Gyp Cip

Abundance often overwhelms, but restriction reaps riches. That’s true in mathematics and science, where you can often understand the whole better by looking at only a part of it first — restriction reaps riches. Egyptian fractions are one example in maths. In ancient Egypt, you could have any kind of fraction you liked so long as it was a reciprocal like 1/2, 1/3, 1/4 or 1/5 (well, there were two exceptions: 2/3 and 3/4 were also allowed).

So when mathematicians speak of “Egyptian fractions”, they mean those fractions that can be represented as a sum of reciprocals. Egyptian fractions are restricted and that reaps riches. Here’s one example: how many ways can you add n distinct reciprocals to make 1? When n = 1, there’s one way to do it: 1/1. When n = 2, there’s no way to do it, because 1 – 1/2 = 1/2. Therefore the summed reciprocals aren’t distinct: 1/2 + 1/2 = 1. After that, 1 – 1/3 = 2/3, 1 – 1/4 = 3/4, and so on. By the modern meaning of “Egyptian fraction”, there’s no solution for n = 2.

However, when n = 3, there is a way to do it:

• 1/2 + 1/3 + 1/6 = 1

But that’s the only way. When n = 4, things get better:

• 1/2 + 1/4 + 1/6 + 1/12 = 1
• 1/2 + 1/3 + 1/10 + 1/15 = 1
• 1/2 + 1/3 + 1/9 + 1/18 = 1
• 1/2 + 1/4 + 1/5 + 1/20 = 1
• 1/2 + 1/3 + 1/8 + 1/24 = 1
• 1/2 + 1/3 + 1/7 + 1/42 = 1

What about n = 5, n = 6 and so on? You can find the answer at the Online Encyclopedia of Integer Sequences (OEIS), where sequence A006585 is described as “Egyptian fractions: number of solutions to 1 = 1/x1 + … + 1/xn in positive integers x1 < … < xn”. The sequence is one of the shortest and strangest at the OEIS:

• 1, 0, 1, 6, 72, 2320, 245765, 151182379

When n = 1, there’s one solution: 1/1. When n = 2, there’s no solution, as I showed above. When n = 3, there’s one solution again. When n = 4, there are six solutions. And the OEIS tells you how many solutions there are for n = 5, 6, 7, 8. But n >= 9 remains unknown at the time of writing.

To understand the problem, consider the three reciprocals, 1/2, 1/3 and 1/5. How do you sum them? They have different denominators, 2, 3 and 5, so you have to create a new denominator, 30 = 2 * 3 * 5. Then you have to adjust the numerators (the numbers above the fraction bar) so that the new fractions have the same value as the old:

• 1/2 = 15/30 = (2*3*5 / 2) / 30
• 1/3 = 10/30 = (2*3*5 / 3) / 30
• 1/5 = 06/30 = (2*3*5 / 5) / 30
• 15/30 + 10/30 + 06/30 = (15+10+6) / 30 = 31/30 = 1 + 1/30

Those three reciprocals don’t sum to 1. Now try 1/2, 1/3 and 1/6:

• 1/2 = 18/36 = (2*3*6 / 2) / 36
• 1/3 = 12/36 = (2*3*6 / 3) / 36
• 1/6 = 06/36 = (2*3*6 / 6) / 36
• 18/36 + 12/36 + 06/36 = (18+12+6) / 36 = 36/36 = 1

So when n = 3, the problem consists of finding three reciprocals, 1/a, 1/b and 1/c, such that for a, b, and c:

• a*b*c = a*b + a*c + b*c

There is only one solution: a = 2, b = 3 and c = 6. When n = 4, the problem consists of finding four reciprocals, 1/a, 1/b, 1/c and 1/d, such that for a, b, c and d:

• a*b*c*d = a*b*c + a*b*d + a*c*d + b*c*d

For example:

• 2*4*6*12 = 576
• 2*4*6 + 2*4*12 + 2*6*12 + 4*6*12 = 48 + 96 + 144 + 288 = 576
• 2*4*6*12 = 2*4*6 + 2*4*12 + 2*6*12 + 4*6*12 = 576

Therefore:

• 1/2 + 1/4 + 1/6 + 1/12 = 1

When n = 5, the problem consists of finding five reciprocals, 1/a, 1/b, 1/c, 1/d and 1/e, such that for a, b, c, d and e:

