Category Archives: Continued Fractions

Worms in Terms of Perms

by in Arithmetic, Continued Fractions, Fractals, Mathematics and tagged , , , , , , , , , | Leave a comment

If you go back far enough, we’re all worms. All us animals, that is. But in a subtler sense, all life is vermiform — animals, plants, fungi, bacteria. DNA is a kind of worm, a string of chemicals encoding the recipe for an animal, plant, fungus or bacterium. And the worms of DNA can be converted into numbers, just as some numbers can be turned into worms:

3/7 = 0·0.428571428571428571428571…
154/183 = 0.841530054644808743169398907…
√2 = 1.414213562373095048801688…
π = 3.1415926535897932384626433…

Those are decimals, but there’s another kind of worm for such numbers. It’s called a continued fraction:

contfrac(3/7) = [0,2,3]
contfrac(154/183) = [0,1,5,3,4,2]
contfrac(√2) = [1,2,2,2,2,2…]
contfrac(π) = [3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2, 1, 84, 2, 1, 1, 15…]

Extracting and enacting continued fractions is very simple. Here’s the extracting:

3/7 → 1/(3/7) = 7/3 = 2+1/3 – 2 = 1/3 → 1(1/3) = 3, ∴ contfrac(3/7) = [0,2,3]
154/183 → 1/(154/183) = 183/154 = 1 + 29/154 – 1 = 29/154 → 1/(29/154) = 154/29 = 5 + 9/29 – 5 = 9/29 → 1/(9/29) = 29/9 = 3 + 2/9 – 3 = 2/9 → 1/(2/9) = 9/2 = 4 + 1/2 – 4 = 1/2 → 1/(1/2) = 2 – 2 = 0, ∴ contfrac(154/183) = [0,1,5,3,4,2]

And here’s the enacting:

[0,2,3] → 3 → 1/3 → 1/3 + 2 = 7/3 → 1/(7/3) = 3/7
[0,1,5,3,4,2] → 2 → 1/2 → 1/2 + 4 = 9/2 → 2/9 + 3 = 29/9 → 9/29 + 5 = 154/29 → 29/154 + 1 = 183/154 → 1/(183/154) = 154/183

Once you’ve got the worm of a continued fraction, you can perm the worm, as it were, generating different fractions like this (I’m dropping the initial [0,…] of the contfracs):

[2,3,4] = contfrac(13/30)
[2,4,3] → 13/29
[3,2,4] → 09/31
[3,4,2] → 09/29
[4,2,3] → 07/31
[4,3,2] → 07/30

Reversing a continued fraction is a kind of permutation, so the fractal below represents one kind of worms in terms of perms:

Variant of a limestone fractal or gryke fractal

I call that graph a fract-L, because it’s shaped like an L and the x axis represents the simplified fractions 1/2, 1/3, 2/3, 1/4, 3/4, 1/5, 2/5, 3/5…, while the y axis represents the fractions you get by reversing the continued fractions of 1/2, 1/3, 2/3…:

contfrac(1/2) = [2] → 1/2
contfrac(1/3) = [3] → 1/3
contfrac(2/3) = [1,2] → 1/3
contfrac(1/4) = [4] → 1/4
contfrac(3/4) = [1,3] → 1/4
contfrac(1/5) = [5] → 1/5
contfrac(2/5) = [2,2] → 2/5
contfrac(3/5) = [1,1,2] → 2/5
contfrac(4/5) = [1,4] → 1/5
contfrac(1/6) = [6] → 1/6
contfrac(5/6) = [1,5] → 1/6
contfrac(1/7) = [7] → 1/7
contfrac(2/7) = [3,2] → 3/7
contfrac(3/7) = [2,3] → 2/7
contfrac(4/7) = [1,1,3] → 2/7
contfrac(5/7) = [1,2,2] → 3/7
contfrac(6/7) = [1,6] → 1/7
contfrac(1/8) = [8] → 1/8
contfrac(3/8) = [2,1,2] → 3/8
contfrac(5/8) = [1,1,1,2] → 3/8
contfrac(7/8) = [1,7] → 1/8
contfrac(1/9) = [9] → 1/9
contfrac(2/9) = [4,2] → 4/9
contfrac(4/9) = [2,4] → 2/9
contfrac(5/9) = [1,1,4] → 2/9
contfrac(7/9) = [1,3,2] → 4/9
contfrac(8/9) = [1,8] → 1/9
[…]

If you perm the worm in other ways, you get other shapes on the fract-L. I looked at continued fractions of fixed length, 4, 5 and 6, and permed them using one of the permutations of [1,2,3,4], [1,2,3,4,5] and [1,2,3,4,5,6]. Here’s a graph for fractions, a/b, and permed fractions, perm(a/b), where length(contfrac(a/b)) = 4:

x = a/b when length(contfrac(a/b)) = 4, y = fraction from contfrac(a/b) permed with [1,3,2,4]

The x axis represents simplified fractions, a/b, when len(cf(a/b)) = 4. The y axis represents the fractions found by applying the perm [1,3,2,4] to contfrac(a/b). That is, the first number of the contfrac stays where it is, the third number moves to position 2, the second number moves to position 3 and the fourth number stays where it is. In short, you simply swap the middle two numbers of contfrac(a/b). Here’s an example:

contfrac(9/43) = [4,1,3,2] → [4,3,1,2] → 11/47, because contfrac(11/47) = [4,3,1,2]

Here are more fract-Ls representing worms in terms of perms:

fract-L for contfrac(a/b) permed by [2,1,3,4]

fract-L for contfrac(a/b) permed by [3,2,1,4]

fract-L for contfrac(a/b) permed by [1,4,2,3,5] (i.e. a/b where len(contfrac(a/b)) = 5)

fract-L for contfrac(a/b) permed by [1,5,3,4,2]

fract-L for contfrac(a/b) permed by [2,1,4,3,5]

fract-L for contfrac(a/b) permed by [3,4,1,2,5]

fract-L for contfrac(a/b) permed by [4,2,3,1,5]

fract-L for contfrac(a/b) permed by [4,2,5,3,1]

fract-L for contfrac(a/b) permed by [4,3,2,1,5]

fract-L for contfrac(a/b) permed by [5,3,4,2,1]

fract-L for contfrac(a/b) permed by [2,1,4,3,5,6] (i.e. a/b where len(contfrac(a/b)) = 6)

fract-L for contfrac(a/b) permed by [2,1,5,4,3,6]

fract-L for contfrac(a/b) permed by [3,2,1,4,5,6]

fract-L for contfrac(a/b) permed by [3,2,1,5,4,6]

fract-L for contfrac(a/b) permed by [3,5,1,4,2,6]

fract-L for contfrac(a/b) permed by [4,2,5,1,3,6]

fract-L for contfrac(a/b) permed by [4,3,2,1,5,6]

fract-L for contfrac(a/b) permed by [4,5,2,3,1,6]

fract-L for contfrac(a/b) permed by [1,3,2,6,5,4,7] (i.e. a/b where len(contfrac(a/b)) = 7)

fract-L for contfrac(a/b) permed by [1,5,2,6,3,4,7]

fract-L for contfrac(a/b) permed by [5,6,3,7,4,1,2]

fract-L for contfrac(a/b) permed by [6,2,3,5,4,1,7]

fract-L for contfrac(a/b) permed by [6,2,5,4,7,3,1]

