If I can make it to £1000 before the end of the month, I’ll be pretty pleased! Donate at pikm.run, or see below for my daily sweaty photos/videos/instagram posts.

Day 9: yep, I'm still doing this #pikmdotrun pic.twitter.com/MpFQvEImgr

— Katie Steckles (@stecks) March 9, 2018

On the 9th, I thought I’d make use of the mathematical properties of π to do a slightly silly one, and made a video:

Day 11: gym again. If you missed it yesterday, here's my video from Day 10: https://t.co/rdiPCJ534N pic.twitter.com/knyzH4L6f9

— Katie Steckles (@stecks) March 11, 2018

Day 13: still going. Thanks to everyone who's supported so far! https://t.co/dtKU5Rj0qH pic.twitter.com/p5qUUNncSD

— Katie Steckles (@stecks) March 13, 2018

I also managed to get in a few more joint runs with running companions:

Day 15: now officially kinda halfway! Buddy gym run again with @elsie_m_ #pikmdotrun pic.twitter.com/VhcVIsghll

— Katie Steckles (@stecks) March 15, 2018

Day 17: logistically complex. Managed to run πkm fully inside the B'ham NEC, as I've been working at @BigBangFair today. GPS watch gave up after 2.04km (no signal). #pikmdotrun pic.twitter.com/eebrhKjUgh

— Katie Steckles (@stecks) March 17, 2018

Day 19: staying in the gym due to the cold weather, on the world's shiniest treadmill #pikmdotrun pic.twitter.com/WO9xaPKaSV

— Katie Steckles (@stecks) March 19, 2018

I was also given an amazing gift by maths/knitting fan Linda Pollard, who came to see me perform at a show. She’s written up the knit of these magnificent π/sum gloves on a Ravelry page. I took the opportunity to test out their warmness on my next outdoor run:

This mild cry for help resulted in plenty of nice replies on Twitter, which has been a real boost – including Aperiodichum Colin Beveridge, who pointed out that my total is around $\pi^4$, a pleasing coincidence.

Day 21: Gym again (but went to a different gym for variety). Found it hard today. Encouragement please. #pikmdotrun pic.twitter.com/lIvuC7yBUZ

— Katie Steckles (@stecks) March 21, 2018

And of course, today’s effort:

Running continues. Watch this space for a final wrap-up and fundraising total at the end of the month.

Katie’s fundraising page at Sport Relief

More information about Sport Relief

You may remember back in September we posted about a mass-participation science experiment, aiming to model the spread of diseases in human populations using a smartphone app. The results of this experiment, presented by the contagiously loveable Hannah Fry, will be presented in a documentary this evening on BBC4. You can also see Hannah chatting about the experiment on this evening’s The One Show.

Contagion! The BBC4 Pandemic, on the BBC watch-o-tron

]]>]]>Robert P. Langlands wins the 2018 Abel Prize “for his visionary program connecting representation theory to number theory.” Congratulations! pic.twitter.com/HBiTJhChe0

— AbelPrize (@abel_prize) March 20, 2018

\begin{array}{l} \color{blue}13, \\ \color{blue}26, \\ \color{blue}39, \\ \color{blue}52 \end{array}

**What happened to $\color{blue}4$‽**

A while ago I was working through the $13$ times table for some boring reason, and I was in the kind of mood to find it really quite vexing that the first digits don’t go $1,2,3,4$. Furthermore, $400 \div 13 \approx 31$, so it takes a long time before you see a 4 at all, and that seemed *really* unfair.

I was being pretty unreasonable in my expectations of basic arithmetic, but I wasn’t completely brain-dead: I smelled an integer sequence! How about

$a(n)$ = least $k$ such that $k \times n$ starts with a $4$.

