I’ve been a fan of your “Is this prime?” game for a while, and after seeing your blog post from last May, I thought I’d say hi and send you some high scores. Until recently, my record was 89 numbers (last March 12), which I think may be the dot in the top right of your “human scores” graph. But I tried playing some more a couple weeks ago, and I found I can go a little faster using my computer’s y/n buttons instead of my phone’s touch screen. It turns out 100 numbers is possible!

Watch in amazement:

But, to the delight of prime fans everywhere, he didn’t stop there:

Today I even got 107 – good to have a prime record again.

Well done, Ravi!

Now is a good time to point out that the data on every attempt ever made at the game is available to download, in case you want to do your own analysis: at time of writing, there have been over 625,000 attempts, and 51 is still the number that catches people out the most.

]]>It was George Boole’s bicentenary in 2015, so the Heslam Trust is a bit slow to reveal its plans to erect a statue of the great man in his home town of Lincoln.

The sculptors, Martin Jennings and Antony Dufort, have come up with a few designs for the statue, and they’d like the public to vote for their favourite.

There’s already a bust of Boole in University College, Cork, installed in plenty of time for the bicentenary. Here’s a picture of me and HRH Poppy Dog standing next to it, last Summer.

Lincoln maths genius to get statue in city – and here are the designs at LincolnshireLive

Proposals for George Boole monument on the City of Lincoln council website

]]>Why should I worry about dying? It’s not going to happen in *my* lifetime!

*Raymond Smullyan, This Book Needs No Title* (1986)

This week, the mathematical community has lost not one but two of its most beloved practitioners. Earlier this week, Swedish statistician **Hans Rosling** passed away aged 68, and today it’s been announced that author and logician **Raymond Smullyan** has also died, aged 97.

** Hans Rosling** was a statistician and doctor, and has appeared in various TV documentaries on statistics, population and global development, as well as being hugely popular in TED talks. He also co-founded the Gapminder foundation, which uses data analysis software Trendalyzer (developed by Rosling’s son Ola) to display and illustrate international statistics in innovative ways.

He’s been awarded many accolades, including being named one of 100 leading global thinkers and one of the 100 most creative people in business. He’s also a member of the Swedish Academy of Sciences and the Swedish Academy of Engineering Sciences, and was instrumental in the establishment of Médecins Sans Frontières in Sweden.

**Raymond Smullyan** was a logician, philosopher and set theorist (and also magician, pianist and puzzle-maker) from New York. He has written books on philosophy, set theory, logic puzzles, Gödel’s incompleteness theorems and even combinatory logic. He is well-known for his ‘Knights and Knaves‘-style logic puzzles, in which you are told that some of the people you’re talking to speak the truth or only tell lies, and you have to work out who is lying based on their statements. The Lady and the Tiger is one famous example, and some incorporate people who only sometimes lie, or lie with a given probability.

Smullyan was a professor of philosophy at Lehman College, City University of New York, and Indiana University. He has published over 30 books, some as recently as last year. On social media, people are honouring his memory by sharing their favourite anecdotes, puzzles, jokes and problems from his oeuvre.

Raymond Smullyan passed away on 6th February aged 97. Hans Rosling died on 7th February at 68, after a year-long battle with pancreatic cancer. They will both be missed by many.

Hans Rosling, statistician and development champion, dies aged 68, at The Guardian

Hans Rosling, physician and statistician, 1948-2017, at the Financial Times

Mathematician and puzzle-maker Raymond Smullyan dead at 97, at International Business Times

]]>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.

]]>The MTaP carnival traditionally starts with a puzzle, game or piece of trivia. As this edition marks the start of a new year, I thought I’d share a new year puzzle I played with my students in the last session before the Christmas break.

Good news everyone – the 2017 game works (though it's easier than previous years, I think). pic.twitter.com/VRyfQ1bWym

— Peter Rowlett (@peterrowlett) December 12, 2016

In fact, this bring me straight onto the submissions this month. Let’s start with some puzzles and games. Denise Gaskins (@letsplaymath) submitted a post about the 2017 Mathematics Game, providing a set of rules similar to the game above that open it up to 1-100. Denise points out that “the goal is adjustable: Young children can start with looking for 1-10, middle grades with 1-25.”

