A FracTeasel on a Fract-L

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 multipled 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

Color-Coded ContFracs

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
4,1,34/13 in b16 = 4/19
5,1,45/14 in b25 = 5/29
6,1,56/15 in b36 = 6/41
7,1,67/16 in b49 = 7/55
8,1,78/17 in b64 = 8/71
9,1,89/18 in b81 = 9/89
A,1,9A/19 in b100 = 10/109 → 10,1,9
B,1,AB/1A in b121 = 11/131 → 11,1,10
C,1,BC/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]
42,1,34/213 in b8 = 4/139 → 34,1,3
4,1,2,3,341/233 in b8 = 33/155
24,1,3,1,224/1312 in b5 = 14/207 → 14,1,3,1,2
1,17,1,2,3117/123 in b14 = 217/227 → 1,21,1,2,3
320,1,23/2012 in b5 = 3/257 → 85,1,2
254,22/542 in b7 = 2/275 → 137,2
3A,33/A3 in b28 = 3/283 → 94,3
3,5,A,235/A2 in b34 = 107/342 → 3,5,10,2
12,1,5,312/153 in b17 = 19/377 → 19,1,5,3
12,1,5,312/153 in b17 = 19/377 → 19,1,5,3
3,1,4,1,4,1,5314/1415 in b8 = 204/781
2,1,36,3,2213/632 in b12 = 303/902 → 2,1,42,3,2
3,2,11,2,2,2321/1222 in b9 = 262/911 → 3,2,10,2,2,2
41,2,1,1,641/2116 in b8 = 33/1102 → 33,2,1,1,6
4H,44/H4 in b65 = 4/1109 → 277,4
249,22/492 in b17 = 2/1311 → 655,2
6,2,1,3,J62/13J in b35 = 212/1349 → 6,2,1,3,19
8,3,3,1,D83/31D in b22 = 179/1487 → 8,3,3,1,13
142,1,1,614/2116 in b9 = 13/1554 → 119,1,1,6
10,1,111,1,1,21011/11112 in b6 = 223/1556 → 6,1,43,1,1,2
204,1,1720/4117 in b8 = 16/2127 → 132,1,15
93,1,89/318 in b27 = 9/2222 → 246,1,8
1,3A,1,1,4,2,213A1/1422 in b12 = 2281/2330 → 1,46,1,1,4,2,2
4340,1,34/34013 in b5 = 4/2383 → 595,1,3
13,1,7,613/176 in b46 = 49/2444 → 49,1,7,6
C7,1,BC/71B in b21 = 12/3119 → 259,1,11
35,3,2,3,1,1,2353/23112 in b6 = 141/3284 → 23,3,2,3,1,1,2
1,2,2,1,O,F122/1OF in b50 = 2602/3715 → 1,2,2,1,24,15
2,1,1,5,55211/555 in b28 = 1597/4065 → 2,1,1,5,145
1P,2,H,21P/2H2 in b47 = 72/5219 → 72,2,17,2
50,14,1,1,1,5501/41115 in b6 = 181/5447 → 30,10,1,1,1,5
5450,1,45/45014 in b6 = 5/6274 → 1254,1,4
3103,1,23/10312 in b9 = 3/6815 → 2271,1,2
4B,1,2,2,C4B/122C in b19 = 87/7631 → 87,1,2,2,12
3G,D,2,33G/D23 in b26 = 94/8843 → 94,13,2,3
3,1,1,A,K,6311/AK6 in b29 = 2553/8996 → 3,1,1,10,20,6
1,2[70],1,3,912[70]/139 in b98 = 9870/9907 → 1,266,1,3,9
14,1,9,A14/19A in b97 = 101/10292 → 101,1,9,10
14,1,9,A14/19A in b97 = 101/10292 → 101,1,9,10
4133,1,14,241/331142 in b5 = 21/11422 → 543,1,9,2
1,E,4,1,M,71E4/1M7 in b100 = 11404/12207 → 1,14,4,1,22,7
LG,5,4L/G54 in b28 = 21/12688 → 604,5,4