There is no obvious reason why the xn should be whole numbers, but the first few terms of the sequence are
1, 2, 3, 5, 10, 28, 154, 3520, 1551880, 267593772160
so you do begin to wonder whether, by some miracle, all the terms are integers.
The truth is, if anything, even more miraculous. Hendrik Lenstra put the equation on a computer, and discovered that the first term that is not an integer is x43. So 42 is the largest integer for which all terms of the sequence, up to and including that one, are integers.
Other sequences of this kind also seem to behave that way - a lot of integers to begin with, but at some point the pattern fails. With the same rule but using sums of cubes, the first term that is not an integer is x89. With fourth powers the first non-integer is x97, with fifth powers it is x214, with sixth powers it is the relatively feeble x19, but with seventh powers we get the astonishing x239. So here is a sequence with a nice pattern, such that the first 238 terms44 are integers, but the 239th is not.
As far as I know, no one really understands why these sequences behave like they do.
False, Not Stated, Not Proved
James Joseph Sylvester was a 19th-century mathematician who specialised in algebra and geometry. He worked a lot of the time with Arthur Cayley, whose day job was in the law. Cayley had a superb memory and knew almost everything that was going on in mathematics. Sylvester was the exact opposite.
On one occasion the American mathematician William Pitt Durfee sent some of his work to Sylvester, only to be informed that the first theorem in it was false, and had never even been stated, let alone proved. Durfee produced a paper whose main objective was to prove the theorem concerned, which it did successfully.
The paper had been written by Sylvester.
James Joseph Sylvester.
Proof That 2 + 2 = 4
By definition,
2 = 1 + 1
3 = 2 + 1
4 = 3 + 1
Therefore,
where (*) is justified by the associative law
(a + b) + c = a + (b + c)
with a = (1 + 1), b = 1, c = 1.
See Note on page 332
Slicing the Doughnut
If you cut this doughnut with three straight slices, what is the largest number of pieces you can create? (You’re not allowed to move the pieces between cuts.)
Answer on page 332
How many pieces with three cuts?
The Kissing Number
If you try to surround a circular coin by coins of the same type, so that all the other coins touch the first one, you quickly discover that exactly six coins fit neatly round the first.
In 2 dimensions the kissing number is 6.
This isn’t exactly news to most of us, but it leads to a concept that turns out to be important in the theory of digital codes, as well as having mathematical interest in its own right. A coin is a circle, which is a 2-dimensional shape, so we’ve just seen that the kissing number in 2-dimensional space is 6. In n-dimensional space, the kissing number is similarly defined to be the largest number of non-overlapping unit (n - 1)-spheres that can touch (‘kiss’) a unit (n - 1)-sphere. Here an (n - 1)-sphere is the natural analogue of a circle (1-sphere) or sphere (2-sphere). The number drops from n to n - 1 because although a sphere, say, lives in 3-dimensional space, its surface has only 2 dimensions. And a circle is a curve (hence 1D) in a 2D space, the plane. The unit (n - 1)-sphere, in fact, comprises all points in n-dimensional space that are distance 1 from some fixed point, the centre of the (n - 1)-sphere.
The exact value of the kissing number is known for very few dimensions: 1, 2, 3, 4, 8 and 24, in fact. In 1D space, which is a line, a 0-sphere is a pair of points spaced 2 units apart (the diameter of a unit n-sphere is 2). So the kissing number in 1D is 2: one on the left, one on the right. We’ve just seen that in 2D space the kissing number is 6. What about higher dimensions?
In 3D space, it is easy to get 12 spheres to kiss a single sphere: you can do it with ping-pong balls and glue dots. But the arrangement is ‘sloppy’, with room to move the spheres around, and with quite a bit of space left between them. Can you fit in a 13th sphere? In 1694, David Gregory, a Scottish mathematician, thought it could be done; no lesser a luminary than Isaac Newton disagreed. The issue was sufficiently delicate that it was not resolved until 1874; it then turned out that Newton was right. So the kissing number in 3D space is 12.