• a*b*c*d*e = a*b*c*d + a*b*c*e + a*b*d*e + a*c*d*e + b*c*d*e

There are 72 solutions and here they are:

• 1/2 + 1/4 + 1/10 + 1/12 + 1/15 = 1 (#1)
• 1/2 + 1/4 + 1/9 + 1/12 + 1/18 = 1 (#2)
• 1/2 + 1/5 + 1/6 + 1/12 + 1/20 = 1 (#3)
• 1/3 + 1/4 + 1/5 + 1/6 + 1/20 = 1 (#4)
• 1/2 + 1/4 + 1/8 + 1/12 + 1/24 = 1 (#5)
• 1/2 + 1/3 + 1/12 + 1/21 + 1/28 = 1 (#6)
• 1/2 + 1/4 + 1/6 + 1/21 + 1/28 = 1 (#7)
• 1/2 + 1/4 + 1/7 + 1/14 + 1/28 = 1 (#8)
• 1/2 + 1/3 + 1/12 + 1/20 + 1/30 = 1 (#9)
• 1/2 + 1/4 + 1/6 + 1/20 + 1/30 = 1 (#10)
• 1/2 + 1/5 + 1/6 + 1/10 + 1/30 = 1 (#11)
• 1/2 + 1/3 + 1/11 + 1/22 + 1/33 = 1 (#12)
• 1/2 + 1/3 + 1/14 + 1/15 + 1/35 = 1 (#13)
• 1/2 + 1/3 + 1/12 + 1/18 + 1/36 = 1 (#14)
• 1/2 + 1/4 + 1/6 + 1/18 + 1/36 = 1 (#15)
• 1/2 + 1/3 + 1/10 + 1/24 + 1/40 = 1 (#16)
• 1/2 + 1/4 + 1/8 + 1/10 + 1/40 = 1 (#17)
• 1/2 + 1/4 + 1/7 + 1/12 + 1/42 = 1 (#18)
• 1/2 + 1/3 + 1/9 + 1/30 + 1/45 = 1 (#19)
• 1/2 + 1/4 + 1/5 + 1/36 + 1/45 = 1 (#20)
• 1/2 + 1/5 + 1/6 + 1/9 + 1/45 = 1 (#21)
• 1/2 + 1/3 + 1/12 + 1/16 + 1/48 = 1 (#22)
• 1/2 + 1/4 + 1/6 + 1/16 + 1/48 = 1 (#23)
• 1/2 + 1/3 + 1/9 + 1/27 + 1/54 = 1 (#24)
• 1/2 + 1/3 + 1/8 + 1/42 + 1/56 = 1 (#25)
• 1/2 + 1/3 + 1/8 + 1/40 + 1/60 = 1 (#26)
• 1/2 + 1/3 + 1/10 + 1/20 + 1/60 = 1 (#27)
• 1/2 + 1/3 + 1/12 + 1/15 + 1/60 = 1 (#28)
• 1/2 + 1/4 + 1/5 + 1/30 + 1/60 = 1 (#29)
• 1/2 + 1/4 + 1/6 + 1/15 + 1/60 = 1 (#30)
• 1/2 + 1/4 + 1/5 + 1/28 + 1/70 = 1 (#31)
• 1/2 + 1/3 + 1/8 + 1/36 + 1/72 = 1 (#32)
• 1/2 + 1/3 + 1/9 + 1/24 + 1/72 = 1 (#33)
• 1/2 + 1/4 + 1/8 + 1/9 + 1/72 = 1 (#34)
• 1/2 + 1/3 + 1/12 + 1/14 + 1/84 = 1 (#35)