That’s not particularly interesting, and someone who comes across this sequence in the OEIS might think “why $4$?” So, I did a bit more thinking and came up with this:

$a(n)$ = least $k$ such that $\{ \text{first digit of } j \times n, \, 0 \leq j \leq k \} = \{ 0,1,2, \dots 9 \}$

I wrote a bit of Python, and in a few minutes I had some numbers:

$n$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|

$a(n)$ | 9 | 45 | 27 | 23 | 18 | 15 | 13 | 12 | 9 | 9 | 9 | 42 | 62 |

And hey, $13$ is a record-setter. I’m really beginning to dislike this number. Anyway, I searched the OEIS for my sequence and it wasn’t there, so I submitted it and it was duly accepted as A249067.

Along the way, OEIS editor Charles R Greathouse IV added this intriguing conjecture:

Conjecture:$a(n) \leq N$ for all $n$. Perhaps $N$ can be taken as $81$.

Why $81$? Maybe look at the graph produced automatically by the OEIS:

The record of $81$ is reached at $a(112)$. And at $a(1112)$. And $a(11112)$. That’s because they’re very slightly bigger than $\frac{1}{9} \times 10^m$, so nine times $1 \dots 12$ is *just* bigger than $9 \dots 9$, i.e. a number starting with a $1$, so it takes nine times nine steps down the times table before you see a number with $9$ as its first digit.

This pattern repeats at every power of $10$, and in fact every pattern in this sequence repeats (more or less) at every power of 10: this animated plot of the sequence with different horizontal scales shows that it’s self-similar:

(The fuzziness in the bigger plots is because each plot just takes a sample of points, and interpolates between them)

So the conjecture *looks* true, and this is my sequence, so I should prove it.

It isn’t surprising that this thing repeats when you multiply by $10$: we’re only looking at the first digit, and obviously the first digit of $n$ is the same as the first digit of $10n$. That doesn’t suffice as a proof of Charles Greathouse’s conjecture though: numbers which don’t end in a $0$ might do something unhelpful.

Fortunately, the day after I thought this sequence up was MathsJam night. I decided I’d set the Charles Grey pub’s brightest minds on the problem. I had a few ideas but I’m not particularly quick at putting thoughts together.

Ji proposed an application of the pigeonhole principle: if you look at the first *two* digits of the numbers you see in $n$’s times table, you can write out everything you might see in a $9 \times 10$ grid:

\begin{array}{}

10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 \\

20 & 21 & 22 & 23 & 24 & 25 & 26 & 27 & 28 & 29 \\

30 & 31 & 32 & 33 & 34 & 35 & 36 & 37 & 38 & 39 \\

40 & 41 & 42 & 43 & 44 & 45 & 46 & 47 & 48 & 49 \\

50 & 51 & 52 & 53 & 54 & 55 & 56 & 57 & 58 & 59 \\

60 & 61 & 62 & 63 & 64 & 65 & 66 & 67 & 68 & 69 \\

70 & 71 & 72 & 73 & 74 & 75 & 76 & 77 & 78 & 79 \\

80 & 81 & 82 & 83 & 84 & 85 & 86 & 87 & 88 & 89 \\

90 & 91 & 92 & 93 & 94 & 95 & 96 & 97 & 98 & 99

\end{array}

The $n$ times table will dance around this grid until all nine rows have been visited. The longest it can do that is by visiting all 80 cells not in the last line. If the process doesn’t visit the same place twice before it hits every row, that means that the latest you can put off visiting the last row is the 81^{st} iteration. So we need to show that you can’t visit the same spot twice before visiting each row once.

Unfortunately, that’s not true. The $12$ times table visits the ’12’ cell at $12 \times 1 = 12$ and again at $12 \times 10 = 120$, before all possible first digits have been seen.

So, we need another explanation.

Katie Steckles and the Manchester MathsJam crowd came up with an alternate explanation: if you can prove $\left\lceil \frac{1}{9} \cdot 10^m \right\rceil$ (that is, $112$, $1112$, $\ldots$) takes 81 steps for all $m \geq 3$, then that’s the maximum, as any $m$-digit number bigger than that will reach $9 \times 10^m$ in at most as many steps, and will definitely have seen all the other initial digits before then, and any $m$-digit number smaller than $\left\lceil \frac{1}{9} \cdot 10^m \right\rceil$ will visit every first digit in the first 9 multiples.

