What does a fractal phallus look like?

*Millions* of people have *axed* this corely key *question*.

The Overlord of the Über-Feral can *answer* it — keyly, corely and *comprehensively* dot dot dot

And here *is* the answer: Phrallic Frolics…

What does a fractal phallus look like?

*Millions* of people have *axed* this corely key *question*.

The Overlord of the Über-Feral can *answer* it — keyly, corely and *comprehensively* dot dot dot

And here *is* the answer: Phrallic Frolics…

Here is a very simple tree with two branches:

Two-branch tree

These are the steps that a simple computer program follows to draw the tree, with a red arrow indicating where the computer’s focus is at each stage:

Two-branch tree stage 1

2-Tree stage 2

2-Tree stage 3

2-Tree stage 4

2-Tree (animated)

If you had to give the computer an explicit instruction at each stage, the instructions might look something like this:

1. Start at node 1, draw a left branch to node 2 and colour the node green.

2. Return to node 1.

3. Draw a right branch to node 3 and colour the node green.

4. Finish.

Now try a slightly less simple tree with branches that fork twice:

Four-branch tree (static)

These are the steps that a simple computer program follows to draw the tree, with a red arrow indicating where the computer’s focus is at each stage:

4-Tree #1

4-Tree #2

4-Tree #3

4-Tree #4

4-Tree #5

4-Tree #6

4-Tree #7

4-Tree #8

4-Tree #9

4-Tree #10

4-Tree #11

4-Tree (animated)

If you had to give the computer an explicit instruction at each stage, the instructions might look something like this:

1. Start at node 1 and draw a left branch to node 2.

2. Draw a left branch to node 3 and colour it green.

3. Return to node 2.

4. Draw a right branch to node 4 and colour it green.

5. Return to node 2.

6. Return to node 1.

7. Draw a right branch to node 5.

8. Draw a left branch to node 6.

9. Draw a left branch to node 7 and colour it green.

10. Return to node 6.

11. Draw a left branch to node 8 and colour it green.

12. Finish.

It’s easy to see that the list of instructions would be much bigger for a tree with branches that fork three times, let alone four times or you. But you don’t need to give a full set of explicit instructions: you can use a program, or a list of instructions using variables. Suppose the tree has branches that fork

1. li = 1, level(1) = 0, level(2) = 0, ... level(f+1) = 0

2. level(li) = level(li) + 1

3. If level(li) = 1, draw a branch to the left and jump to step 7

4. If level(li) = 2, draw a branch to the right and jump to step 7

5. li = li - 1 (note that this line is reached if the tests fail in lines 3 and 4)

6. If li > 0, jump to step 2, otherwise jump to step 11

7. If li = f, draw a green node and jump to step 5

9. li = li + 1

10. Jump to step 2

11. Finish.

By changing the value of

16-Tree (static)

16-Tree (animated)

With simple adjustments, the program can be used for other shapes whose underlying structure can be represented symbolically as a tree. The program is in fact a fractalizer, that is, it draws a fractal. So if you use a version of the program to draw fractals based on right-triangles, you can say you are “tright treeing” (tright = triangle-that-is-right).

Here is some tright treeing. Start with a simple isoceles right-triangle. It can be divided into smaller isoceles right-triangles by finding the midpoint of the hypotenuse, then repeating:

Right-triangle rep-2 stage 1

Right-triangle #2

Tright #3

Tright #4

Tright #5

Tright #6

Tright #7

Tright #7 (no internal lines)

You can distort the isoceles right-triangle in interesting ways by finding the midpoint of a side other than the hypotenuse, like this:

Right-triangle (distorted) #1

Distorted tright #2

Distorted tright #3

Distorted tright #4

Distorted tright #5

Distorted tright #6

Distorted tright #7

Distorted tright #8

Distorted tright #9

Distorted tright #10

Distorted tright #11

Distorted tright #12

Distorted tright #13

Distorted tright (animated)

Here’s a different right-triangle. When you divide it regularly, it looks like this:

Right-triangle rep-3 stage 1

Rep-3 Tright #2

3-Tright #3

3-Tright #4

3-Tright #5

3-Tright #6

3-Tright #7

3-Tright #8

3-Tright #9

3-Tright (one colour)