• 1/2 + 1/4 + 1/6 + 1/14 + 1/84 = 1 (#36)
• 1/2 + 1/3 + 1/8 + 1/33 + 1/88 = 1 (#37)
• 1/2 + 1/3 + 1/10 + 1/18 + 1/90 = 1 (#38)
• 1/2 + 1/3 + 1/7 + 1/78 + 1/91 = 1 (#39)
• 1/2 + 1/3 + 1/8 + 1/32 + 1/96 = 1 (#40)
• 1/2 + 1/3 + 1/9 + 1/22 + 1/99 = 1 (#41)
• 1/2 + 1/4 + 1/5 + 1/25 + 1/100 = 1 (#42)
• 1/2 + 1/3 + 1/7 + 1/70 + 1/105 = 1 (#43)
• 1/2 + 1/3 + 1/11 + 1/15 + 1/110 = 1 (#44)
• 1/2 + 1/3 + 1/8 + 1/30 + 1/120 = 1 (#45)
• 1/2 + 1/4 + 1/5 + 1/24 + 1/120 = 1 (#46)
• 1/2 + 1/5 + 1/6 + 1/8 + 1/120 = 1 (#47)
• 1/2 + 1/3 + 1/7 + 1/63 + 1/126 = 1 (#48)
• 1/2 + 1/3 + 1/9 + 1/21 + 1/126 = 1 (#49)
• 1/2 + 1/3 + 1/7 + 1/60 + 1/140 = 1 (#50)
• 1/2 + 1/4 + 1/7 + 1/10 + 1/140 = 1 (#51)
• 1/2 + 1/3 + 1/12 + 1/13 + 1/156 = 1 (#52)
• 1/2 + 1/4 + 1/6 + 1/13 + 1/156 = 1 (#53)
• 1/2 + 1/3 + 1/7 + 1/56 + 1/168 = 1 (#54)
• 1/2 + 1/3 + 1/8 + 1/28 + 1/168 = 1 (#55)
• 1/2 + 1/3 + 1/9 + 1/20 + 1/180 = 1 (#56)
• 1/2 + 1/3 + 1/7 + 1/54 + 1/189 = 1 (#57)
• 1/2 + 1/3 + 1/8 + 1/27 + 1/216 = 1 (#58)
• 1/2 + 1/4 + 1/5 + 1/22 + 1/220 = 1 (#59)
• 1/2 + 1/3 + 1/11 + 1/14 + 1/231 = 1 (#60)
• 1/2 + 1/3 + 1/7 + 1/51 + 1/238 = 1 (#61)
• 1/2 + 1/3 + 1/10 + 1/16 + 1/240 = 1 (#62)
• 1/2 + 1/3 + 1/7 + 1/49 + 1/294 = 1 (#63)
• 1/2 + 1/3 + 1/8 + 1/26 + 1/312 = 1 (#64)
• 1/2 + 1/3 + 1/7 + 1/48 + 1/336 = 1 (#65)
• 1/2 + 1/3 + 1/9 + 1/19 + 1/342 = 1 (#66)
• 1/2 + 1/4 + 1/5 + 1/21 + 1/420 = 1 (#67)
• 1/2 + 1/3 + 1/7 + 1/46 + 1/483 = 1 (#68)
• 1/2 + 1/3 + 1/8 + 1/25 + 1/600 = 1 (#69)
• 1/2 + 1/3 + 1/7 + 1/45 + 1/630 = 1 (#70)
• 1/2 + 1/3 + 1/7 + 1/44 + 1/924 = 1 (#71)
• 1/2 + 1/3 + 1/7 + 1/43 + 1/1806 = 1 (#72)

