Professor Stewart's Hoard of Mathematical Treasures Read Online Free Page A

ignoring symmetrically related ones: a silly one of size 1 and a sensible one of size 3.

    The only possible normal magic hexagons, of size 1 and 3, and an abnormal hexagon of size 7.
    That’s true for ‘normal’ magic hexagons, where the numbers are consecutive integers starting 1, 2, 3, . . . . But it turns out that there are more possibilities if you allow ‘abnormal’ ones, where the numbers remain consecutive but start further along, say 3, 4, 5, . . . . The largest known abnormal magic hexagon was found by Zahray Arsen in 2006. It has size 7, the numbers run from 2 to 128, and the magic constant - the sum of the numbers in any row or slanting line - is 635. Arsen has also discovered abnormal magic hexagons of size 4 and 5. See en.wikipedia.org/wiki/Magic_hexagon

The Collatz-Syracuse-Ulam Problem
    Simple questions need not be easy to answer. Here’s a famous example. You can explore it with pencil and paper, or a calculator, but what it does in general baffles even the world’s greatest mathematicians. They think they know the answer, but no one can prove it. It goes like this.
    Think of a number. Now apply the following rules over and over again:
    • If the number is even, divide it by 2.
    • If the number is odd, multiply it by 3 and add 1.
    What happens?
    I thought of 11. This is odd, so the next number is 3×11 + 1 = 34. That’s even, so I divide by 2 to get 17. This is odd, and leads to 52. After that the numbers go 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1. From there, we get 4, 2, 1, 4, 2, 1 indefinitely. So usually we add a third rule:
    • If you reach 1, stop.
    In 1937, Lothar Collatz asked whether this procedure always reaches 1, no matter what number you start with. More than seventy years later, we still don’t know the answer. There are several other names for this problem: the Syracuse problem, the 3 n + 1 problem, the Ulam problem. It is often posed as a conjecture which states that the answer is yes, and that’s what most mathematicians expect.

    The fates of the numbers 1-20, and anything else they lead to.
    One thing that makes the Collatz-Syracuse-Ulam problem or conjecture hard is that the numbers don’t always get smaller as you proceed. The chain starting with 15 gets up to 160 before eventually subsiding. Little old 27 positively explodes:
    It takes 111 steps to get to 1. But it does get there eventually.
    This kind of thing makes you wonder whether there might be some particular number for which the process is even more explosive, and heads off to infinity. The numbers will go up and down a lot, of course. Any odd number leads to an increase, but the number can’t increase twice in succession: when n is odd, 3n + 1 is even, so the next step after that is to divide by 2. But the result at that stage is still bigger than n; in fact, it’s(3 n + 1). However, if this is also even, we get to something smaller than n, namely(3 n + 1). So what happens is quite delicate.
    If no number explodes to infinity, the other possibility is that there might be some other cycle which some numbers hit instead of 4 → 2 → 1. It has been proved that any such cycle must contain at least 35,400 terms.
    Up to 100 million, the number that takes longest to reach 1 is 63,728,127, which requires 949 steps.
    Computer calculations show that every starting number up to at least 19×2 58 ≈ 5.48×10 18 eventually hits 1. This is impressively large, and a lot of theoretical input has to go into the computation - you don’t just check the numbers one by one. But the example of Skewes’ number (see page 46) shows that 10 18 isn’t really very big, as such things go, so the computer evidence
isn’t as convincing as it might seem. Everything we know about the question conspires to indicate that if there is an exceptional number that doesn’t hit 1, it will be absolutely gigantic.
    Probability calculations suggest that the probability that some number heads off to infinity is zero. However, such calculations are