When you distort the divisions, you can create interesting fractals (click on images for larger versions):

Distorted 3-Tright

Distorted 3-Tright

Distorted 3-Tright

Distorted 3-Tright

Distorted 3-Tright

Distorted 3-Tright

Distorted 3-Tright (animated)

And when four of the distorted right-triangles (rep-2 or rep-3) are joined in a diamond, you can create shapes like these:

Creating a diamond #1

Creating a diamond #2

Creating a diamond #3

Creating a diamond #4

Creating a diamond (animated)

Rep-3 right-triangle diamond (divided)

Rep-3 right-triangle diamond (single colour)

Distorted rep-3 right-triangle diamond

Distorted 3-tright diamond

Distorted 3-tright diamond

Distorted 3-tright diamond

Distorted 3-tright diamond

Distorted 3-tright diamond

Distorted 3-tright diamond

Distorted 3-tright diamond

Distorted 3-tright diamond

Distorted 3-tright diamond

Distorted 3-tright diamond

Distorted 3-tright diamond (animated)

Distorted rep-2 right-triangle

Distorted 2-tright diamond

Distorted 2-tright diamond

Distorted 2-tright diamond

Distorted 2-tright diamond

Distorted 2-tright diamond (animated)

Discovering something that’s new to you in recreational maths is good. But so is re-discovering it by a different route. I’ve long been *passionate* about what happens when a point is allowed to jump repeatedly halfway towards the randomly chosen vertices of a square. If the point can choose any vertex any number of times, the interior of the square fills slowly and completely with points, like this:

Point jumping at random halfway towards vertices of a square

However, if the point is banned from jumping towards the same vertex twice or more in a row, an interesting fractal appears:

Fractal #1 — ban on jumping towards vertex v_{i} twice or more

If the point can’t jump towards the vertex one place clockwise of the vertex it’s just jumped towards, this fractal appears:

Fractal #2 — ban on jumping towards vertex v_{i+1}

If the point can’t jump towards the vertex two places clockwise of the vertex it’s just jumped towards, this fractal appears (two places clockwise is also two places anticlockwise, i.e. the banned vertex is diagonally opposite):

Fractal #3 — ban on jumping towards vertex v_{i+2}

Now I’ve discovered a new way to create these fractals. You take a filled square, divide it into smaller squares, then remove some of them in a systematic way. Then you do the same to the smaller squares that remain. For fractal #1, you do this:

Fractal #1, stage #1

Stage #2

Stage #3

Stage #4

Stage #5

Stage #6

Stage #7

Stage #8

Fractal #1 (animated)

For fractal #2, you do this:

Fractal #2, stage #1

Stage #2

Stage #3

Stage #4

Stage #5

Stage #6

Stage #7

Stage #8

Fractal #2 (animated)

For fractal #3, you do this:

Fractal #3, stage #1

Stage #2

Stage #3

Stage #4

Stage #5

Stage #6

Stage #7

Stage #8

Fractal #3 (animated)

If the sub-squares are coloured, it’s easier to understand how, say, fractal #1 is created:

Fractal #1 (coloured), stage #1

Stage #2

Stage #3

Stage #4

Stage #5

Stage #6

Stage #7

Stage #8

Fractal #1 (coloured and animated)

The fractal is actually being created in quarters, with one quarter rotated to form the second, third and fourth quarters:

Fractal #1, quarter

Here’s an animation of the same process for fractal #3:

Fractal #3 (coloured and animated)

So you can create these fractals either with a jumping point or by subdividing a square. But in fact I discovered the subdivided-square route by looking at a variant of the jumping-point route. I wondered what would happen if you took a point inside a square, allowed it to trace all possible routes towards the vertices without marking its position, then imposed the restriction for Fractal #1 on its final jump, namely, that it couldn’t jump towards the vertex it jumped towards on its previous jump. If the point is marked after its final jump, this is what appears (if the routes chosen had been truly random, the image would be similar but messier):

Fractal #1, restriction on final jump

Then I imposed the same restriction on the point’s final two jumps:

Fractal #1, restriction on final 2 jumps

And final three jumps:

Fractal #1, restriction on final 3 jumps

And so on:

Fractal #1, restriction on final 4 jumps

Fractal #1, restriction on final 5 jumps

Fractal #1, restriction on final 6 jumps

Fractal #1, restriction on final 7 jumps

Here are animations of the same process applied to fractals #2 and #3:

Fractal #2, restrictions on final 1, 2, 3… jumps

Fractal #3, restrictions on final 1, 2, 3… jumps

The longer the points are allowed to jump before the final restriction is imposed on their

Fractal #1, packed points #1

Packed points #2

Packed points #3

Eventually, the individual points will form a solid mass, like this:

Fractal #1, solid mass of points

Fractal #1, packed points (animated)

Previously pre-posted (please peruse):

• Square Routes

• Square Routes Revisited

• Square Routes Re-Revisited

• Square Routes Re-Re-Revisited

10111 in base 2

212 in base 3

113 in base 4

43 in base 5

35 in base 6

32 in base 7

27 in base 8

25 in base 9

23 in base 10

21 in base 11

1B in base 12

1A in base 13

19 in base 14

18 in base 15

17 in base 16

16 in base 17

15 in base 18

14 in base 19

13 in base 20

12 in base 21

11 in base 22

10 in base 23

N in all bases >= 24

√23 = 4.7958315__23__31…

Twice before on Overlord-in-terms-of-Core-Issues-around-Maximal-Engagement-with-Key-Notions-of-the-Über-Feral, I’ve interrogated issues around pursuit curves. Imagine four mice or four beetles each sitting on one corner of a square and looking towards the centre of the square. If each mouse or beetle begins to run towards the mouse or beetle to its left, it will follow a curving path that takes it to the centre of the square, like this:

vertices = 4, pursuit = +1

The paths followed by the mice or beetles are pursuit curves. If you arrange eight mice clockwise around a square, with a mouse on each corner and a mouse midway along each side, you get a different set of pursuit curves:

v = 4 + 1 on the side, p = +1

Here each mouse is pursuing the mouse two places to its left:

v = 4+s1, p = +2

And here each mouse is pursuing the mouse three places to its left:

v = 4+s1, p = +3

Now try a different arrangement of the mice. In the square below, the mice are arranged clockwise in rows from the bottom right-hand corner. That is, mouse #1 begins on the bottom left-hand corner, mouse #2 begins between that corner and the centre, mouse #3 begins on the bottom left-hand corner, and so on. When each mice runs towards the mouse three places away, these pursuit curves appear:

v = 4 + 1 internally, p = +3

Here are some more:

v = 4 + i1, p = +5

v = 4 + i2, p = +1

v = 4 + i2, p = +2

So far, all the mice have eventually run to the centre of the square, but that doesn’t happen here:

v = 4 + i2, p = 4

Here are more pursuit curves for the v4+i2 mice, using an animated gif:

v = 4 + i2, p = various (animated — open in new tab for clearer image)

And here are more pursuit curves that don’t end in the centre of the square:

v = 4 + i4, p = 4

v = 4 + i4, p = 8

v = 4 + i4, p = 12

v = 4 + i4, p = 16

But the v4+i4 pursuit curves more usually look like this:

v = 4 + i4, p = 7

Now try adapting the rules so that mice don’t run directly towards another mouse, but towards the point midway between two other mice. In this square, the odd- and even-numbered mice follow different rules. Mice #1, #3, #5 and #7 run towards the point midway between the mice one and two places away, while ice #2, #4, #6 and #8 run towards the point midway between the mice two and seven places away:

v = 4 + s1, p(1,3,5,7) = 1,2, p(2,4,6,8) = 2,7

I think the curves are very elegant. Here’s a slight variation:

v = 4 + s1, p1 = 1,3, p2 = 2,7

Now try solid curves:

v = 4 + s1, p1 = 1,3, p2 = 2,7 (red)

v = 4 + s1, p1 = 1,3, p2 = 2,7 (yellow-and-blue)