All the sums start with 1/2 except for one:

• 1/2 + 1/5 + 1/6 + 1/12 + 1/20 = 1 (#3)
• 1/3 + 1/4 + 1/5 + 1/6 + 1/20 = 1 (#4)

Here are the solutions in another format:

(2,4,10,12,15), (2,4,9,12,18), (2,5,6,12,20), (3,4,5,6,20), (2,4,8,12,24), (2,3,12,21,28), (2,4,6,21,28), (2,4,7,14,28), (2,3,12,20,30), (2,4,6,20,30), (2,5,6,10,30), (2,3,11,22,33), (2,3,14,15,35), (2,3,12,18,36), (2,4,6,18,36), (2,3,10,24,40), (2,4,8,10,40), (2,4,7,12,42), (2,3,9,30,45), (2,4,5,36,45), (2,5,6,9,45), (2,3,12,16,48), (2,4,6,16,48), (2,3,9,27,54), (2,3,8,42,56), (2,3,8,40,60), (2,3,10,20,60), (2,3,12,15,60), (2,4,5,30,60), (2,4,6,15,60), (2,4,5,28,70), (2,3,8,36,72), (2,3,9,24,72), (2,4,8,9,72), (2,3,12,14,84), (2,4,6,14,84), (2,3,8,33,88), (2,3,10,18,90), (2,3,7,78,91), (2,3,8,32,96), (2,3,9,22,99), (2,4,5,25,100), (2,3,7,70,105), (2,3,11,15,110), (2,3,8,30,120), (2,4,5,24,120), (2,5,6,8,120), (2,3,7,63,126), (2,3,9,21,126), (2,3,7,60,140), (2,4,7,10,140), (2,3,12,13,156), (2,4,6,13,156), (2,3,7,56,168), (2,3,8,28,168), (2,3,9,20,180), (2,3,7,54,189), (2,3,8,27,216), (2,4,5,22,220), (2,3,11,14,231), (2,3,7,51,238), (2,3,10,16,240), (2,3,7,49,294), (2,3,8,26,312), (2,3,7,48,336), (2,3,9,19,342), (2,4,5,21,420), (2,3,7,46,483), (2,3,8,25,600), (2,3,7,45,630), (2,3,7,44,924), (2,3,7,43,1806)


Note

Strictly speaking, there are two solutions for n = 2 in genuine Egyptian fractions, because 1/3 + 2/3 = 1 and 1/4 + 3/4 = 1. As noted above, 2/3 and 3/4 were permitted as fractions in ancient Egypt.

Fract-Hills

The Farey sequence is a fascinating sequence of fractions that divides the interval between 0/1 and 1/1 into smaller and smaller parts. To find the Farey fraction a[i] / b[i], you simply find the mediant of the Farey fractions on either side:

• a[i] / b[i] = (a[i-1] + a[i+1]) / (b[i-1] + b[i+1])

Then, if necessary, you reduce the numerator and denominator to their simplest possible terms. So the sequence starts like this:

• 0/1, 1/1

To create the next stage, find the mediant of the two fractions above: (0+1) / (1+1) = 1/2

• 0/1, 1/2, 1/1

For the next stage, there are two mediants to find: (0+1) / (1+2) = 1/3, (1+1) / (2+3) = 2/3

• 0/1, 1/3, 1/2, 2/3, 1/1

Note that 1/2 is the mediant of 1/3 and 2/3, that is, 1/2 = (1+2) / (3+3) = 3/6 = 1/2. The next stage is this:

• 0/1, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 1/1

Now 1/2 is the mediant of 2/5 and 3/5, that is, 1/2 = (2+3) / (5+5) = 5/10 = 1/2. Further stages go like this:

• 0/1, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 1/1

• 0/1, 1/6, 1/5, 2/9, 1/4, 3/11, 2/7, 3/10, 1/3, 4/11, 3/8, 5/13, 2/5, 5/12, 3/7, 4/9, 1/2, 5/9, 4/7, 7/12, 3/5, 8/13, 5/8, 7/11, 2/3, 7/10, 5/7, 8/11, 3/4, 7/9, 4/5, 5/6, 1/1

• 0/1, 1/7, 1/6, 2/11, 1/5, 3/14, 2/9, 3/13, 1/4, 4/15, 3/11, 5/18, 2/7, 5/17, 3/10, 4/13, 1/3, 5/14, 4/11, 7/19, 3/8, 8/21, 5/13, 7/18, 2/5, 7/17, 5/12, 8/19, 3/7, 7/16, 4/9, 5/11, 1/2, 6/11, 5/9, 9/16, 4/7, 11/19, 7/12, 10/17, 3/5, 11/18, 8/13, 13/21, 5/8, 12/19, 7/11, 9/14, 2/3, 9/13, 7/10, 12/17, 5/7, 13/18, 8/11, 11/15, 3/4, 10/13, 7/9, 11/14, 4/5, 9/11, 5/6, 6/7, 1/1

The Farey sequence is actually a fractal, as you can see more easily when it’s represented as an image:

Farey fractal stage #1, representing 0/1, 1/2, 1/1

Farey fractal stage #2, representing 0/1, 1/3, 1/2, 2/3, 1/1

Farey fractal stage #3, representing 0/1, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 1/1

Farey fractal stage #4, representing 0/1, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 1/1

