A006720

Somos-4 sequence: $a(0)=a(1)=a(2)=a(3)=1$; for $n \geq 4$, $a(n)=(a(n-1)a(n-3)+a(n-2)^2)/a(n-4)$.1, 1, 1, 1, 2, 3, 7, 23, 59, 314, 1529, 8209, 83313, 620297, 7869898, 126742987, 1687054711, 47301104551, 1123424582771, 32606721084786, 1662315215971057, 61958046554226593, 4257998884448335457, 334806306946199122193, ...

# You're reading: cp’s mathem-o-blog

### An enneahedron for Herschel

The building where I work is named after Alexander Stewart Herschel. I suspect this is because it used to be the home of the physics department, since he was an astronomer, but it works for us too because he also has a pretty cool graph named after him.

Helpfully, it’s called the Herschel graph. It’s the smallest non-Hamiltonian polyhedral graph – you can’t draw a path on it that visits each vertex exactly once, but you can make a polyhedron whose vertices and edges correspond with the graph exactly. It’s also bipartite – you can colour the vertices using two colours so that edges only connect vertices of different colours. The graph’s automorphism group – its symmetries – is $D_6$, the symmetry group of the hexagon. That means that there’s threefold rotational symmetry, as well as a couple of lines of reflection. It’s hard to see the threefold symmetry in the usual diagram of the graph, but it’s there!

Anyway, at the start of the summer, one of the lecturers here, Dr Michael White, told me about this graph and asked if we could work out how to construct the corresponding polyhedron. Making *a* polyhedron is quite simple – take the diagram on the Wikipedia page, pinch the middle and pull up – but it would be really nice if you could make a polyhedron which has the same symmetries as the graph.

### Julia Robinson and Hilbert’s Tenth Problem, by George Csicsery

Over on Google+, David Roberts just posted this trailer (via Антид Ото) to a lovely documentary about Julia Robinson and her contributions towards answering Hilbert’s tenth problem.

[youtube url=https://www.youtube.com/watch?v=e4x9XKNAYjU]

David Hilbert’s tenth problem was to find an algorithm to solve diophantine equations, that is, to find roots of polynomials with integer coefficients. We now know that the problem is unsolvable in general, and Julia Robinson did a lot of the work to get there; she wrote that she “couldn’t bear to die without knowing the answer.”

David asked if anyone knows of any present-day female mathematicians of similar standing to Julia Robinson. Apart from President of the IMU Ingrid Daubechies and people who are active on Twitter my knowledge of top mathmos is quite poor, so I thought I’d open the question up to The Aperiodical’s readers.

The full DVD of *Julia Robinson and Hilbert’s Tenth Problem* is available from George Csicsery’s ZALA Films site, and it looks like there are a few copies on Amazon as well.

### Integer Sequence Review Mêlée Hyper-Battle DX 2000, THE GRAND FINALE

Welcome to the Field of Dreams. Talking of which: why can’t you grow wheat in $\mathbb{Z}/6\mathbb{Z}$?

Anyway, we’re finally here: the Grand Finale of our tournament to find the Integest Sequence 2013. Here’s a reminder of the sequences vying for the title:

- From Bracket 1: A002210, the decimal expansion of Khintchine’s constant.
- From Bracket 2: A001220, the Wieferich primes.
- From Bracket 3: A001462, Golomb’s sequence.
- From Bracket 4: A023811, the largest metadromes in base $n$.
- From the first round of reviews, we picked the one whose score we fiddled the least: A010727, all the 7s.
- And the wildcard is A058883, the wild numbers.

It’s a been a long, hard battle. We’ve seen some good sequences, some bad sequences, and an awful lot of plagiarised GIFs. So, without further ado, it’s time to start the

### Integer sequence reviews on Numberphile (or vice versa)

There’s no new integer sequence review this week, because David and I are taking a break before the Grand Finale Ultimate Showdown of Dreams next week. To tide you over, top chap Brady Haran has recorded a Numberphile video with Tony Padilla explaining each of the six sequences in the final in his Enthusiastic Maths Outreach™ voice.

If you haven’t made your mind up yet, maybe the video will sway you. *Or will it sow doubt into your previously made-up mind*???!?!?!?!!?!?! Anyway, it’s a very good video.

[youtube url=http://www.youtube.com/watch?v=VDD6FDhKCYA]

The Integest Sequence 2013 will be announced next week in a glitzy celebrity gala event. There’s still time to vote for your favourite sequence, and there’s still time for us to decide how much attention we’ll pay to your vote. Everything’s still to play for!

### Links

Integest Sequence 2013 Public Vote

### Ipso Post Facto Navigato

### Integer Sequence Review Mêlée Hyper-Battle DX 2000 (Bracket 4)

Last week, A001462 – Golomb’s sequence – booked its place in the final. In retaliation for last week’s palaver, this week Christian has picked all the sequences. Unfortunately, the British Summer is happening today so we’re failing a bit, intellectually.

With that in mind, it’s time for round 4 of…

Here are the rules: we’re judging each sequence on four axes: **Aesthetics, Completeness, Explicability,** and **Novelty**. We’re reviewing six sequences each week for four weeks, picking a winner from each. Then, we’ll pick one sequence from the ones we reviewed individually before this thing started, plus a wildcard. Finally, a single sequence will be crowned the **Integest Sequence 2013! **

### Integer Sequence Review Mêlée Hyper-Battle DX 2000 (Bracket 3)

Last week, A001220 – the Wieferich primes – booked its place in the final. This week, David has picked six sequences all on his own to form Bracket 3 of…

Here are the rules: we’re judging each sequence on four axes: **Aesthetics, Completeness, Explicability,** and **Novelty**. We’re reviewing six sequences each week for four weeks, picking a winner from each. Then, we’ll pick one sequence from the ones we reviewed individually before this thing started, plus a wildcard. Finally, a single sequence will be crowned the **Integest Sequence 2013! **