Manan Shah (@shahlock) offers Alphanumeric Puzzle #1, saying “I thought readers might get a kick out of this”. It’s a type of puzzle known as a cryptarithmetic puhezzle.

Megan Schmidt (@veganmathbeagle) submitted Math Game: Draw 10 by Annie Forest. She says “Annie always has thought provoking ideas to share and this one is no different. She brings the math and the excitement for those days before Christmas when engagement is tough to come by.”

Next up, Measurement Games and Activities, a list of measurement/estimation games submitted by Crystal (@Tri_Learning), who says “The ability to estimate and measure are important skills for every day life and many professions. Find measurement games and activities for all ages.”

And if you still haven’t had enough games, Denise Gaskins (@letsplaymath) has it covered in My Favorite Math Games. Denise says “I like to use games as a warm-up with my co-op math circle. Here’s a collection of my favorites.”

Next, there are several posts offering stories and mathematical investigations.

This entertaining post by James Propp tells a tale centred around the (Prouhet-)Thue-Morse Sequence which covers religious law, fussy eaters, fractals, music, poetry, chess, rowing and more: Avoiding chazakah with the Prouhet-Thue-Morse sequence.

1226221 Is A Fascinating Number was submitted by Manan Shah (@shahlock), who says “What’s so fascinating about 1226221? It’s a palindrome, but that’s not all!”

I saw this fun post by Ben Orlin which asks the question: how many apparently different looking but actually the same sudoku puzzles can be generated from a single sudoku: 1.2 Trillion Ways to Play the Same Sudoku.

This being a mathematics teacher blog carnival, it may not come as a surprise that most submissions were tips and approaches to bringing a little fun to the classroom.

First, Tips for Teaching Students How to Identify Functions was submitted by Mrs. E (@mrseteachesmath), who says “This is a fun lesson idea for teaching students how to identify functions. It keeps kids engaged and a song helps them remember the definition of a function.”

This second post submitted by Mrs. E is about using paper folding to help students discover and think about geometry constructions: Altitudes and Angle Bisectors Paper Folding Activity.

And this third from Mrs. E is about using a puzzle/pattern-spotting approach to introducing logarithms: How I Teach Intro to Logs.

Crystal (@Tri_Learning) submitted Our Favorite Math Manipulatives, about using physical objects to help students learn, for students at various levels.

Next, the submitter didn’t leave their name, but we have Tell me everything you know about… by Jen McAleer, which offers some ideas and in-class activities to improve engagement, first by asking a loosely-defined question and second by holding a quick competition.

Amber Thomas submitted Playing with Our Food, Nutrition, and Fraction Line Plots, saying:

I found this fun nutrition website that I fondly refer to as “Smash My Food.” The kids see how much oil, salt, and sugar is in food (by watching it get squeezed out). It’s good, gross fun! Where the math comes in, is I take those amounts and create fractions on a line plot. You know, that pesky common core standard that math textbooks don’t cover at all (CCSS Math 4.MD.B.4)? Yeah, that one. Kids find the information relevant and relate-able, and I’ve gotten good feedback from others who have tried teaching fraction line plots this way.

Rupesh Gesota submitted Understanding v/s Answer-getting, a post about trying to teach understanding, even if it doesn’t use the standard set of rules. Rupesh says “This article shows some interesting (non-standard) ways of approaching and solving couple of mathematical problems. These are the ways developed by the students themselves based on understanding rather than procedures/ rules instructed by the teacher…”

Finally, in one last burst of teaching-related fun, Jo Morgan (@mathsjem) submitted the results of her World Cup of Maths, a Twitter competition to find UK maths teachers’ favourite topic to teach.

To round off this carnival, although I should disclaim that I am mentioned in it, this post by mathsbyagirl offers a guide to getting started listening to mathematical podcasts in 5 Maths Podcasts.