And some variants:

v = 4 + s1, p1 = 1,7, p2 = 1,2

v = 4 + s1, p1 = 2,3, p2 = 2,5

v = 4 + s1, p1 = 5,6, p2 = 1,3

v = 4 + s1, p1 = 5,6, p2 = 1,4

v = 4 + s1, p1 = 5,6, p2 = 1,6

Elsewhere other-posted:

Pre-previously on Overlord-in-terms-of-Core-Issues-around-Maximal-Engagement-with-Key-Notions-of-the-Über-Feral, I interrogated issues around this shape, the horned triangle:

Horned Triangle (more details)

Now I want to look at the tricorn (from Latin

Tricorn, or three-horned triangle

These are the stages that make up the tricorn:

Tricorn (stages)

Tricorn (animated)

And there’s no need to stop at triangles. Here is a four-horned square, or quadricorn:

Quadricorn

Quadricorn (animated)

Quadricorn (coloured)

And a five-horned pentagon, or quinticorn:

Quinticorn, or five-horned pentagon

Quinticorn (anim)

Quinticorn (col)

And below are some variants on the shapes above. First, the reversed tricorn:

Reversed Tricorn

Reversed Tricorn (anim)

Reversed Tricorn (col)

The nested tricorn:

Nested Tricorn (anim)

Nested Tricorn (col)

Nested Tricorn (red-green)

Nested Tricorn (variant col)

The nested quadricorn:

Nested Quadricorn (anim)

Nested Quadricorn

Nested Quadricorn (col #1)

Nested Quadricorn (col #2)

Finally (and ferally), the pentagonal octopus or pentapus:

Pentapus (anim)

Pentapus

Pentapus #2

Pentapus #3

Pentapus #4

Pentapus #5

Pentapus #6

Pentapus (col anim)

Elsewhere other-engageable:

• The Art Grows Onda — the horned triangle and Katsushika Hokusai’s painting *The Great Wave off Kanagawa* (c. 1830)

Pre-previously on Overlord-In-Terms-of-Issues-Around-Engagement-with-the-Über-Feral, I’ve looked at various ways of creating fractals by restricting the moves of a point jumping towards the vertices of a polygon. For example, the point can be banned from jumping towards the same vertex twice in a row. This time, I want to look at fractals created not by restriction, but by compulsion. If the point jumps towards vertex *v* and then tries to jump towards vertex *v* again, it will be forced to jump towards vertex *v*+1 instead, and so on.

You could call *v* → *v*+1 a forced increment or finc. So these are finc fractals. In some cases, restriction and compulsion create the same fractals, but I’ve found some new fractals using compulsion. Consider the fractal created by the rule *v*_{[-2]}+1, *v*_{[-1]} → +0,+1, where the subscripts refer to the history of jumps: *v*_{[-2]} is the jump-before-last, *v*_{[-1]} is the last jump. If the new vertex, *v*_{[0]}, chosen is the same as *v*_{[-2]}+1 (e.g., *v*_{[0]} = 2 = *v*_{[-2]}+1 = 1+1), then the forced increment is 0, i.e., the point is allowed to choose that jump. However, if *v*_{[0]} = *v*_{[-1]}, then the forced increment is 1 and the point must jump towards *v*_{[-1]}+1.

Here is the fractal in question:

*v*_{[-2]}+1, *v*_{[-1]} → +0,+1 (black-and-white)

*v*_{[-2]}+1, *v*_{[-1]} → +0,+1 (colour)

1,0 → +0,+1 (animated)

1,0 → +1,+0 (bw)

1,0 → +1,+0 (col)

1,0 → +1,+0 (anim)

1,0 → +1,+1 (bw)

1,0 → +1,+1 (col)

1,0 → +1,+1 (animated)

0,1 → +2,+1 (anim)

0,1 → +3,+1

1,0 → +0,+1

1,0 → +1,+0

1,1 → +0,+1

1,1 → +1,+2

1,1 → +1,+3

1,1 → +2,+1

1,2 → +0,+3

1,3 → +0,+1

2,2 → +0,+1

But suppose the history of jumps records not actual jumps, but the jumps the point wanted to make instead. In some cases, the jump made will be the same as the jump originally chosen, but in other cases it won’t. Here are some fractals using this method:

0 → +2

0 → +3

2 → +1

2 → +2

Imagine a game with six players, numbered #1 to #6, and one six-sided die. Someone rolls the die and the player who matches the number wins the game. That is, if the die rolls 1, player #1 wins; if the die rolls 2, player #2 wins; and so on. With a fair die, this is a fair game, because each player has exactly a 1/6 chance of winning. You could call it a simultaneous game, because all players are playing at once. It has one rule:

• If the die rolls *n*, then player #*n* wins.