Farey fractal stage #5

Farey fractal stage #6

Farey fractal stage #7

Farey fractal stage #8

Farey fractal stage #9

Farey fractal stage #10

Farey fractal (animated)

That looks like the slope of a hill to me, so you could call it a Farey fract-hill. But Farey fract-hills or Farey fractals aren’t confined to the unit interval, 0/1 to 1/1. Here are Farey fractals for the intervals 0/1 to n/1, n = 1..10:

Farey fractal for interval 0/1 to 1/1

Farey fractal for interval 0/1 to 2/1, beginning 0/1, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 1/1, 5/4, 4/3, 7/5, 3/2, 8/5, 5/3, 7/4, 2/1

Farey fractal for interval 0/1 to 3/1, beginning 0/1, 1/3, 1/2, 2/3, 1/1, 5/4, 4/3, 7/5, 3/2, 8/5, 5/3, 7/4, 2/1, 7/3, 5/2, 8/3, 3/1

Farey fractal for interval 0/1 to 4/1, beginning
0/1, 1/3, 1/2, 2/3, 1/1, 4/3, 3/2, 5/3, 2/1, 7/3, 5/2, 8/3, 3/1, 10/3, 7/2, 11/3, 4/1

Farey fractal for interval 0/1 to 5/1, beginning 0/1, 1/1, 5/4, 10/7, 5/3, 7/4, 2/1, 7/3, 5/2, 8/3, 3/1, 13/4, 10/3, 25/7, 15/4, 4/1, 5/1

Farey fractal for interval 0/1 to 6/1, beginning 0/1, 1/2, 1/1, 4/3, 3/2, 5/3, 2/1, 5/2, 3/1, 7/2, 4/1, 13/3, 9/2, 14/3, 5/1, 11/2, 6/1

Farey fractal for interval 0/1 to 7/1, beginning 0/1, 7/5, 7/4, 2/1, 7/3, 21/8, 14/5, 3/1, 7/2, 4/1, 21/5, 35/8, 14/3, 5/1, 21/4, 28/5, 7/1

Farey fractal for interval 0/1 to 8/1, beginning 0/1, 1/2, 1/1, 3/2, 2/1, 5/2, 3/1, 7/2, 4/1, 9/2, 5/1, 11/2, 6/1, 13/2, 7/1, 15/2, 8/1

Farey fractal for interval 0/1 to 9/1, beginning 0/1, 1/1, 3/2, 2/1, 3/1, 7/2, 4/1, 13/3, 9/2, 14/3, 5/1, 11/2, 6/1, 7/1, 15/2, 8/1, 9/1

Farey fractal for interval 0/1 to 10/1, beginning 0/1, 5/4, 5/3, 2/1, 5/2, 3/1, 10/3, 15/4, 5/1, 25/4, 20/3, 7/1, 15/2, 8/1, 25/3, 35/4, 10/1

The shape of the slope is determined by the factorization of n:

n = 12 = 2^2 * 3

n = 16 = 2^4

n = 18 = 2 * 3^2

n = 20 = 2^2 * 5

n = 25 = 5^2

n = 27 = 3^3

n = 32 = 2^5

n = 33 = 3 * 11

n = 42 = 2 * 3 * 7

n = 64 = 2^6

n = 65 = 5 * 13

n = 70 = 2 * 5 * 7

n = 77 = 7 * 11

n = 81 = 3^4

n = 96 = 2^5 * 3

n = 99 = 3^2 * 11

n = 100 = 2^2 * 5^2

Farey fractal-hills, n = various

For Revver and Fevver

This shape reminds me of the feathers on an exotic bird:

feathers

(click or open in new window for full size)


feathers_anim

(animated version)


The shape is created by reversing the digits of a number, so you could say it involves revvers and fevvers. I discovered it when I was looking at the Halton sequence. It’s a sequence of fractions created according to a simple but interesting rule. The rule works like this: take n in base b, reverse it, and divide reverse(n) by the first power of b that is greater thann.