In 3 dimensions the kissing number is 12.
A similar story holds in 4D space, where it’s relatively easy to find an arrangement of 24 kissing 3-spheres, but there’s enough room left so that maybe a 25th might fit in. This gap was eventually sorted out by Oleg Musin in 2003: the answer is 24.
In most other dimensions, mathematicians know that some particular number of kissing spheres is possible, because they can find such an arrangement, and that some generally much larger number is impossible, for various indirect reasons. These numbers are called the lower bound and upper bound for the kissing number, and it must lie between them.
In just two cases beyond 4D, the known lower and upper bounds coincide, and their common value is therefore the kissing number. These dimensions are 8 and 24, where the kissing numbers are, respectively, 240 and 196,650. In these dimensions there exist two highly symmetric lattices, higher-dimensional analogues of grids of squares or more generally grids of parallelograms. These special lattices are known as E8 (or the Coxeter-Todd lattice) and the Leech lattice, and spheres can be placed at suitable lattice points. By an almost miraculous coincidence, the provable upper bounds for the kissing number in these dimensions are the same as the lower bounds provided by these special lattices.
The current state of play can be summed up in a table, where I’ve used boldface for those dimensions where an exact answer is known:
The best known lower bounds, for all dimensions up to 40 and a few larger ones, can be found at:
www.research.att.com/~njas/lattices/kiss.html
The kissing number for regular arrangements, in which the centres of the spheres all lie on a lattice, is known exactly for 1-9 dimensions, as well as 24. In 1, 2, 3, 4, 8 and 24 dimensions, it is the value shown in the table. For 5, 6, 7 and 9 dimensions it is, respectively, 40, 72, 126 and 272. (The table entry 306 in 9 dimensions does not refer to a regular arrangement.)
Tippe Top Twister
The two positions of a tippe top.
The toy known as a tippe (or alternatively ‘tippy’) top is made by slicing a bit off a sphere and adding a cylindrical ‘stalk’. When you spin it - fast enough - it turns upside down. Most of us have played with a tippe top at some point, but here’s a question we possibly haven’t thought about. Suppose that, when you first spin the top, while it is still the usual way up, you spin it clockwise, looking down from above. This is the natural direction for right-handers.
When it turns over, in which direction does it spin?
Answer on page 333
When Is a Knot Not Knotted?
Topologists study things like knots, and they try to work out whether two knots are ‘topologically equivalent’, that is, can be deformed into each other. Or not. To do that, they invent cunning ‘invariants’, which are equal for equivalent knots, but may or may not be equal for two knots that are not equivalent. So knots with different invariants are definitely topologically different, but knots with the same invariants may or may not be topologically different.
It’s a knotty issue. Most of the useful invariants aren’t perfect: they’re a bit like using ‘odd/even’ to distinguish people’s ages. If Eva’s age is even and Ollie’s age is odd, then we know their ages must be different, even if we don’t know what their ages are. But if Evangeline’s age is even and Everett’s age is even, then their ages might be the same (for instance, 24 and 24) or maybe not (24 and 52). So in this case we can’t tell.
Sometimes topologists get lucky, and the invariant is good enough to tell them when a knot is not actually knotted, even if it can’t reliably
distinguish all different knots. A case in point is the so-called ‘knot group’, one of the first knot invariants discovered. I mention all this not because of the topology, which is highly technical, but because in 1972, in the mathematical fanzine Manifold, it gave rise to a poem that summed up what was good and bad about the knot group. It bore the title Knode:
A knot and
Another
knot may
not be the
same knot, though
the knot group of
the knot and the
other knot’s
knot group
differ not; BUT
if the knot group
of a knot
is the knot group
of the not
knotted
knot, then
the knot is
not
knotted.
The Origin of the Factorial Symbol
The early symbol for ‘factorial n’, which is
n×(n - 1)×(n - 2)× ‧‧‧ × 3 × 2 × 1
was
but this was tricky to print. So in 1808, the French mathematician Christian Kramp decided to change it to
which was easy to typeset. The old-fashioned version quickly went out of vogue, one of a number of examples where the practicalities of printing have affected mathematical symbolism.
