The Fibonacci sequence is beautiful like clockwork. There’s a perfectly clear, rigorously defined mechanism ticking out an entirely predictable result for ever:
But I also like sequences that you might call definitely arbitrary. That is, there’s a perfectly clear, rigorously defined mechanism, but the results seem arbitrary — not predictable at all:
What’s the formula there? That sequence is defined at the OEIS as “The number of three-term Egyptian fractions of rational numbers x/y, 0 < x/y < 1, ordered as below. The sequence is the number of (p,q,r) such that x/y = 1/p + 1/q + 1/r where p, q, and r are integers with p < q < r.” For example: “The sixth rational number is 3/4 [and] 3/4 = 1/2 + 1/5 + 1/20 = 1/2 + 1/6 + 1/12 = 1/3 + 1/4 + 1/5, so a(6)=3.”
Numbers have thin skins. And they’re easily replaced. Take 71624133. Here it is permuting its pellicles:
71624133 in base 10 = 100010001001110010111000101 in base 2 = 11222202212211200 in b3 = 10101032113011 in b4 = 121313433013 in b5 = 11035053113 in b6 = 1526536500 in b7 = 421162705 in b8 = 158685750 in b9 = 374802A9 in b11 = 1BBA1199 in b12 = 11AB9B59 in b13 = 9726137 in b14 = 644BE73 in b15 = F3855B7 in b16
But if digits are the skin of 71624133, what are its bones? Well, you could say the skeleton of a number, something that doesn’t change from base to base, is its prime factorization:
71624133 = 32 × 72 × 162413
But the primes themselves are numbers, so they’re wearing pellicles too. And it turns out that, in base 10, the pellicles of the prime factors of 71624133 match the pellicle of 71624133 itself:
You can find them at the Online Encyclopedia of Integer Sequences under A121342, “Composite numbers that are a concatenation of their distinct prime divisors in some order.” But what about pairs of primal pellicles, that is, pairs of numbers where the prime factors of each form the pellicle of the other?
You get the rational fractions ordered by denominator in their simplest form: 1/2, 1/3, 2/3, 1/4, 3/4, 1/5, 2/5, 3/5… There are no pairs like 2/4 and 5/35, because those can be simplified: 2/4 → 1/2; 15/35 → 3/7. You can get the same set of rational fractions by listing every successive pair in this sequence, the Stern-Brocot sequence:
But the fractions don’t come ordered by denominator this time. In fact, they seem to come at random: 1/2, 1/3, 2/3, 1/4, 3/5, 2/5, 3/4, 1/5, 4/7, 3/8, 5/7, 2/7, 5/8… But they’re not random at all. There’s a complicated way of generating them and a simple way. An amazingly simple way, I think:
Moshe Newman proved that the fraction a(n+1)/a(n+2) can be generated from the previous fraction a(n)/a(n+1) = x by 1/(2*floor(x) + 1 – x). The successor function f(x) = 1/(floor(x) + 1 – frac(x)) can also be used. — A002487, “Stern-Brocot Sequence”, at the OEIS
In another form, the Stern-Brocot sequence is generated by what’s called the Calkin-Wilf Tree. Now suppose you use the Stern-Brocot sequence to supply the x co-ordinate of an L-graph whose arms run from 0 to 1. And you use the Calkin-Wilf Tree to supply the y co-ordinate of the L-tree. What do you get? As I described in “I Like Gryke”, you get this fractal:
Limestone fractal
I call it a limestone fractal or pavement fractal or gryke fractal, because it reminds me of the fissured patterns you see in the limestone pavements of the Yorkshire Dales:
But what happens when you plot the (x,y) of the Stern-Brocot sequence and the Calkin-Wilf Tree on a circle instead? You get an interestingly distorted limestone fractal:
Limestone fractal on circle
You can also plot the (x,y) around the perimeter of a polygon, then stretch the polygon into a circle. Here’s a square:
Limestone fractal on square
⇓
Limestone square stretched to circle
And here are a pentagon, hexagon, heptagon and octagon — note the interesting perspective effects:
Limestone fractal on pentagon
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Limestone pentagon stretched to circle
Limestone fractal on hexagon
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Limestone hexagon stretched to circle
Limestone fractal on heptagon
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Limestone heptagon stretched to circle
Limestone fractal on octagon
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Limestone octagon stretched to circle
And finally, here are animations of limestone polygons stretching to circles:
Limestone square stretched to circle (animated at EZgif)
Limestone pentagon to circle (animated)
Limestone hexagon to circle (animated)
Limestone heptagon to circle (animated)
Limestone octagon to circle (animated)
Previously Pre-Posted (Please Peruse)
• I Like Gryke — a first look at the limestone fractal
If you want a good example of how, in math, something very simple can quickly get very deep, just look at partitions. Here are the partitions of 1 to 5, that is, the ways 1 to 5 can be expressed as a sum of integers smaller than or equal to themselves:
It’s very easy to understand the concept of partitions, but very difficult to understand how partitions behave. For example, here is numbpart(n), the count of partitions for 1, 2, 3,…
1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101, 135, 176, 231, 297, 385, 490, 627, 792, 1002, 1255, 1575, 1958, 2436, 3010, 3718, 4565, 5604, 6842, 8349, 10143, 12310, 14883, 17977, 21637, 26015, 31185, 37338, 44583, 53174, 63261, 75175, 89134, 105558, 124754, 147273, 173525, 204226, … A000041 at the Online Encyclopedia of Integer Sequences, “a(n) is the number of partitions of n (the partition numbers)”
What’s the formula for numbpart(n)? That’s a tricky question. And what’s the formula for the curves produced by counting the various lengths of partitions(n)? That’s another tricky question, but one thing is easy to see. As n gets bigger, the graph of countlen(partitions(n)) acquires a strange, lopsided beauty. Here are the partitions of 8, with the count of how many partitions of a particular length there are:
“The uniquely unrepresentative ‘Egyptian’ fraction.” That’s what David Wells calls 2/3 = 0·666… in The Penguin Dictionary of Curious and Interesting Numbers (1986). Why unrepresentative”? Wells goes on to explain: “the Egyptians used only unit fractions, with this one exception. All other fractional quantities were expressed as sums of unit fractions.”