For example, suppose n = 6 and b = 2. In base 2, 6 = 110 and reverse(110) = 011 = 11 = 3. The first power of 2 that is greater than 6 is 2^3 or 8. Therefore, halton(6) in base 2 equals 3/8. Here is the same procedure applied to n = 1..20:

1: halton(1) = 1/10[2] → 1/2
2: halton(10) = 01/100[2] → 1/4
3: halton(11) = 11/100[2] → 3/4
4: halton(100) = 001/1000[2] → 1/8
5: halton(101) = 101/1000[2] → 5/8
6: halton(110) = 011/1000 → 3/8
7: halton(111) = 111/1000 → 7/8
8: halton(1000) = 0001/10000 → 1/16
9: halton(1001) = 1001/10000 → 9/16
10: halton(1010) = 0101/10000 → 5/16
11: halton(1011) = 1101/10000 → 13/16
12: halton(1100) = 0011/10000 → 3/16
13: halton(1101) = 1011/10000 → 11/16
14: halton(1110) = 0111/10000 → 7/16
15: halton(1111) = 1111/10000 → 15/16
16: halton(10000) = 00001/100000 → 1/32
17: halton(10001) = 10001/100000 → 17/32
18: halton(10010) = 01001/100000 → 9/32
19: halton(10011) = 11001/100000 → 25/32
20: halton(10100) = 00101/100000 → 5/32…

Note that the sequence always produces reduced fractions, i.e. fractions in their lowest possible terms. Once 1/2 has appeared, there is no 2/4, 4/8, 8/16…; once 3/4 has appeared, there is no 6/8, 12/16, 24/32…; and so on. If the fractions are represented as points in the interval [0,1], they look like this:

line1_1_2

point = 1/2


line2_1_4

point = 1/4


line3_3_4

point = 3/4


line4_1_8

point = 1/8


line5_5_8

point = 5/8


line6_3_8

point = 3/8


line7_7_8

point = 7/8


line_b2_anim

(animated line for base = 2, n = 1..63)


It’s apparent that Halton points in base 2 will evenly fill the interval [0,1]. Now compare a Halton sequence in base 3:

1: halton(1) = 1/10[3] → 1/3
2: halton(2) = 2/10[3] → 2/3
3: halton(10) = 01/100[3] → 1/9
4: halton(11) = 11/100[3] → 4/9
5: halton(12) = 21/100[3] → 7/9
6: halton(20) = 02/100 → 2/9
7: halton(21) = 12/100 → 5/9
8: halton(22) = 22/100 → 8/9
9: halton(100) = 001/1000 → 1/27
10: halton(101) = 101/1000 → 10/27
11: halton(102) = 201/1000 → 19/27
12: halton(110) = 011/1000 → 4/27
13: halton(111) = 111/1000 → 13/27
14: halton(112) = 211/1000 → 22/27
15: halton(120) = 021/1000 → 7/27
16: halton(121) = 121/1000 → 16/27
17: halton(122) = 221/1000 → 25/27
18: halton(200) = 002/1000 → 2/27
19: halton(201) = 102/1000 → 11/27
20: halton(202) = 202/1000 → 20/27
21: halton(210) = 012/1000 → 5/27
22: halton(211) = 112/1000 → 14/27
23: halton(212) = 212/1000 → 23/27
24: halton(220) = 022/1000 → 8/27
25: halton(221) = 122/1000 → 17/27
26: halton(222) = 222/1000 → 26/27
27: halton(1000) = 0001/10000 → 1/81
28: halton(1001) = 1001/10000 → 28/81
29: halton(1002) = 2001/10000 → 55/81
30: halton(1010) = 0101/10000 → 10/81

And here is an animated gif representing the Halton sequence in base 3 as points in the interval [0,1]:

line_b3_anim


Halton points in base 3 also evenly fill the interval [0,1]. What happens if you apply the Halton sequence to a two-dimensional square rather a one-dimensional line? Suppose the bottom left-hand corner of the square has the co-ordinates (0,0) and the top right-hand corner has the co-ordinates (1,1). Find points (x,y) inside the square, with x supplied by the Halton sequence in base 2 and y supplied by the Halton sequence in base 3. The square will gradually fill like this:

square1

x = 1/2, y = 1/3


square2

x = 1/4, y = 2/3


square3

x = 3/4, y = 1/9


square4

x = 1/8, y = 4/9


square5

x = 5/8, y = 7/9


square6

x = 3/8, y = 2/9


square7

x = 7/8, y = 5/9


square8

x = 1/16, y = 8/9


square9

x = 9/16, y = 1/27…


square_anim

animated square


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