Post-Performative Post-Scriptum

Much as I hate the phrase “in terms of”, I was happy to use it in the title of this post. After all, it isn’t ugly but assonant there. And it began life in mathematics, where it still has its proper meaning rather than being pretentious and prolix:

How did this complex preposition come into being? The OED [Oxford English Dictionary] reveals that it has been in use since the mid-18c. as a mathematical expression “said of a series…stated in terms involving some particular (my emphasis) quantity”, and illustrates this technical usage by citing examples from the work of Herbert Spencer (1862), J. F. W. Herschel (1866), and other writers. From this technical use came at first a trickle and, after the 1940s, a flood of imitative uses by non-mathematicians. — “Terminal Trinity”

Elsewhere Other-Engageable

• A Fracteasel on a Fract-L — an earlier look at continued fractions and fractal fract-Ls

A FracTeasel on a Fract-L

by in Arithmetic, Continued Fractions, Fractals, Mathematics and tagged , , , , , , , , , , , , | Leave a comment

Here are two new fractals, both of which remind me of the seedheads of the wildflower known as a teasel, Dipsacus fullonum:

A FracTeasel fractal

Dried seedheads of teasel, Dipsacus fullonum (Wikipedia)

Another FracTeasel fractal (embedded in the first)

Flowering seedhead of teasel, Dipsacus fullonum (Wikipedia)

How do you create the two FracTeasels? Let’s look first at the fractal they’re inspired by. In “Back to Frac’” I talked about this fractional fractal, a variant of what I call the limestone fractal:

Variant of a limestone fractal or gryke fractal

It’s a fractal on a fract-L, that is, the x and y co-ordinates of the red L represent pairs of fractions generating decimals between 0 and 1. The x represents the fractions a1/b1 = 1/n to (n-1)/n in simplest form: 1/2, 1/3, 2/3, 1/4, 3/4, 1/5, 2/5, 3/5, 4/5, 1/6, 5/6, 1/7, 2/7, 3/7, 4/7, 5/7, 6/7, 1/8, 3/8, 5/8, 7/8,…

And what about the y? It represents the fraction found by taking the continued fraction of a1/b1, reversing it, and generating a new fraction, a2/b2, from the reversal. For example, here’s the continued fraction of a1/b1 = 3/23 = 0.1304347826…:

contfrac(3/23) = 7,1,2

The continued fraction of a1/b1 = 3/23 is used like this to reconstruct a1/b1:

7,1,2

0 → 1 / (0 + 2) = 1/2 → 1 / (1/2 + 1) = 2/3 → 1 / (7 + 2/3) = 3/23

Now reverse the continued fraction, 7,1,2 → 2,1,7, and generate a2/b2:

2,1,7

0 → 1 / (0 + 7) = 1/7 → 1 / (1/7 + 1) = 7/8 → 1 / (2 + 7/8) = 8/23 = 0.3478260869565…

The limestone fractal above appears when a1/b1 → a2/b2 for a1/b1 = 1/2, 1/3, 2/3, 1/4, 3/4, 1/5, 2/5, 3/5, 4/5, 1/6, 5/6, 1/7, 2/7, 3/7, 4/7, 5/7, 6/7, 1/8, 3/8, 5/8, 7/8,… But you can do other things to contfrac(a1/b1) beside just reversing it. What about the permutations of contfrac(a1/b1), for example? If length(contfrac(a1/b1)) = n, the permutations can generate up to n! (factorial n) new a2/b2 for the y co-ordinate (if all the numbers of contfrac(a1/b1) are different, you’ll get n! permutations). The resultant fractal is the first of the FracTeasels above (note that a2/b2 isn’t multiplied by two):

FracTeasel #1 from fract-L for y = perm(contfrac(a1/b1))

If you think about it, you’ll see that the fractal from permed contfrac(a1/b1) contains the fractal from reversed contfrac(a1/b1). It also contains the second FracTeasel:

FracTeasel #2

How so? Because the second FracTeasel — let’s call it the stemmed FracTeasel — is created by shifting some numbers in contfrac(a1/b1) and leaving others alone. For example:

contfrac(940/1089) = 1, 6, 3, 4, 5, 2 → 1, 4, 3, 2, 5, 6 = contfrac(1008/1243)

So the function is finding one particular permutation of contfrac(a1/b1) to generate a2/b2, not all permutations. And so the function creates the stemmed FracTeasel, which carries an infinite number of seedheads on the same stem. To show that, here’s an animated gif zooming in on the bend of the fract-L for the stemmed FracTeasel:

Zooming the FracTeasel (animated at ezGif)

Elsewhere Other-Accessible…

• I Like Gryke — a first look at the limestone fractal
• Lime Time — more on the limestone fractal

Matchin’ Fraction

by in Arithmetic, Base Behavior, Continued Fractions, Mathematics, Narcissistic numbers and tagged , , , , | Leave a comment

I wondered whether contfrac(a/b), the continued fraction of a/b, ever matched the digits of a and b in some base. The answer was yes. But I haven’t found any examples in base 10:

3,1,2 = contfrac(3/12) in base 9 = contfrac(3/11) in base 10
4,1,3 = contfrac(4/13) in b16 = contfrac(4/19) in b10
5,1,4 ← 5/14 in b25 = 5/29
6,1,5 ← 6/15 in b36 = 6/41
25,2 ← 2/52 in b9 = 2/47 → 23,2
7,1,6 ← 7/16 in b49 = 7/55
8,1,7 ← 8/17 in b64 = 8/71
9,1,8 ← 9/18 in b81 = 9/89
A,1,9 ← A/19 in b100 = 10/109 → 10,1,9
42,1,3 ← 4/213 in b8 = 4/139 → 34,1,3
4,1,2,3,3 ← 41/233 in b8 = 33/155
1,17,1,2,3 ← 117/123 in b14 = 217/227 → 1,21,1,2,3
3A,3 ← 3/A3 in b28 = 3/283 → 94,3
3,5,A,2 ← 35/A2 in b34 = 107/342 → 3,5,10,2
3,1,4,1,4,1,5 ← 314/1415 in b8 = 204/781
2,1,36,3,2 ← 213/632 in b12 = 303/902 → 2,1,42,3,2
3,2,11,2,2,2 ← 321/1222 in b9 = 262/911 → 3,2,10,2,2,2
4H,4 ← 4/H4 in b65 = 4/1109 → 277,4
6,2,1,3,J ← 62/13J in b35 = 212/1349 → 6,2,1,3,19
8,3,3,1,D ← 83/31D in b22 = 179/1487 → 8,3,3,1,13
93,1,8 ← 9/318 in b27 = 9/2222 → 246,1,8
1,3A,1,1,4,2,2 ← 13A1/1422 in b12 = 2281/2330 → 1,46,1,1,4,2,2
C7,1,B ← C/71B in b21 = 12/3119 → 259,1,11
1,2,2,1,O,F ← 122/1OF in b50 = 2602/3715 → 1,2,2,1,24,15
2,1,1,5,55 ← 211/555 in b28 = 1597/4065 → 2,1,1,5,145
3,1,1,A,K,6 ← 311/AK6 in b29 = 2553/8996 → 3,1,1,10,20,6
1,2[70],1,3,9 ← 12[70]/139 in b98 = 9870/9907 → 1,266,1,3,9
1,E,4,1,M,7 ← 1E4/1M7 in b100 = 11404/12207 → 1,14,4,1,22,7
LG,5,4 ← L/G54 in b28 = 21/12688 → 604,5,4
G4,1,F ← G/41F in b64 = 16/16463 → 1028,1,15

Back to Frac’

by in Arithmetic, Continued Fractions, Fractals, Mathematics and tagged , , , , , , , , , | Leave a comment

Here’s a second serendipitous fractal:

A serendipitous fractal on a fract-L

It looks like (and is related to) the limestone fractal and I found it similarly serendipitously. This time I was looking at continued fractions, a simple yet subtle and seductive way of representing non-integer numbers like 2/3 and 7/9 (or √2 and π). To generate a continued fraction from a/b < 1, you divide a/b into 1 and take away the integer part. Then you repeat with the remainder until nothing is left (or, as with irrationals like 1/√2 and 1/π, you've calculated long enough for your needs). The integers at each stage are the numbers of the continued fraction. Here is the working for contfrac(2/3), the continued fraction of 2/3:

int(1/(2/3)) = int(3/2) = int(1.5) = 1
3/2 – 1 = 1/2
int(1/(1/2)) = int(2) = 2
2 – 2 = 0

contfrac(2/3) = 1, 2

By working backwards with (1, 2), you can use the continued fraction to reconstruct the original number a/b. Start with a/b = 0/1:

1 / (0/1 + 2) = 1 / ((0+2*1)/2) = 1 / (2/1) = 1/2
1 / (1/2 + 1) = 1 / ((1+2*1)/2) = 1 / (3/2) = 2/3

And here’s the working for contfrac(7/9), the continued fraction of 7/9:

int(1/(7/9)) = int(9/7) = int(1.285714…) = 1
9/7 – 1 = 2/7
int(1/(2/7)) = int(7/2) = int(3.5) = 3
7/2 – 3 = 1/2
int(1/(1/2)) = int(2) = 2
2 – 2 = 0

contfrac(7/9) = 1, 3, 2

And here’s the reconstruction of 7/9 from its continued fraction, starting again with a/b = 0/1:

1 / (0/1 + 2) = 1 / ((0+2*1)/2) = 1 / (2/1) = 1/2
1 / (1/2 + 3) = 1 / ((1+2*3)/2) = 1 / (7/2) = 2/7
1 / (2/7 + 1) = 1 / ((2+7*1)/7) = 1 / (9/7) = 7/9

From that simple algorithm arise subtle and seductive things. Look at some continued fractions, cf(a/b), for a/b in simplest form (giving only the first few reciprocals, 1/b, because cf(1/b) = b). Interesting patterns appear, e.g. when a/b uses adjacent or nearly adjacent Fibonacci numbers:

cf(1/3) = 3 = cf(0.333333333…)
cf(2/3) = 1,2 = cf(0.666666666…)
cf(1/4) = 4 = cf(0.25)
cf(3/4) = 1,3 = cf(0.75)
cf(1/5) = 5 = cf(0.2)
cf(2/5) = 2,2 = cf(0.4)
cf(3/5) = 1,1,2 = cf(0.6)
cf(4/5) = 1,4 = cf(0.8)
cf(5/6) = 1,5 = cf(0.833333333…)
cf(2/7) = 3,2 = cf(0.285714285…)
cf(3/7) = 2,3 = cf(0.428571428…)
cf(4/7) = 1,1,3 = cf(0.571428571…)
cf(5/7) = 1,2,2 = cf(0.714285714…)
cf(6/7) = 1,6 = cf(0.857142857…)
cf(3/8) = 2,1,2 = cf(0.375)
cf(5/8) = 1,1,1,2 = cf(0.625)
cf(7/8) = 1,7 = cf(0.875)
cf(2/9) = 4,2 = cf(0.222222222…)
cf(4/9) = 2,4 = cf(0.444444444…)
cf(5/9) = 1,1,4 = cf(0.555555555…)
cf(7/9) = 1,3,2 = cf(0.777777777…)
cf(8/9) = 1,8 = cf(0.888888888…)
cf(3/10) = 3,3 = cf(0.3)
cf(7/10) = 1,2,3 = cf(0.7)
cf(9/10) = 1,9 = cf(0.9)
cf(2/11) = 5,2 = cf(0.181818181…)
cf(3/11) = 3,1,2 = cf(0.272727272…)
cf(4/11) = 2,1,3 = cf(0.363636363…)
cf(5/11) = 2,5 = cf(0.454545454…)
cf(6/11) = 1,1,5 = cf(0.545454545…)
cf(7/11) = 1,1,1,3 = cf(0.636363636…)
cf(8/11) = 1,2,1,2 = cf(0.727272727…)
cf(9/11) = 1,4,2 = cf(0.818181818…)
cf(10/11) = 1,10 = cf(0.909090909…)
cf(5/12) = 2,2,2 = cf(0.416666666…)
cf(7/12) = 1,1,2,2 = cf(0.583333333…)
cf(11/12) = 1,11 = cf(0.916666666…)
cf(2/13) = 6,2 = cf(0.153846153…)
cf(3/13) = 4,3 = cf(0.230769230…)
cf(4/13) = 3,4 = cf(0.307692307…)
cf(5/13) = 2,1,1,2 = cf(0.384615384…)
cf(6/13) = 2,6 = cf(0.461538461…)
cf(7/13) = 1,1,6 = cf(0.538461538…)
cf(8/13) = 1,1,1,1,2 = cf(0.615384615…)
cf(9/13) = 1,2,4 = cf(0.692307692…)
cf(10/13) = 1,3,3 = cf(0.769230769…)
cf(11/13) = 1,5,2 = cf(0.846153846…)
cf(12/13) = 1,12 = cf(0.923076923…)
cf(3/14) = 4,1,2 = cf(0.214285714…)
cf(5/14) = 2,1,4 = cf(0.357142857…)
cf(9/14) = 1,1,1,4 = cf(0.642857142…)
cf(11/14) = 1,3,1,2 = cf(0.785714285…)
cf(13/14) = 1,13 = cf(0.928571428…)
cf(2/15) = 7,2 = cf(0.133333333…)
cf(4/15) = 3,1,3 = cf(0.266666666…)
cf(7/15) = 2,7 = cf(0.466666666…)
cf(8/15) = 1,1,7 = cf(0.533333333…)
cf(11/15) = 1,2,1,3 = cf(0.733333333…)
cf(13/15) = 1,6,2 = cf(0.866666666…)
cf(14/15) = 1,14 = cf(0.933333333…)
cf(3/16) = 5,3 = cf(0.1875)
cf(5/16) = 3,5 = cf(0.3125)
cf(7/16) = 2,3,2 = cf(0.4375)