Juniper Green
‘Let’s play a number game,’ said Mathophila.
Innumeratus, ever a sucker, took the bait. ‘What kind of game?’
Mathophila placed cards numbered 1-100 face up on the table.45 ‘I’ll show you the rules.’ She wrote down:
• Players take turns to choose one card. The chosen card is removed and cannot be used again.
• Apart from the opening move, the chosen number must either be an exact divisor of the previous one, or an exact multiple of it.
• The first player who cannot obey the rules loses.
‘OK,’ said Innumeratus. ‘You go first.’
‘Well, actually—’ Mathophila began, but then stopped. ‘Oh, very well.’ She picked up card 97 and discarded it.
Innumeratus, after some counting on fingers, said ‘That’s prime, isn’t it?’ When she nodded, he added ‘So I’ve got to choose card 1.’
‘Yes. The only other divisor is 97, and that’s gone. The smallest multiple is 194, and that’s too big.’
So Innumeratus picked up card 1, and discarded it.
Mathophila grinned, and picked up 89. ‘You lose.’
‘That’s prime, too?’ asked Innumeratus, who could sometimes be quite bright.
‘Yes.’
‘So I have to choose 1 again ... Oh. I can’t, it’s already gone.’ He paused. ‘That’s a silly game. The first player always wins.’
‘Yes, we call it the double-whammy tactic.’
Innumeratus thought for a moment. ‘OK, let me start. Now I’ll choose a prime.’ And he picked up card 47.
Mathophila, disdaining the 1, chose 94 instead.
‘Oops,’ said Innumeratus. ‘I hadn’t thought of that.’
‘The double whammy only works for big primes. Bigger than 50, which is half of 100.’
‘Right. So I have to choose 2 now. Because if I chose 1, you’d choose 97 again. Or 89. And I’d lose.’ So he chose 2. And eventually he lost. ‘It’s still a silly game,’ he protested. ‘I should have started with 97.’
‘True. But you were the one who insisted on playing before I told you the fourth rule, which is designed to prevent double whammies.’ And she wrote:
• The opening move must be an even number.
‘Now it’s a sensible game,’ said Mathophila. And they played quite a long game, without much regard for tactics, which illustrates the rules nicely.
I suggest that at this point you stop reading, make a set of cards, and play the game for a while. I’m going to ask you to figure out a winning strategy, and it helps to have played the game. Anyway, it’s a lot of fun.
Done that? Now we can get theoretical. Let’s look at a simplified version where the cards run from 1 to 40. That will get you started.
Some opening moves lose very quickly. For example:
An opening move of 34 suffers the same fate.
Some numbers are best avoided altogether - like 1 in the 100-card game. Suppose Mathophila is silly enough to play 5. Then Innumeratus strikes back:
Note that 25 must still be available when needed here, despite any previous moves, because it can be chosen only if the previous player plays 1 or 5.
There’s a hint of a winning strategy here. Mathophila knows she’s in trouble if she chooses 5, so she could try to force Innumeratus to choose 5 instead. Can she do this? Well, if Innumeratus chooses 7 then she can choose 35, and then Innumeratus has to choose 1 or 5, both of which lose.
Yes, but can she force Innumeratus to play 7? Well, if Innumeratus chooses 3 then Mathophila can choose 21, forcing Innumeratus to choose 7. Yes, but how does she make Innumeratus choose 3? Well, if he chooses 13, then Mathophila chooses 39 ...
Mathophila can keep building hypothetical sequences of moves, all of which force Innumeratus’s reply at every stage and lead to his inevitable defeat. The big question is: can she trap Innumeratus into such a sequence?
At some stage someone has to choose an even number, so we need to think about card 2. This is crucial because if Innumeratus chooses 2 then Mathophila can choose 26, forcing Innumeratus into the trap of playing 13. So now we come to the crunch: how can Mathophila force Innumeratus to choose 2?
