huffington Post

Tetraphobic residents of Richmond Hill, Ontario have succeeded in pressuring local lawmakers to ban the number four in the addresses of new houses built in the area.

The story makes sense (in a way) once you discover that the suburb has a large Chinese population (21.4% according to Wikipedia). The number four is considered deeply unlucky in Chinese culture, since it sounds similar to the Chinese word for ‘death’. Some buildings in Hong Kong omit the fourth floor, and sometimes any floor number including the digit 4 is skipped – although it’s possible this is done in part to make apartments sound like they are on a higher floor, and hence more valuable.

1. $\Pr \left( X \leq 104 \mid X \sim \operatorname{Geo}\left( 0.0000034 \right)  \right) = 1 – (1 – 0.0000034)^{104} \approx 0.0003648 \approx \frac{1}{2471}$