After investigating some of those patterns, I wondered what happened when you reversed the continued fraction cf(a/b) and used those reversed numbers backward (that is, used the numbers of cf(a/b) forward) to generate another and different a/b. And a/b will always be different unless cf(a/b) is a palindrome, like cf(5/12) = 2,2,2 or cf(5/13) = 2,1,1,2 or cf(4/15) = 3,1,3. Note that a continued fraction never ends in 1, so that when reversing, say, cf(5/8) = (1, 1, 1, 2), you need an adjustment from (2, 1, 1, 1) to (2, 1, 1+1) = (2, 1, 2). Here’s a little of what happens when you reverse cf(a1/b1) to generate a2/b2:

cf(1/2) = 2 → 2 = cf(1/2)
1/2 = 0.5 : 0.5 = 1/2
cf(1/3) = 3 → 3 = cf(1/3)
1/3 = 0.333333333 : 0.333333333 = 1/3
cf(2/3) = 1, 2 → 2, 1 → 3 = cf(1/3)
2/3 = 0.666666666 : 0.333333333 = 1/3
cf(3/4) = 1, 3 → 3, 1 → 4 = cf(1/4)
3/4 = 0.75 : 0.25 = 1/4
cf(2/5) = 2, 2 → 2, 2 = cf(2/5)
2/5 = 0.4 : 0.4 = 2/5
cf(3/5) = 1, 1, 2 → 2, 1, 1 → 2, 2 = cf(2/5)
3/5 = 0.6 : 0.4 = 2/5
cf(4/5) = 1, 4 → 4, 1 → 5 = cf(1/5)
4/5 = 0.8 : 0.2 = 1/5
cf(5/6) = 1, 5 → 5, 1 → 6 = cf(1/6)
5/6 = 0.833333333 : 0.166666666 = 1/6
cf(2/7) = 3, 2 → 2, 3 = cf(3/7)
2/7 = 0.285714286 : 0.428571428 = 3/7
cf(3/7) = 2, 3 → 3, 2 = cf(2/7)
3/7 = 0.428571429 : 0.285714286 = 2/7
cf(4/7) = 1, 1, 3 → 3, 1, 1 → 3, 2 = cf(2/7)
4/7 = 0.571428571 : 0.285714286 = 2/7
cf(5/7) = 1, 2, 2 → 2, 2, 1 → 2, 3 = cf(3/7)
5/7 = 0.714285714 : 0.428571429 = 3/7
cf(6/7) = 1, 6 → 6, 1 → 7 = cf(1/7)
6/7 = 0.857142857 : 0.142857143 = 1/7
cf(3/8) = 2, 1, 2 → 2, 1, 2 = cf(3/8)
0.375 : 0.375
cf(5/8) = 1, 1, 1, 2 → 2, 1, 1, 1 → 2, 1, 2 = cf(3/8)
0.625 : 0.375
cf(7/8) = 1, 7 → 7, 1 → 8 = cf(1/8)
0.875 : 0.125
cf(2/9) = 4, 2 → 2, 4 = cf(4/9)
0.222222222 : 0.444444444
cf(4/9) = 2, 4 → 4, 2 = cf(2/9)
0.444444444 : 0.222222222

And if you plot x = a1/b1 and y = (a2/b2 * 2) on a fract-L, that is, a graph whose horizontal and vertical arms represent 0 to 1, you get the fractal right at the beginning:

Fract-L for x = a1/b1 and y = (a2/b2 * 2), where a2/b2 is generated from reversed(cf(a1/b1))

You need to use (a2/b2 * 2) because a2/b2 from reversed(cf(a1/b1)) is always <= 0.5, so using raw a2/b2 generates this graph:

Fract-L for x = a1/b1 and y = a2/b2 (i.e. a2/b2 is unadjusted)

Why is it always true that a2/b2 <= 0.5? For two reasons. First, a/b > 0.5 always generate continued fractions that start with 1, like cf(2/3) = 1, 2 or cf(3/4) = 1, 3 or cf(3/5) = 1, 1, 2. Second, as previously mentioned, no continued fraction ends with 1. Therefore a reversed cf(a1/b1), where the final number, n > 1, moves to the beginning, will never begin with 1 and the a2/b2 generated from reversed(cf(a1/b1)) will always be less than 0.5 (or equal to it in the solitary case of cf(1/2) = 2).