There’s some evidence for this: the $m+3$-digit numbers that take 81 steps seem to be the ones between $11\ldots112$ and $112499\ldots99$.

I don’t know if there’s a clever way of showing that $\left\lceil \frac{1}{9} \cdot 10^m \right\rceil$ takes 81 steps, but would it convince you if I said that $112$ takes that long, and adding more $1$s in the middle can’t make it any worse? Anyway, that’s good enough for me.

I think I can now answer my question: *exactly how bad is the $13$ times table?* Let’s compute the record-setters for A24097: the numbers that take longer than any smaller number to see every possible leading digit:

$n$ | 1 | 2 | 13 | 112 |
---|---|---|---|---|

$a(n)$ | 9 | 45 | 62 | 81 |

$13$ is a record-setter in the sequence, which means it’s pretty bad, but it’s not the worst: we’ve shown above that $112$ takes the longest possible number of steps to see every digit. And the number $2$ comes under scrutiny for taking way longer than its neighbours. So really, $13$ is just unlucky to find itself in such company.

If you’re interested in the working-out I did for this post, I’ve put my Jupyter notebook online.

]]>The Online Encyclopedia of Integer Sequences just keeps on growing: at the end of last month it added its 300,000^{th} entry.

Especially round entry numbers are set aside for particularly nice sequences to mark the passing of major milestones in the encyclopedia’s size; this time, we have four nice sequences starting at A300000. These were sequences that were originally submitted with indexes in the high 200,000s but were bumped up to get the attention associated with passing this milestone.

Here they are:

1, 10, 99, 999, 9990, 99900, 999000, 9990000, 99900000, 999000000, 9990000000, 99899999991, 998999999919, 9989999999190, 99899999991900, 998999999918991, 9989999999189910, 99899999991899109, 998999999918991090, 9989999999189910900, 99899999991899108991, 998999999918991089910, 998999999918991089910

The number formed by concatenating the first three digits in the sequence is $110 = 1 + 10 + 99$. This has a Golomb sequence vibe about it, though it’s a bit more straightforward to generate.

This sequence was submitted by Eric Angelini, a Belgian TV producer who has added countless sequences to the OEIS, usually generated like this by picking a constraint and working out what the sequence would need to look like in order to obey it.

1, 0, 0, 2, 0, 3, 4, 4, 3, 4, 5, 6, 4, 5, 6, 4, 5, 6, 5, 6, 6, 5, 7, 6, 5, 7, 6, 6, 7, 6, 7, 7, 6, 7, 7, 6, 7, 7, 8, 7, 7, 8, 7, 8, 8, 7, 8, 8, 7, 8, 9, 8, 8, 9, 8, 8, 9, 8, 9, 9, 8, 9, 9, 8, 9, 9, 9, 10, 9, 9, 10, 9, 9, 10, 9, 10, 10, 9, 10, 10, 9, 10, 10, 10, 10, 10, 11, 10, 10, 11, 10, 10, 11, 10, 11, 11, 10, 11, 11, 10

I’m amazed this one wasn’t already in! Seems like exactly the kind of thing that would appear in something like *Dudeney’s Amusements*. There’s an associated paper on the arXiv, by Ales Drapal and Carlo Hamalainen, which notes that some of the earliest work on triangle dissections was done by Bill Tutte, of Bletchley Park fame.

The entry page contains some fab plaintext-art drawings of solutions for a few different $n$.

1, 2, 4, 3, 6, 5, 9, 16, 14, 20, 7, 15, 8, 12, 18, 31, 26, 27, 40, 30, 49, 38, 19, 10, 23, 53, 11

The definition of this one is a bit opaque if you’re not in the right frame of mind, but it’s really neat. If you plot the sequence, as the OEIS can automatically do for you, you get this:

Or, if you want to do this in your head, think of the set of points $(n, a(n))$.