Now try a different game with six players and one die. Player #1 rolls the die. If he gets 1, he wins the game. If not, then he leaves the game and player #2 rolls the die. If he gets 2, he wins the game. If not, then he leaves the game and player #3 rolls the die. And so on. You could call this a sequential game, because the players are playing in sequence. It has two rules:

• If player #*n* rolls *n* on the die, then he wins.

• If player #*n* doesn’t roll *n*, then player *n*+1 rolls the die.

Is it a fair game? No, definitely not. Player #1 has the best chance of winning. 1/6 or 16.6% of the time he rolls 1 and wins the game. 5/6 of the time, he rolls 2, 3, 4, 5 or 6 and passes the die to player #2. Now player #2 has a 1/6 chance of rolling a 2 and winning. But he has the opportunity to roll the die only 5/6 of the time, so his chance of winning the game is 1/6 * 5/6 = 5/36 = 13.8%. However, if player #2 rolls a 1, 3, 4, 5 or 6, then he loses and player #3 rolls the die. But player #3 has that opportunity only 5/6 * 5/6 = 25/36 of the time. So his chance of winning is 1/6 * 25/36 = 11.57%. And so on.

To put it another way, if the six players play 46656 = 6^6 games under the sequential rules, then on average:

• Player #1 wins 7776 games

• Player #2 wins 6480 games

• Player #3 wins 5400 games

• Player #4 wins 4500 games

• Player #5 wins 3750 games

• Player #6 wins 3125 games

• 15625 games end without a winner.

In other words, player #1 is 20% more likely to win than player #2, 44% more likely than player #3, 72.8% more likely than player #4, 107% more likely than player #5, and 148.8% more likely than player #6. Furthermore, player #2 is 20% more likely to win than player #3, 44% more likely than player #4, 72.8% more likely than player #5, and so on.

But there is a simple way to make the sequential game perfectly fair, so long as it’s played with a fair die. At least, I’ve thought of a simple way, but there might be more than one.

To make the sequential game fair, you add an extra rule:

1. If player #*n* rolls *n* on the die, he wins the game.

2. If player #*n* rolls a number greater than *n*, he loses and the die passes to player *n*+1.

3. If player #*n* rolls a number less than *n*, then he rolls again.

Let’s run through a possible game to see that it’s fair. Player #1 rolls first. He has a 1/6 chance of rolling a 1 and winning the game. However, 5/6 of the time he loses and passes the die to player #2. If player #2 rolls a 1, he rolls again. In other words, player #2 is effectively playing with a five-sided die, because all rolls of 1 are ignored. Therefore, he has a 1/5 chance of winning the game at that stage.

But hold on: a 1/5 chance of winning is better than a 1/6 chance, which is what player #1 had. So how is the game fair? Well, note the qualifying phrase at the end of the previous paragraph: *at that stage*. The game doesn’t always reach that stage, because if player #1 rolls a 1, the game is over. Player #2 rolls only if player doesn’t roll 1, which is 5/6 of the time. Therefore player #2’s chance of winning is really 1/5 * 5/6 = 5/30 = 1/6.

However, 4/5 of the time player #2 rolls a 3, 4, 5 or 6 and the die passes to player #3. If player #3 rolls a 1 or 2, he rolls again. In other words, player #3 is effectively playing with a four-sided die, because all rolls of 1 and 2 are ignored. Therefore, he has a 1/4 chance of winning the game at that stage.