Now let's look at the development of the fractal as a1/b1 uses larger and larger denominators:

Fract-L for x = a1/b1 and y = (a2/b2 * 2) for a1/b1 <= 6/7

Fract-L for for a1/b1 <= 14/15

Fract-L for a1/b1 <= 30/31

Fract-L for a1/b1 <= 62/63

Fract-L for a1/b1 <= 126/127

Fract-L for a1/b1 <= 254/255

Fract-L for a1/b1 <= 357/358

Fract-L for a1/b1 <= 467/468

Animated fract-L for x = a1/b1 and y = (a2/b2 * 2) (animated at ezGif)

The fractal changes subtly when you restrict the b1 of a1/b1 in some way, say using multiples of 2, 3, 4, 5…:

Fract-L for x = a1/b1 and y = (a2/b2 * 2) for b1 = n = 2, 3, 4, 5, 6, 7, 8…

Fract-L for b1 = 2n = 2, 4, 6, 8, 10…

Fract-L for b1 = 3n = 3, 6, 9, 12, 15…

Fract-L for b1 = 4n

Fract-L for b1 = 5n

Fract-L for b1 = 6n

Animated fract-L for b1 = 1n..12n (animated at ezGif)

Finally, here are fract-Ls when b1 is a triangular, square, hexagonal or octagonal number:

Fract-L for x = a1/b1 and y = (a2/b2 * 2) for triangular(b1) = 3, 6, 10, 15, 21, 28,…

Fract-L for square(b1) = 4, 9, 16, 25, 36, 49,…

Fract-L for hexagonal(b1) = 6, 15, 28, 45, 66, 91,…

Fract-L for octagonal(b1) = 8, 21, 40, 65, 96, 133,…

Elsewhere Other-Accessible…

• Back to Drac’ — a parallel pun for a pre-previous fractal
• I Like Gryke — a first look at the limestone fractal
• Lime Time — more on the limestone fractal

Color-Coded ContFracs

by in Arithmetic, Base Behavior, Continued Fractions, Mathematics and tagged , , , , , | Leave a comment

Continued fractions are cool… Too cool for school. Or too cool for my school, at least. Because I never learnt about them there. Now that I have learnt about them, they’ve helped me wade a little further into the immeasurable Mare Mathematicum. Or Mare Matris Mathematicæ. I’m almost ankle-deep now, rather than just toe-deep. (I wish.)

But apart from aiding my understanding, continued fractions have always enhanced my entertainment. I can use them to find pretty but (probably) puny patterns like these:


[3,1,2] = contfrac(3/12) in base 9 = contfrac(3/11) in base 10
4,1,3 ← 4/13 in b16 = 4/19
5,1,4 ← 5/14 in b25 = 5/29
6,1,5 ← 6/15 in b36 = 6/41
7,1,6 ← 7/16 in b49 = 7/55
8,1,7 ← 8/17 in b64 = 8/71
9,1,8 ← 9/18 in b81 = 9/89
A,1,9 ← A/19 in b100 = 10/109 → 10,1,9
B,1,A ← B/1A in b121 = 11/131 → 11,1,10
C,1,B ← C/1B in b144 = 12/155 → 12,1,11

Those patterns with square numbers carry on for ever, I assume. I also assume that the similar patterns below do too, though I’m not sure if every base contains an infinite number of them. Maybe some bases don’t contain any at all. I haven’t found any in base 10 so far:


[25,2] = contfrac(2/52) in base 9 = contfrac(2/47) in base 10 = [23,2]
42,1,3 ← 4/213 in b8 = 4/139 → 34,1,3
4,1,2,3,3 ← 41/233 in b8 = 33/155
24,1,3,1,2 ← 24/1312 in b5 = 14/207 → 14,1,3,1,2
1,17,1,2,3 ← 117/123 in b14 = 217/227 → 1,21,1,2,3
320,1,2 ← 3/2012 in b5 = 3/257 → 85,1,2
254,2 ← 2/542 in b7 = 2/275 → 137,2
3A,3 ← 3/A3 in b28 = 3/283 → 94,3
3,5,A,2 ← 35/A2 in b34 = 107/342 → 3,5,10,2
12,1,5,3 ← 12/153 in b17 = 19/377 → 19,1,5,3
12,1,5,3 ← 12/153 in b17 = 19/377 → 19,1,5,3
3,1,4,1,4,1,5 ← 314/1415 in b8 = 204/781
2,1,36,3,2 ← 213/632 in b12 = 303/902 → 2,1,42,3,2
3,2,11,2,2,2 ← 321/1222 in b9 = 262/911 → 3,2,10,2,2,2
41,2,1,1,6 ← 41/2116 in b8 = 33/1102 → 33,2,1,1,6
4H,4 ← 4/H4 in b65 = 4/1109 → 277,4
249,2 ← 2/492 in b17 = 2/1311 → 655,2
6,2,1,3,J ← 62/13J in b35 = 212/1349 → 6,2,1,3,19
8,3,3,1,D ← 83/31D in b22 = 179/1487 → 8,3,3,1,13
142,1,1,6 ← 14/2116 in b9 = 13/1554 → 119,1,1,6
10,1,111,1,1,2 ← 1011/11112 in b6 = 223/1556 → 6,1,43,1,1,2
204,1,17 ← 20/4117 in b8 = 16/2127 → 132,1,15
93,1,8 ← 9/318 in b27 = 9/2222 → 246,1,8
1,3A,1,1,4,2,2 ← 13A1/1422 in b12 = 2281/2330 → 1,46,1,1,4,2,2
4340,1,3 ← 4/34013 in b5 = 4/2383 → 595,1,3
13,1,7,6 ← 13/176 in b46 = 49/2444 → 49,1,7,6
C7,1,B ← C/71B in b21 = 12/3119 → 259,1,11
35,3,2,3,1,1,2 ← 353/23112 in b6 = 141/3284 → 23,3,2,3,1,1,2
1,2,2,1,O,F ← 122/1OF in b50 = 2602/3715 → 1,2,2,1,24,15
2,1,1,5,55 ← 211/555 in b28 = 1597/4065 → 2,1,1,5,145
1P,2,H,2 ← 1P/2H2 in b47 = 72/5219 → 72,2,17,2
50,14,1,1,1,5 ← 501/41115 in b6 = 181/5447 → 30,10,1,1,1,5
5450,1,4 ← 5/45014 in b6 = 5/6274 → 1254,1,4
3103,1,2 ← 3/10312 in b9 = 3/6815 → 2271,1,2
4B,1,2,2,C ← 4B/122C in b19 = 87/7631 → 87,1,2,2,12
3G,D,2,3 ← 3G/D23 in b26 = 94/8843 → 94,13,2,3
3,1,1,A,K,6 ← 311/AK6 in b29 = 2553/8996 → 3,1,1,10,20,6
1,2[70],1,3,9 ← 12[70]/139 in b98 = 9870/9907 → 1,266,1,3,9
14,1,9,A ← 14/19A in b97 = 101/10292 → 101,1,9,10
14,1,9,A ← 14/19A in b97 = 101/10292 → 101,1,9,10
4133,1,14,2 ← 41/331142 in b5 = 21/11422 → 543,1,9,2
1,E,4,1,M,7 ← 1E4/1M7 in b100 = 11404/12207 → 1,14,4,1,22,7
LG,5,4 ← L/G54 in b28 = 21/12688 → 604,5,4