A unit fraction is 1 divided by a higher integer: 1/2, 1/3, 1/4, 1/5 and so on. Modern mathematicians are interested in those sums of unit fractions that produce integers, like this:
And I looked for Egyptian fractions whose denominators summed to rep-digits like 111 and 666 (denominators are the bit below the stroke of 1/3 or 2/3, where the bit above is called the numerator):
Today is 23rd June, so here’s a Fascinating Fibonacci Fact for Phiday. First, list the rational fractions < 1 in simplified form and mark the Fibonacci fractions:
The 11th, 12th and 23rd day of a month can be called a φ-day (pronounced fy-day). Why so? Because those numbers are consecutive entries in the famous Fibonacci sequence, which offers better and better approximations to a mathematical constant called φ = (√5 + 1) / 2 = 1.6180339887498948…:
Each number after the second is the sum of the preceding two (so 11, 12, 23… could be the start of a similar sequence). When you divide fib(i) by fib(i-1), you get these approximations to φ:
Today is the 23rd and not just a φ-day but a Friday (or φriday). So here’s one of the interesting results I’ve recently found while playing with the Fibonacci sequence. As any recreational mathematician kno, you can also find the Fibonacci sequence — and φ — with this little algorithm:
“The most merciful thing in the world, I think, is the inability of the human mind to correlate all its contents.” So said HPL in “The Call of Cthulhu” (1926). But I’d still like to correlate the contents of mine a bit better. For example, I knew that φ, the golden ratio, is the most irrational of all numbers, in that it is the slowest to be approximated with rational fractions. And I also knew that continued fractions, or CFs, were a way of representing both rationals and irrationals as a string of numbers, like this:
But I didn’t correlate those two contents of my mind: the maximal irrationality of φ and the way continued fractions work.
That’s why I was surprised when I was looking at the continued fractions of 2..(n-1) / n for 3,4,5,6,7… That is, I was looking at the continued fractions of 2/3, 3/4, 2/5, 3/5, 4/5, 5/6, 2/7, 3/7… (skipping fractions like 2/4, 2/6, 3/6 etc, because they’re reducible: 2/4 = ½, 2/6 = 1/3, 3/6 = ½ etc). I wondered which fractions set successive records for the length of their continued fractions as one worked through ½, 2/3, 3/4, 2/5, 3/5, 4/5, 5/6, 2/7, 3/7… And because I hadn’t correlated the contents of my mind, I was surprised at the result. I shouldn’t have been, of course:
Which n1/n2 set records for the length of their continued fractions (with n2 > n1)? It’s the successive Fibonacci fractions, fib(i)/fib(i+1), of course. I didn’t anticipate that answer because I didn’t understand φ and continued fractions properly. And I still don’t, because I’ve been surprised again today looking at palindromic CFs like these:
Now it’s the successive Fibonacci skip-one fractions, fib(i)/fib(i+2), that set records for the length of their palindromic continued fractions. But I think you’d have to be very good at maths not to be surprised by that result.
After that, I continued to be compelled by the Call of CFulhu and started to look at the CFs of Fibonacci skip-n fractions in general. That’s contfrac(fib(i)/fib(i+n)) for n = 1,2,3,… And I’ve found more interesting patterns, as I’ll describe in a follow-up post.
It’s a very simple function that raises a very difficult question. An unanswered question, in fact. Take any whole number. If it’s odd, multiply it by 3 and add 1. If it’s even, divide it by 2. Repeat until you reach 1. That’s the hailstone function, because the numbers rise and fall like hailstones being formed in a cloud. Here are some examples:
But is this function truly a hailstone function? That is, does every number fall finally to earth and reach 1? So far, for every number tested, the answer has been yes. But do all numbers reach 1? The Collatz conjecture says they do. But no-one can prove it. Or disprove it. All it would take is one number failing to fall to earth. Mathematicians don’t think there is one, but numbers can take a surprising length of time to get to the ground. Here’s 27:
27 takes 111 steps to reach 1. And the 111 made me think of another question. If the function hail(n) returns the number of steps required for n to reach 1, then hail(27) = 111. But what about hail(n) = 666? That is, what is the first number that requires 666 steps to reach 1? I say “first number”, because one very big number is guaranteed to take 666 steps:
Put another way, 666 = hail(2^666), because for any power of 2, hail(2^p) = p. But is there a smaller number, which I’ll call satan, for which hail(satan) = 666? Here’s a tantalizing taster of the task:
Here’s a question I haven’t answered: if satanic numbers are those n satisfying hail(n) = 666, how many satanic numbers are there? We’ve already seen two of them: 666 = hail(2^666) = hail(26597116). But how many more are there? Not infinitely many, because for n > 2^666, hail(n) > 666. In fact, after satan = 26597116, the next three satanic numbers arrive very quickly:
So there are four consecutive satanic numbers. But it isn’t unusual for a run of consecutive numbers to have the same hail(). Here’s a graph of the values of hail(n) for n = 1,2,3… (running left-to-right, down-up, with 1,2,3… in the lower lefthand corner). When n is divisible by 10, hail(n) is represented in red; when n is odd and divisible by 5, hail(n) is green. Note how many runs of identical hail(n) there are:
Graph for hail(n)
Here are successive records for runs of identical hail(n):
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