If you enjoyed this and the MTaP carnival in general, consider also reading the Carnival of Mathematics blog carnival which is coordinated by Katie Steckles here at The Aperiodical. Both Carnivals are always open for submissions and always happy to hear from potential future hosts. Submit a post you’ve enjoyed (whether you wrote it or not), and consider hosting a carnival at your blog!

]]>I’ll start at the beginning.

I’ve always loved maths, but I wasn’t aware of the number of YouTube maths channels there were. During the months of February and March 2016, I started following some of them (Brady Haran’s Numberphile, James Grime and Matt Parker among others). On July 13th, Matt published the shortest maths video he has ever made:

Maybe it’s a short video, but it got me truly mired in those numbers, as I’ve loved them since I read *The Number Devil* when I was 8. I only needed some pens, some paper, my calculator (Casio fx-570ES) and if I needed extra help, my laptop to write some code. And I had that quite near me, as I had just got home from tutoring high school students in maths.

I’ll start explaining now how I focused on this puzzle trying to figure out a solution.

The first thing that I thought of was to take my calculator and enter $ \sqrt{\frac{x(x+1)}{2}} $, letting $ x $ be an integer greater than $ 8 $, as $ \sqrt{\frac{8(8+1)}{2}}=6 $ (and that’s the one Matt used to introduce the challenge). That way I found the next one, $ 1225 $, for $ x=49 $. After trying with some more values for $ x $, I got tired and I decided to write it in PHP. In this way I found the first few terms of the sequence: $0$, $1$, $36$, $1225$, $41616$, $1413721$, $48024900$, $1631432881$…

I went on naming, square numbers as $ S $, triangular numbers as $ T $ and square triangular numbers as $ ST $. I then thought of addressing this problem, $ ST_{n}=S_{i}=T_{j} $, by expressing $ S $ and $ T $ as series:

\[ S_{i}=i^2=\sum_{x=0}^{i-1} 2x+1 \] \[ T_{j}=\frac{j \cdot (j+1)}{2}=\sum_{x=0}^{j} x \]

As each square number is a sum of consecutive odd numbers and each triangular number is a sum of consecutive numbers, I focused on the different elements in each sum, noticing that parity of triangular numbers is important and that there has to be a sum of consecutive odd numbers that equals a sum of consecutive even numbers. For example:

\[ ST_{2}=36=S_{6}=1+3+5+7+\underline{9}+\underline{11} \]

\[ ST_{2}=36=T_{8}=1+\underline{2}+3+\underline{4}+5+\underline{6}+7+\underline{8} \]

Underlined numbers are all different summands between both sums, so they add to the same result: $ 9+11=2+4+6+8=20 $. Considering the sum of consecutive even numbers, we always obtain twice a triangle, $ 2+4+6+8=2 \cdot (1+2+3+4)=2 \cdot T_{4} $. Knowing that and doing the same thing for $ 1225$, $41616$, $\ldots$ I looked for a pattern in the series and I found that for every $ST_n = S_{i} = T_{j} $:

\[ \sum_{x={\frac{j+(j \bmod 2)}{2}}}^{i-1} 2x+1 = \sum_{x=0}^{{\frac{j-(j \bmod 2)}{2}}} 2x = 2 \cdot T_{\frac{j-(j \bmod 2)}{2}}\]

For $ n \gt 1 $, I noticed that $ \frac{j+(j \bmod 2)}{2} $ is a square number and it divides $ ST_{n} $, resulting in another square number that equals $ j+((j+1) \bmod 2) $, so every square triangular number is a product of two square numbers:

\[ \begin{align} ST_{2} &= 36 = S_6 = T_8 \\ &= \frac{8+0}{2} \cdot (8+1) = 2^2 \cdot 3^2 \end{align} \\ \sum_{x=\frac{8+0}{2}}^{6-1} \left( 2x+1 \right) = \sum_{x=0}^{ \frac{8-0}{2}} \left( 2x \right) = 2 \cdot T_{4} = 20\]

\[ \begin{align} ST_3 &= 1225 = S_{35} = T_{49} \\ &= \frac{49+1}{2} \cdot (49+0) = 5^2 \cdot 7^2 \end{align} \\ \sum_{x=\frac{49+1}{2}}^{35-1} \left( 2x+1 \right) = \sum_{x=0}^{\frac{49-1}{2}} \left( 2x \right) = 2 \cdot T_{24} = 600 \]