Mathematicoynte

by in Arithmetic, Continued Fractions, Fractals, Mathematics and tagged , , , , , , , , | Leave a comment

Pre-previously, I looked at a fractal phallus. Now I want to look at a fractal fanny (in the older British sense). In fact, it’s a fractional fractal fanny. Take a look at these fractions:


1/10, 1/9, 1/8, 1/7, 1/6, 1/5, 2/10, 2/9, 1/4, 2/8, 2/7, 3/10, 1/3, 2/6, 3/9, 3/8, 2/5, 4/10, 3/7, 4/9, 1/2, 2/4, 3/6, 4/8, 5/10, 5/9, 4/7, 3/5, 6/10, 5/8, 2/3, 4/6, 6/9, 7/10, 5/7, 3/4, 6/8, 7/9, 4/5, 8/10, 5/6, 6/7, 7/8, 8/9, 9/10

They’re all the fractions for 1/2..(n-1)/n, n = 10, sorted by increasing size. But obviously some of them are the same: 1/2 = 2/4 = 3/6 = 5/10, 1/3 = 2/6 = 3/9, 1/4 = 2/8, and so on. If you remove the duplicates, you get this set of reduced fractions:


1/10, 1/9, 1/8, 1/7, 1/6, 1/5, 2/9, 1/4, 2/7, 3/10, 1/3, 3/8, 2/5, 3/7, 4/9, 1/2, 5/9, 4/7, 3/5, 5/8, 2/3, 7/10, 5/7, 3/4, 7/9, 4/5, 5/6, 6/7, 7/8, 8/9, 9/10

Now here are the reduced fractions for 1/2..(n-1)/n, n = 30:


1/30, 1/29, 1/28, 1/27, 1/26, 1/25, 1/24, 1/23, 1/22, 1/21, 1/20, 1/19, 1/18, 1/17, 1/16, 1/15, 2/29, 1/14, 2/27, 1/13, 2/25, 1/12, 2/23, 1/11, 2/21, 1/10, 3/29, 2/19, 3/28, 1/9, 3/26, 2/17, 3/25, 1/8, 3/23, 2/15, 3/22, 4/29, 1/7, 4/27, 3/20, 2/13, 3/19, 4/25, 1/6, 5/29, 4/23, 3/17, 5/28, 2/11, 5/27, 3/16, 4/21, 5/26, 1/5, 6/29, 5/24, 4/19, 3/14, 5/23, 2/9, 5/22, 3/13, 7/30, 4/17, 5/21, 6/25, 7/29, 1/4, 7/27, 6/23, 5/19, 4/15, 7/26, 3/11, 8/29, 5/18, 7/25, 2/7, 7/24, 5/17, 8/27, 3/10, 7/23, 4/13, 9/29, 5/16, 6/19, 7/22, 8/25, 9/28, 1/3, 10/29, 9/26, 8/23, 7/20, 6/17, 5/14, 9/25, 4/11, 11/30, 7/19, 10/27, 3/8, 11/29, 8/21, 5/13, 7/18, 9/23, 11/28, 2/5, 11/27, 9/22, 7/17, 12/29, 5/12, 8/19, 11/26, 3/7, 13/30, 10/23, 7/16, 11/25, 4/9, 13/29, 9/20, 5/11, 11/24, 6/13, 13/28, 7/15, 8/17, 9/19, 10/21, 11/23, 12/25, 13/27, 14/29, 1/2, 15/29, 14/27, 13/25, 12/23, 11/21, 10/19, 9/17, 8/15, 15/28, 7/13, 13/24, 6/11, 11/20, 16/29, 5/9, 14/25, 9/16, 13/23, 17/30, 4/7, 15/26, 11/19, 7/12, 17/29, 10/17, 13/22, 16/27, 3/5, 17/28, 14/23, 11/18, 8/13, 13/21, 18/29, 5/8, 17/27, 12/19, 19/30, 7/11, 16/25, 9/14, 11/17, 13/20, 15/23, 17/26, 19/29, 2/3, 19/28, 17/25, 15/22, 13/19, 11/16, 20/29, 9/13, 16/23, 7/10, 19/27, 12/17, 17/24, 5/7, 18/25, 13/18, 21/29, 8/11, 19/26, 11/15, 14/19, 17/23, 20/27, 3/4, 22/29, 19/25, 16/21, 13/17, 23/30, 10/13, 17/22, 7/9, 18/23, 11/14, 15/19, 19/24, 23/29, 4/5, 21/26, 17/21, 13/16, 22/27, 9/11, 23/28, 14/17, 19/23, 24/29, 5/6, 21/25, 16/19, 11/13, 17/20, 23/27, 6/7, 25/29, 19/22, 13/15, 20/23, 7/8, 22/25, 15/17, 23/26, 8/9, 25/28, 17/19, 26/29, 9/10, 19/21, 10/11, 21/23, 11/12, 23/25, 12/13, 25/27, 13/14, 27/29, 14/15, 15/16, 16/17, 17/18, 18/19, 19/20, 20/21, 21/22, 22/23, 23/24, 24/25, 25/26, 26/27, 27/28, 28/29, 29/30

Can you see the fractal fanny? Not unless you’re superhuman. But any normal human can see the fractal fanny when you turn those reduced and sorted fractions, a/b, into a graph, where y = b and x = n for a/bn:

graph for b of reduced a/b = 1/2..29/30, sorted by size of a/b

(click for larger)

If you don’t reduce the fractions, you get this distorted coynte:

graph for b of all fractions 1/2..29/30, sorted by a/b

And you can use other variables for y, like the sum of the continued fraction of a/b:

graph for sum(contfrac(a/b)) of reduced fractions 1/2..29/30, sorted by a/b

graph for cfsum of all fractions 1/2..29/30, sorted by a/b

And the product of the continued fraction of a/b:

graph for prod(contfrac(a/b)) of reduced fractions 1/2..29/30, sorted by a/b

graph for cfmul of all fractions 1/2..29/30, sorted by a/b

And you can sort by the size of other variables, like the number of factors of b:

graph for a+b of all fractions 1/2..29/30, sorted by factornum(b)