She has to play an even number, and the more divisors this has, the more choices Innumeratus has, and Innumeratus might escape the trap. Anyway, the analysis gets complicated, too. Keep it simple. Suppose Mathophila opens with 22, twice a (smallish) prime. Then Innumeratus either chooses 2 and falls into Mathophila’s trap - the long sequence of forced moves just outlined - or he chooses 11. If Mathophila plays 1 she loses, so she chooses 33 instead. Now 11 has already been used, so Innumeratus is forced to choose 3 - and the trap is sprung. We already know how Mathophila can win when he does that. So Mathophila must win if she starts with 22.
That’s probably a bit confusing by now, so here’s a summary of Mathophila’s winning strategy. The two sets of columns deal with the two alternatives available to Innumeratus. For simplicity, I’ve assumed throughout that both players avoid 1, since it is an instant loss. With this choice eliminated, virtually every move is forced.
There is at least one other opening move for Mathophila that also lets her force a win: if she chooses 26, then the same kind of game develops, but with a few of the moves interchanged.
The crucial features of Mathophila’s strategy are the primes 11 and 13. Her opening move is twice such a prime: 22 or 26. It forces Innumeratus to reply either with 2 - at which point Mathophila is home and dry - or the prime. Then Mathophila replies with three times the prime, forcing Innumeratus to go to 3 - and she’s home and dry again.
So Mathophila escapes trouble because as well as twice the prime, there is exactly one other multiple of such a prime in the range being played, namely 33 or 39. This provides her with an escape route. Call these the medium primes - they lie between one-third and one-quarter of the number of cards. If Mathophila chooses twice a medium prime, then Innumeratus must choose that prime. Then she chooses three times that prime, forcing Innumeratus to play the number 3.
Here are two questions for you:
• Can Mathophila win by any other strategy?
• Is there an analogous winning strategy for the 100-card version, and who wins?
More ambitiously, consider the game JG-n with the same rules, using an arbitrary whole number n of cards. Since no draws are allowed, and every game stops after finitely many moves, game theory implies that either Mathophila has a winning strategy, or Innumeratus does.
• With perfect strategy, who wins JG-n, assuming Mathophila goes first?
Certainly the answer depends on n. Mathophila wins when n is 3 or 8,
whereas Innumeratus wins when n = 1, 2, 4, 5, 6, 7, 9. What about n = 100? What about all the values of n from 10 to 99? Can you solve the whole thing?
Answers on page 333
Mathematical Metajoke
An engineer, a physicist, and a mathematician found themselves in a joke, very similar to many that you will have heard before, but did not immediately realise where they were.46 After a hasty back-of-the-envelope calculation, the engineer worked out what had happened and began to chuckle. Soon after, the physicist intuited where they were, based on a loose analogy with a particle confined in a box, and began laughing uproariously. The mathematician, however, seemed not to find their situation remotely funny. Eventually the others asked why.
‘I saw immediately that I was in an anecdote of some kind,’ he replied. ‘But it was only after I noticed characteristic structural features that I could be sure the anecdote was a joke. However, this joke is far too trivial a consequence of the general case to have any amusement value.’
Beyond the Fourth Dimension
Physicists are seeking a Theory of Everything (ToE) that will unify the two pillars of modern physics, relativity and quantum mechanics, while fixing certain inconsistencies between these two theories. The search has led them to speculate that our familiar 3-dimensional (3D) space is not actually 3D at all, but 10D or maybe 11D. The extra dimensions provide a place for fundamental particles to vibrate in (like a violin string), thereby giving rise to quantum numbers such as spin and charge (which are like the notes produced by the violin string). Now, you might think that it would be difficult for everyone to have been so wrong for so long about something so basic as the dimensionality of space. And in any case, surely space is space, and it can’t have 10 dimensions because there isn’t any room to fit 7 more in once we’ve sorted out the first 3.
Ian Stewart Page 20