\[ \begin{align} ST_4 &= 41616 = S_{204} = T_{288} \\ &= \frac{288+0}{2} \cdot \left( 288+1 \right) = 12^2 \cdot 17^2 \end{align} \\ \sum_{x=\frac{288+0}{2}}^{204-1} \left( 2x+1 \right) = \sum_{x=0}^{\frac{288-0}{2}} \left( 2x \right) = 2 \cdot T_{144} = 20880 \]

\[ \begin{align} ST_5 &= 1413721 = S_{1189} = T_{1681} \\ &= \frac{1681+1}{2} \cdot (1681+0) = 29^2 \cdot 41^2 \end{align} \\ \sum_{x=\frac{1681+1}{2}}^{1189-1} \left( 2x+1 \right) = \sum_{x=0}^{\frac{1681-1}{2}} \left( 2x \right) = 2 \cdot T_{840} = 706440 \]

At this point I was freaking out, because I was feeling I was so close to a formula for these numbers – I was getting something quite amazing but I couldn’t be sure because my method was so chaotic. Then I tried to find a general term for a sequence $ a $ which has those factors in increasing order, that is $0$, $1$, $1$, $2$, $3$, $5$, $7$, $12$, $17$, $29$, $41$, $70$, $99$, $169$, $239$…

I simply defined a sequence $ a $ considering $ a_{0}=0 $, $ a_{1}=1 $ and $ a_{2}=1 $ because I noticed how the numbers were growing and I thought it might be similar to Fibonacci numbers, but I had no idea what sequence this was. It was quite difficult to get a general formula for the terms in this sequence, but after a while, I finally managed to find it. Having $ n > 2 $:

\[ a_{n}=a_{n-1-((n-1) \bmod 2)}+2^{(n-1) \bmod 2} \cdot a_{n-2-((n-1) \bmod 2)} \]

Nice fact: $ a_{13} = 13^2 $. I guess this only happens for $ n=13 $, but I’m not sure.

Finally I was able to write $ i $ and $ j $ in $ ST_{n}=S_{i}=T_{j} $ for a given $ n>0 $, considering $ ST_{0}=0 $:

\[ i=a_{2n-1} \cdot a_{2n} \hspace{5pt} ; \hspace{4pt} n > 0 \] \[ j=2 \cdot (a_{2n-1})^2 – ({a_{2n-1}} \bmod 2) \hspace{5pt} ; \hspace{4pt} n > 0 \]

I had finally solved it! I thought I should review everything to check I had everything right. After re-reading it several times, it seemed everything was alright. While I was looking again at the equivalent sum of consecutive odd numbers and sum of consecutive even numbers, I noticed this was twice a triangle. That fact got my attention, and I thought: ‘what if I subtract those two triangles from each corresponding square triangular number?’ I imagined it graphically, with each $ ST_{n} $ for $ n>2 $ and I obtained these results: $ 16$, $625$, $20736$, $707281$, $24010000$, $815730721$, $\ldots$

I recognised the first two results – I knew they were $ 2^4 $ and $ 5^4 $, but I couldn’t imagine they were all going to be fourth powers: $ 2^4$, $5^4$, $12^4$, $29^4$, $70^4$, $169^4$…

‘Wow, this is truly nice,’ I thought. After discovering this pattern I checked for $ ST_{1} $ and it worked too as it’s trivial, so I realised I had conjectured by chance some more formulae based on the sequence $ a $:

\[ ST_{n}=(a_{2n-1})^4+\sum_{x=(a_{2n-1})^2}^{(a_{2n-1} \cdot a_{2n})-1} \left( 2x+1 \right) \hspace{5pt} ; \hspace{4pt} n > 0 \]

\[ ST_{n}=(a_{2n-1})^4+\sum_{x=0}^{(a_{2n-1})^2 – (a_{2n-1} \bmod 2)} \left( 2x \right) \hspace{5pt} ; \hspace{4pt} n > 0 \]

\[ ST_{n}=(a_{2n-1})^4+2 \cdot T_{(a_{2n-1})^2 – (a_{2n-1} \bmod 2)} \hspace{5pt} ; \hspace{4pt} n > 0 \]

I noticed then that I only needed the terms of the sequence $ a $ in odd positions to write the last two formulae, but I didn’t (and don’t) know how to prove everything by induction.