A 1/4 chance of winning is better than a 1/5 chance and a 1/6 chance, but the same reasoning applies as before. Player #3 rolls the die only 5/6 * 4/5 = 20/30 = 2/3 of the time, so his chance of winning is really 1/4 * 2/3 = 2/12 = 1/6.

However, 3/4 of the time player #2 rolls a 4, 5 or 6 and the die passes to player #4. If player #4 rolls a 1, 2 or 3, he rolls again. In other words, player #4 is effectively playing with a three-sided die, because all rolls of 1, 2 and 3 are ignored. Therefore, he has a 1/3 chance of winning the game at that stage. 1/3 > 1/4 > 1/5 > 1/6, but the same reasoning applies as before. Player #4 rolls the die only 5/6 * 4/5 * 3/4 = 60/120 = 1/2 of the time, so his chance of winning is really 1/3 * 1/2 = 1/6.

And so on. If the die reaches player #5 and he gets a 1, 2, 3 or 4, then he rolls again. He is effectively rolling with a two-sided die, so his chance of winning is 1/2 * 5/6 * 4/5 * 3/4 * 2/3 = 120/720 = 1/6. If player #5 rolls a 6, he loses and the die passes to player #6. But there’s no need for player #6 to roll the die, because he’s bound to win. He rolls again if he gets a 1, 2, 3, 4 or 5, so eventually he must get a 6 and win the game. If player #5 loses, then player #6 automatically wins.

It’s obvious that this form of the game will get slower as more players drop out, because later players will be rolling again more often. To speed the game up, you can refine the rules like this:

1. If Player #1 rolls a 1, he wins the game. Otherwise…

2. If player #2 rolls a 2, he wins the game. If he rolls a 1, he rolls again. Otherwise…

3. Player #3 rolls twice and adds his scores. If the total is 3, 4 or 5, he wins the game. Otherwise…

4. Player #4 rolls once. If he gets 1 or 2, he wins the game. Otherwise…

5. Player #5 rolls once. If he gets 1, 2 or 3, he wins the game. Otherwise…

6. Player #6 wins the game.

Only player #2 might have to roll more than twice. Player #3 has to roll twice because he needs a way to get a 1/4 chance of winning. If you roll two dice, there are:

• Two ways of getting a total of 3: roll #1 is 1 and roll #2 is 2, or vice versa.

• Three ways of getting a total of 4 = 1+3, 3+1, 2+2.

• Four ways of getting 5 = 1+4, 4+1, 2+3, 3+2.

This means player #3 has 2 + 3 + 4 = 9 ways of winning. But there are thirty-six ways of rolling one die twice. Therefore player #3 has a 9/36 = 1/4 chance of winning. Here are the thirty-six ways of rolling one die twice, with asterisks marking the winning totals for player #3:

01. (1,1)

02. (1,2)*

03. (2,1)*

04. (1,3)*

05. (3,1)*

06. (1,4)*

07. (4,1)*

08. (1,5)

09. (5,1)

10. (1,6)

11. (6,1)

12. (2,2)*

13. (2,3)*

14. (3,2)*

15. (2,4)

16. (4,2)

17. (2,5)

18. (5,2)

19. (2,6)

20. (6,2)

21. (3,3)

22. (3,4)

23. (4,3)

24. (3,5)

25. (5,3)

26. (3,6)

27. (6,3)

28. (4,4)

29. (4,5)

30. (5,4)

31. (4,6)

32. (6,4)

33. (5,5)

34. (5,6)

35. (6,5)

36. (6,6)

• φασὶ γοῦν Ἵππαρχον τὸν Πυθαγόρειον, αἰτίαν ἔχοντα γράψασθαι τὰ τοῦ Πυθαγόρου σαφῶς, ἐξελαθῆναι τῆς διατριβῆς καὶ στήλην ἐπ’ αὐτῷ γενέσθαι οἷα νεκρῷ. — Κλήμης ὁ Ἀλεξανδρεύς, *Στρώματα*.

• They say, then, that Hipparchus the Pythagorean, being guilty of writing the tenets of Pythagoras in plain language, was expelled from the school, and a pillar raised for him as if he had been dead. — Clement of Alexandria, *The Stromata*, 2.5.9.57.3-4

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