Now, if you pick any polynomial of degree $k$, there’s no subset of $k+2$ of the points on the scatter plot that lie on that polynomial. It’s a ‘duck-and-dive’ sequence – it always picks the smallest number that won’t be on any of the $2^{n-1}$ polynomials defined by the sequence leading up to $a(n)$.

The OEIS entry contains a conjecture that this sequence is a permutation of the natural numbers. It’s easily shown that it contains no duplicates – otherwise, if the number $m$ is repeated, there’d be two elements lying on the line $y=m$, a degree-0 polynomial. What’s not obvious is that every number will eventually turn up. It’d be pretty wild if some numbers never did – and that’d form a new sequence, too!

1, 1, 1, 1, 1, 3, 2, 1, 6, 15, 2, 1, 10, 52, 55, 2, 1, 15, 129, 389, 184, 2, 1, 21, 266, 1563, 2539, 648, 2, 1, 28, 487, 4642, 16445, 16604, 2111, 2, 1, 36, 820, 11407, 69863, 169034, 105365, 6352, 2, 1, 45, 1297, 24600, 228613, 1016341, 1686534, 654030, 17337, 2

I don’t like “triangle read by rows” entries, purely because the OEIS’s web interface doesn’t make them easy to read. It’s debatable whether sequences generated by two parameters are even ‘sequences’, but that’s not a fight worth having, because there are some truly fab bits of maths hiding in the OEIS’s triangles.

This one looks at what you can do by starting with the list of numbers $1,2, \ldots, n$, and repeatedly picking a block of adjacent numbers and reversing their order. It’s like a generalised version of the Oval Track puzzle.

]]>

\[ n > 2 \]

An unexpected bit of controversy involving mathematical notation hit the internet last week, when China's government briefly blocked all Chinese internet users from viewing any page or message containing the letter *n*.

Apparently, those in charge of the Great Firewall feared that those who disapproved of Xi Jinping removing the two-term limit on his presidency of China would use the letter *n* to refer to the now-arbitrary number of terms for which he can remain in power.

There's some more context in a post by Victor Mair on Language Log, and in the Guardian.

]]>The London Mathematical Society are organising an event later this month in honour of the late Fields Medalist Maryam Mirzakhani. It's at the University of Warwick on 22nd March, and will include talks outlining some of Mirzakhani's work, followed by a drinks reception and dinner. The event is part of a larger EPSRC symposium on Teichmüller dynamics.

]]>This is nice for International Women's Day. Filmmaker Irina Linke and mathematician Eugénie Hunsicker have put together this montage of women in maths from all around the world.

]]>This month I'm doing a completely irrational sponsored run for Sport Relief, aiming to raise £100π by running πkm per day, every day in March. I'm one week in, and here's the story so far.

Given the first few days of my challenge coincided with one of the most ridiculous periods of cold weather we've seen in a while, I wasn't quite willing to brave the outdoors yet, so my first few days were done on a treadmill in the gym.

On the fourth day, the weather broke, so I took advantage of the fact that my parents live near a lake whose perimeter works out to almost exactly πkm. I dragged my dad round it – and even made a short video:

On day 5, I dragged another of my family members for an outdoor run:

And finally, today I managed my first solo outdoor run, literally running an errand to fetch some workshop materials from the Museum of Science and Industry across town. By careful route choice, I managed to hit my target within sight of the front doors:

The best news of all is that on day four, I managed to hit my fundraising target! I reached £314.15 on 4th March, and promptly decided to extend the challenge by upping the target. My new goal is** £3141.59**, which I'm almost certainly not going to reach, but it gives me something to shoot for. I've got the rest of the month!

You can make a contribution (all to the brilliant cause that is BBC Sport Relief), and continue to follow along with my social media updates/proof, by heading to my fundraising page at pikm.run.

]]>The next issue of the Carnival of Mathematics, rounding up blog posts from the month of February, and compiled by Ben, is now online at Math Off The Grid.

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.

]]>