So I drew the two first square triangular numbers as rectangular triangular shapes, and I discovered that each fourth power is graphically a square of squares with two identical triangles attached to two of its sides in such a way that the square’s corner between them either is or isn’t part of the whole shape’s hypotenuse, depending on the parity of the total number of elements.

Having spent the most part of two days working on this, I searched on the internet for everything I could find about these numbers. I was so surprised when I couldn’t find anything related to these patterns, except for one of the formulas that I found, $ ST_{n}=(a_{2n-1} \cdot a_{2n})^2 \hspace{3pt} ; \hspace{2pt} n > 0 $, and it was a contribution by Hugh Darwen, approved in the OEIS on February 23, 2012. But the rest of the work I’d done… I couldn’t find anything like it. I looked for papers that may have something related to this, but again… nothing. Maybe I should have looked harder, because I couldn’t believe I’d actually been able to notice something that nobody has noticed before… I found out that sequence $ a $ is almost A002965, but not exactly, as one initial term is missing. And I discovered too that $ a $ terms in odd positions correspond to Pell numbers, which I didn’t know before… I found out about the Silver ratio and other stuff that I find amazing! I’m glad that I saw Matt’s video in the first place.

Then I decided to sign up for an account on the OEIS to publish these formulae in a draft, and to write about this in my OEIS profile page. When I received an email from the OEIS telling me that Neil Sloane had approved my changes to sequence A001110… I was shocked. Really. I wasn’t expecting anything like that. I find all of the work I’ve done to be kind of basic, because I took as a starting point something I learned in a book I read when I was 8… Is this possible? It seems to be.

After that contribution, I wanted to generalise the formula $ ST_{n}=(a_{2n-1})^4+2 \cdot T_{(a_{2n-1})^2 – (a_{2n-1} \bmod 2)} \hspace{5pt} ; \hspace{4pt} n > 0 $, and since that moment I’ve been contributing more sequences and formulae. In fact, I first generalised this formula in sequence A275496 for every fourth power (every square of squares). I also tessellated each whole triangle and I found another formula for square triangular numbers. Letting $ P_{n} $ be the $n$th Pell number:

\[ ST_{n}={{P_{n}}^2 \cdot T_{2 \cdot P_{n} – 1} + 2 \cdot P_{n} \cdot T_{P_{n} – (n \bmod 2)}} \hspace{5pt} ; \hspace{4pt} n > 0 \]

After that, I generalized all these inquiries publishing sequence A276914, also making an illustration for the OEIS where I explain the main pattern.

I hope Matt sees all this (he already knows a bit about this, as I’ve written to him by email). I’ve been making more inquiries about other numbers, such as centered polygonal numbers (A276916 are centered square numbers based in A276914, by the way). I’ve posted some images on Twitter about really really nice patterns related to prime numbers and stuff.

Maybe I can write another post in future about some of my number games!

*We showed Dani’s article to Matt Parker. He says:*

]]>Congratulations to Dani for such a fantastic explanation of the mathematics he found in triangle-square numbers! It was nice to see how he started the same way I did when I first investigated these numbers but then very quickly explored new directions I’d never considered.I love it when people play around with mathematics because of something they have seen online (or noticed in normal life), but it is always extra-amazing if they then take the time to share their experiences with other people. So thanks again to Dani for emailing me about what he was doing, and writing it up for The Aperiodical.

Having spoken at the MathsJam annual conference in November 2016 about my previous phone spreadsheet on multiples of nine, I was contacted by a member of the audience with another interesting number fact they’d used a phone spreadsheet to investigate: my use of `=MID()`

to pick out individual digits had inspired them, and I thought I’d share it here in another of these columns (LOL spreadsheet jokes).