And so on:

graph for a of reduced fractions 1/2..29/30, sorted by a/b

graph for a of reduced fractions 1/2..29/30, sorted by a/b

graph for a of all fractions 1/2..29/30, sorted by a/b

graph for a of all fractions 1/2..29/30, sorted by length(contfrac(a/b))

graph for a of all fractions 1/2..29/30, sorted by factornum(b)

graph for a of all fractions 1/2..29/30, sorted by gcd(a/b)

graph for a+b of all fractions 1/2..29/30, sorted by a/b

graph for a+b of reduced fractions 1/2..29/30, sorted by a/b

graph for a+b of all fractions 1/2..29/30, sorted by a+b

graph for a+b of all fractions 1/2..29/30, sorted by cflen(a/b)

graph for a+b of all fractions 1/2..29/30, sorted by gbd(a,b)

graph for b of all fractions 1/2..29/30, sorted by a+b

graph for b of all fractions 1/2..29/30, sorted by cflen(a/b)

graph for b of all fractions 1/2..29/30, sorted by factnum(b)

graph for b of all fractions 1/2..29/30, sorted by gcd(a,b)

graph for b-a of all fractions 1/2..29/30, sorted by a/b

graph for b-a of reduced fractions 1/2..29/30, sorted by a/b

graph for b-a of all fractions 1/2..29/30, sorted by a+b

graph for b-a of all fractions 1/2..29/30, sorted by factnum(b)

graph for cfmul of all fractions 1/2..29/30, sorted by a

graph for cfsum of all fractions 1/2..29/30, sorted by a

Previously Pre-Posted (Please Peruse)

• Phrallic Frolics — a look at fractal phalluses, a.k.a. phralluses

The Call of CFulhu

by in Arithmetic, √2, Continued Fractions, Fibonacci Sequence, H.P. Lovecraft, Integer Sequences, φ, Mathematics, Phi, Square root of 2 and tagged , , , , , | Leave a comment

“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:

contfrac(10/7) = [1; 2, 3]
10/7 = 1 + 1/(2 + 1/3)
10/7 = 1.428571428571…

contfrac(3/5) = [0; 1, 1, 2]
4/5 = 0 + 1/(1 + 1/(1 + 1/2))
4/5 = 0.8

contfrac(11/8) = [1; 2, 1, 2]
11/8 = 1 + 1/(2 + 1/(1 + 1/2))
11/8 = 1.375

contfrac(4/7) = [0; 1, 1, 3]
4/7 = 0 + 1/(1 + 1/(1 + 1/3))
4/7 = 0.57142857142…

contfrac(17/19) = [0; 1, 8, 2]
17/19 = 0 + 1/(1 + 1/(8 + 1/2))
17/19 = 0.8947368421052…

contfrac(8/25) = [0; 3, 8]
8/25 = 0 + 1/(3 + 1/8)
8/25 = 0.32

contfrac(√2) = [1; 2, 2, 2, 2, 2, 2, 2…] = [1; 2]

√2 = 1 + 1/(2 + 1/(2 + 1/(2 + 1/(2 + 1/(2 + 1/(2 + 1/2 + …))))))

√2 = 1.41421356237309504…

contfrac(φ) = [1; 1, 1, 1, 1, 1, 1, 1, 1…]

φ = 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/1 + …)))))))

φ = 1.6180339887498948…

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:

contfrac(1/2) = [0; 2] (cfl=1)
1/2 = 0 + 1/2
1/2 = 0.5

contfrac(2/3) = [0; 1, 2] (cfl=2)
2/3 = 0 + 1/(1 + 1/2)
2/3 = 0.666666666…

contfrac(3/5) = [0; 1, 1, 2] (cfl=3)
3/5 = 0 + 1/(1 + 1/(1 + 1/2))
3/5 = 0.6

contfrac(5/8) = [0; 1, 1, 1, 2] (cfl=4)
5/8 = 0 + 1/(1 + 1/(1 + 1/(1 + 1/2)))
5/8 = 0.625

contfrac(8/13) = [0; 1, 1, 1, 1, 2] (cfl=5)
8/13 = 0 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/2))))
8/13 = 0.615384615…

contfrac(13/21) = [0; 1, 1, 1, 1, 1, 2] (cfl=6)
13/21 = 0 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/2)))))
13/21 = 0.619047619…

contfrac(21/34) = [0; 1, 1, 1, 1, 1, 1, 2] (cfl=7)
21/34 = 0 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/2))))))
21/34 = 0.617647059…

contfrac(34/55) = [0; 1, 1, 1, 1, 1, 1, 1, 2] (cfl=8)
contfrac(55/89) = [0; 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=9)
contfrac(89/144) = [0; 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=10)
contfrac(144/233) = [0; 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=11)
contfrac(233/377) = [0; 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=12)
contfrac(377/610) = [0; 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=13)
contfrac(610/987) = [0; 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=14)
contfrac(987/1597) = [0; 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=15)
contfrac(1597/2584) = [0; 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=16)
contfrac(2584/4181) = [0; 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=17)
contfrac(4181/6765) = [0; 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=18)
[…]

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:

contfrac(2/5) = [0; 2, 2] (cfl=2)
2/5 = 0 + 1/(2 + 1/2)
2/5 = 0.4

contfrac(3/8) = [0; 2, 1, 2] (cfl=3)
3/8 = 0 + 1/(2 + 1/(1 + 1/2))
3/8 = 0.375

contfrac(3/10) = [0; 3, 3] (cfl=2)
3/10 = 0 + 1/(3 + 1/3)
3/10 = 0.3

contfrac(5/12) = [0; 2, 2, 2] (cfl=3)
5/12 = 0 + 1/(2 + 1/(2 + 1/2))
5/12 = 0.416666666…

contfrac(5/13) = [0; 2, 1, 1, 2] (cfl=4)
5/13 = 0 + 1/(2 + 1/(1 + 1/(1 + 1/2)))
5/13 = 0.384615384…

contfrac(4/15) = [0; 3, 1, 3] (cfl=3)
4/15 = 0 + 1/(3 + 1/(1 + 1/3))
4/15 = 0.266666666…

contfrac(7/16) = [0; 2, 3, 2] (cfl=3)
7/16 = 0 + 1/(2 + 1/(3 + 1/2))
7/16 = 0.4375

contfrac(4/17) = [0; 4, 4] (cfl=2)
4/17 = 0 + 1/(4 + 1/4)
4/17 = 0.235294117…

Again, I wondered which of these fractions set successive records for the length of their palindromic continued fractions. Here’s the answer:

contfrac(1/2) = [0; 2] (cfl=1)
1/2 = 0 + 1/2
1/2 = 0.5

contfrac(2/5) = [0; 2, 2] (cfl=2)
2/5 = 0 + 1/(2 + 1/2)
2/5 = 0.4

contfrac(3/8) = [0; 2, 1, 2] (cfl=3)
3/8 = 0 + 1/(2 + 1/(1 + 1/2))
3/8 = 0.375

contfrac(5/13) = [0; 2, 1, 1, 2] (cfl=4)
5/13 = 0 + 1/(2 + 1/(1 + 1/(1 + 1/2)))
5/13 = 0.384615384…

contfrac(8/21) = [0; 2, 1, 1, 1, 2] (cfl=5)
8/21 = 0 + 1/(2 + 1/(1 + 1/(1 + 1/(1 + 1/2))))
8/21 = 0.380952380…

contfrac(13/34) = [0; 2, 1, 1, 1, 1, 2] (cfl=6)
13/34 = 0 + 1/(2 + 1/(1 + 1/(1 + 1/(
1 + 1/(1 + 1/2)))))
13/34 = 0.382352941..

contfrac(21/55) = [0; 2, 1, 1, 1, 1, 1, 2] (cfl=7)
21/55 = 0 + 1/(2 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/2))))))
21/55 = 0.381818181…

contfrac(34/89) = [0; 2, 1, 1, 1, 1, 1, 1, 2] (cfl=8)
contfrac(55/144) = [0; 2, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=9)
contfrac(89/233) = [0; 2, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=10)
contfrac(144/377) = [0; 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=11)
contfrac(233/610) = [0; 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=12)
contfrac(377/987) = [0; 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=13)
contfrac(610/1597) = [0; 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=14)
contfrac(987/2584) = [0; 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=15)
contfrac(1597/4181) = [0; 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=16)
contfrac(2584/6765) = [0; 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=17)
[…]

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.

Message from Mater

by in Arithmetic, √2, Continued Fractions, Irrational numbers, π, φ, Mathematics, Square root of 2, Ulam Spirals and tagged , , , , , , , , | Leave a comment

As any recreational mathematician kno, the Ulam spiral shows the prime numbers on a spiral grid of integers. Here’s a Ulam spiral with 1 represented in blue and 2, 3, 5, 7… as white blocks spiralling anti-clockwise from the right of 1:

The Ulam spiral of prime numbers

Ulam spiral at higher resolution

I like the Ulam spiral and whenever I’m looking at new number sequences I like to Ulamize it, that is, display it on a spiral grid of integers. Sometimes the result looks good, sometimes it doesn’t. But I’ve always wondered something beforehand: will this be the spiral where I see a message appear? That is, will I see a message from Mater Mathematica, Mother Maths, the omniregnant goddess of mathematics? Is there an image or text embedded in some obscure number sequence, revealed when the sequence is Ulamized and proving that there’s divine intelligence and design behind the universe? Maybe the image of a pantocratic cat will appear. Or a text in Latin or Sanskrit or some other suitably century-sanctified language.

That’s what I wonder. I don’t wonder it seriously, of course, but I do wonder it. But until 22nd March 2025 I’d never seen any Ulam-ish spiral that looked remotely like a message. But 22nd May is the day I Ulamed some continued fractions. And I saw something that did look a little like a message. Like text, that is. But I might need to explain continued fractions first. What are they? They’re a fascinating and beautiful way of representing both rational and irrational numbers. The continued fractions for rational numbers look like this in expanded and compact format:

5/3 = 1 + 1/(1 + ½) = 1 + ⅔
5/3 = [1; 1, 2]

19/7 = 2 + 1/(1 + 1/(2 + ½)) = 2 + 4/7
19/7 = [2; 1, 2, 2]

2/3 = 0 + 1/(1 + 1/2)
2/3 = [0; 1, 2] (compare 5/3 above)

3/5 = 0 + 1/(1 + 1/(1 + 1/2))
3/5 = [0; 1, 1, 2]

5/7 = 0 + 1/(1 + 1/(2 + 1/2))
5/7 = [0; 1, 2, 2] (compare 19/7 above)

13/17 = 0 + 1/(1 + 1/(3 + 1/4))
13/17 = [0; 1, 3, 4]

30/67 = 0 + 1/(2 + 1/(4 + 1/(3 + ½)))
30/67 = [0; 2, 4, 3, 2]

The continued fractions of irrational numbers are different. Most importantly, they never end. For example, here are the infinite continued fractions for φ, √2 and π in expanded and compact format:

φ = 1 + (1/(1 + 1/(1 + 1/(1 + …)))φ = [1; 1]

√2 = 1 + (1/(2 + 1/(2 + 1/(2 + …)))
√2 = [1; 2]

π = 3 + 1/(7 + 1/(15 + 1/(1 + 1/(292 + 1/(1 + 1/(1 + 1/(1 + 1/(2 + 1/(1 + 1/(3 +…))))))))))
π = [3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3…]

As you can see, the continued fraction of π doesn’t fall into a predictable pattern like those for φ and √2. But I’ve already gone into continued fractions further than I need for this post, so let’s return to the continued fractions of rationals. I set up an Ulam spiral to show patterns based on the continued fractions for 1/1, ½, ⅓, ⅔, 1/4, 2/4, 3/4, 1/5, 2/5, 3/5, 4/5, 1/6, 2/6, 3/6… (where the fractions are assigned to 1,2,3… and 2/4 = ½, 2/6 = ⅓ etc). For example, if the continued fraction contains a number higher than 5, you get this spiral:

Spiral for continued fractions containing at least number > 5

With tests for higher and higher numbers in the continued fractions, the spirals start to thin and apparent symbols start to appear in the arms of the spirals:

Spiral for contfrac > 10

Spiral for contfrac > 15

Spiral for contfrac > 20

Spiral for contfrac > 25

Spiral for contfrac > 30

Spiral for contfrac > 35

Spiral for contfrac > 40

Spirals for contfrac > 5..40 (animated at EZgif)

Here are some more of these spirals at increasing magnification:

Spiral for contfrac > 23 (#1)

Spiral for contfrac > 23 (#2)

Spiral for contfrac > 23 (#3)

Spiral for contfrac > 13

Spiral for contfrac > 15 (off-center)

Spiral for contfrac > 23 (off-center)

And here are some of the symbols picked out in blue:

Spiral for contfrac > 15 (blue symbols)

Spiral for contfrac > 23 (blue symbols)

But they’re not really symbols, of course. They’re quasi-symbols, artefacts of the Ulamization of a simple test on continued fractions. Still, they’re the closest I’ve got so far to a message from Mater Mathematica.