Twin primes, which are pairs of prime numbers with a difference of two, are well-studied. It’s not known whether there are infinitely many such pairs (although we’ve made some progress on that), but it’s suspected. The gaps between pairs get larger as you go up the number line, and there’s a few interesting bits known about them, including Brun’s Theorem. Today I’m going to investigate a thing about their products.

We can start by populating a spreadsheet with a list of pairs of twin primes, starting with the classic \((3,\!5)\) and proceeding from there. This data was obtained from the amazing resource at primes.utm.edu, maintained by known prime-basher Chris Caldwell. Putting this list of primes in the first column, and populating the second column with the number two higher gives us a lovely list of twin prime pairs.

In the next column, we calculate the product of each pair:

And in the fourth column, we can use a similar `MID()`

/ `VALUE()`

based construction to find the sum of the digits in this number:

Giving the not-especially-interesting result:

Or is it? In the fifth column, we can take `=MOD(`

and see what happens:*this value*,9)

With the exception of the first row, and let’s be honest the first one’s always weird anyway, you’ll see that all of these numbers have the value $8$ modulo $9$.

For an explanation of this, done in a bar after a few pints of course, we can thank sometime Aperiodical contributor and general maths guy Colin Beveridge, who contributes the following; if you’d like to try and work it out yourself, we’ve made these bullet points appear one at a time when you click, so you can use as many as you need and work the rest out.

- All twin prime pairs other than $(3,5)$ are of the form $6n +1$ and $6n – 1$ (it says it on Wikipedia so it must be true; left as a pleasing exercise for the reader)
- The product will therefore be of the form $36n^2 – 1$ (classic difference of two squares)
- This means the first part of this number will be divisible by $9$, as $36$ is divisible by $9$
- The whole thing must therefore be congruent to $8$ modulo $9$, as it’s a multiple of $9$ minus one.
- This means the sum of the digits of the number will also be $8$ modulo $9$.

A fun diversion for everyone. Thanks to Simon Allen, who sent me the email, and to Colin for his neat explanation, relayed via Simon.

If you’ve seen any nice number facts I can investigate using a spreadsheet on my phone, please send them in!

]]>On Wednesday 4th January, an error was discovered in the proof. Harald Helfgott (of the University of Göttingen in Germany and France’s National Center for Scientific Research), who studied the paper for several months, discovered that the algorithm was not *quasipolynomial* ($\displaystyle{ 2^{\mathrm{O}((\log n)^{c})} }$ for some fixed $c>0$) as claimed, but merely *subexponenential*: growing faster than a polynomial but still significantly slower than exponential growth).

Adorably, Babai posted this message on his website:

I apologize to those who were drawn to my lectures on this subject solely because of the quasipolynomial claim, prematurely magnified on the internet in spite of my disclaimers. I believe those looking for an interesting combination of group theory, combinatorics, and algorithms need not feel disappointed.

But maths is all about the drama, so on Monday 9th January Babai announced a fix for the error, and it’s now back on the quasipolynomial table. This has now been confirmed (as of 14th Jan) by Harald Helfgott himself at the Bourbaki seminar in Paris. Amusingly, Helfgott had only been studying the paper in such detail in order to give the seminar, and it was this close scrutiny which allowed him to discover the mistake.

Announcement on Babai’s website

Fixing the UPCC Case of Split-or-Johnson – Babai’s paper detailing the fix (PDF)

Graph Isomorphism Vanquished — Again, at Quanta Magazine

Bourbaki Seminar – Harald Helfgott, on YouTube

]]>I’m so far from understanding the mind of a mathematical genius that it’s simply inconceivable that you could tell a person an apparently random number and he could intuit or deduce the kind of fact that he deduced about that taxi license number. I mean, I can’t run a four-minute mile, but I once ran a five-minute mile, and I can extrapolate from my own experience, in a way understand how someone might just be a lot better than me at something that, in an inferior way, I can also do. But Ramanujan isn’t like that. It’s as though this man were a different species, not just a superior example of the same species. Can you learn to do this kind of thing? Could I, if I had applied myself? Or is it that goddess again, is it really just genius?

Answers on a postcard!

]]>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.

]]>