Allus Pour, Horic

*As a rotating animated gif (optimized at ezGIF).


Performativizing Paronomasticity

The title of this incendiary intervention, which is perhaps my most contrived title yet, is a paronomasia on Shakespeare’s “Alas, poor Yorick!” (Hamlet, Act 5, scene 1). “Allus” is a northern form of “always”, “pour” has its standard meaning, and “Horic” is from the Greek ὡρῐκός, hōrikos, which strictly speaking means “in one’s prime, blooming”. However, it could also be interpreted as meaning “hourly”. So the paronomasia means “Always pour, O Hourly One!” (i.e. hourglass).

Squooh You

Here’s an interesting little puzzle:

Winnie-the-Pooh and Piglet set out to visit one another. They leave their houses at the same time and walk along the same road. But Piglet is absorbed in counting the birds overhead, and Winnie-the-Pooh is composing a new “hum,” so they pass one another without noticing. One minute after the meeting, Winnie-the-Pooh is at Piglet’s house, and 4 minutes after the meeting Piglet is at Winnie-the-Pooh’s. How long has each of them walked? — “A puzzle by S. Sefibekov” viâ Futility Closet

If you’re good at maths, you should see the answer in an intuitive instant. I’m not good at maths, so it took me much longer, because I didn’t understand what was going on. But I can explain the answer like this. Pooh is obviously walking faster than Piglet. Therefore, Pooh and Piglet can’t have met after one minute, because that would mean Pooh takes one minute to walk the distance walked by Piglet in one minute.

So let’s suppose Pooh and Piglet met after two minutes. If Pooh takes one minute to walk the distance walked by Piglet in two minutes, then Pooh is walking twice as fast as Piglet. Does that work? Yes, because Piglet walks Pooh’s distance in four minutes, which is twice as long as Pooh took. Therefore Piglet is walking twice as slowly as Pooh. It’s symmetrical and we can conclude that they did indeed meet after two minutes. Pooh then walks another minute, for three minutes in total, and Piglet walks another four minutes, for six minutes in total.

But guessing is not a good way to find the answer to the puzzle. Let’s try to reason it through properly. Suppose that Pooh and Piglet meet after one unit of time, during which Piglet has walked one unit of distance and Pooh has walked x units of distance, where x > 1. In other words, Pooh is walking x times faster than Piglet. The distances they walk before meeting are therefore in the ratio:

1 : x

Next, note that Pooh will cover the distance Piglet has already walked in 1 unit / x = 1 minute, while Piglet covers the distance Pooh has already walked in x / 1 = 4 minutes. The times they take are therefore in the ratio:

1 / x : x → 1 : x^2 → 1 : 4

And if 1 : x^2 is 1 : 4 (the ratio of the minutes they walk after meeting), then 1 : x (the ratio of the distances they walk before meeting) = 1 : √(x^2) = 1 : √4 = 1 : 2. Pooh is therefore walking 2x faster than Piglet and Piglet is walking 2x slower than Pooh. If Pooh covers Piglet’s distance in 1 minute, Piglet must have taken 2 minutes to walk that distance. And if Piglet covers Pooh’s distance in 4 minutes, Pooh must have taken 2 minutes to walk that distance.

Therefore, when they meet, each of them has been walking for 2 minutes. Pooh therefore walks 2 + 1 = 3 minutes in total and Piglet walks 2 + 4 = 6 minutes in total.

The result can be generalized for all relative speeds. Suppose that Pooh and Piglet meet after m1 minutes and that Pooh then takes m2 minutes to walk the distance Piglet walked in m1 minutes, while Piglet takes m3 minutes to walk the distance Pooh walked in m1 minutes. The time they walk before they meet, m1 minutes, is therefore supplied by this simple equation:

m1 = √(m3 / m2)

And you can call √(m3 / m2), the square root of m3 / m2, the squooh function:

m1 = √(m3 / m2) = squooh(m2,m3)

Now suppose the distance between Pooh’s and Piglet’s houses houses is 12 units of distance and that Piglet always walks at 1 unit a minute. If Pooh walks at the same speed as Piglet, i.e. 1 unit a minute, then:

After they meet, Pooh walks 6 more min = m2, Piglet walks 6 more min = m3.

How long do they walk before they meet?

m1 = m3 / m2 = 1, √1 * 6 = 6

They meet after 6 min.

Now suppose that after they meet, Pooh walks 2 more min, Piglet walks 8 more min.

Therefore, m3 / m2 = 4, √4 * 2 = 2 * 2 = 4 = m1

Therefore they walk for 4 min before they meet and Pooh walks 2x faster than Piglet.

After they meet, Pooh walks 1 more min, Piglet walks 9 more min (m3 / m2 = 9, √9 * 1 = 3)

Therefore they walk for 3 min before they meet and Pooh walks 3x faster than Piglet.

After they meet, Pooh walks 0·6 more min, Piglet walks 9·6 more min (m3 / m2 = 16, √16 * 0·6 = 4 * 0·6 = 2·4)

Therefore they walk for 2·4 min before they meet and Pooh walks 4x faster than Piglet:

After they meet, Pooh walks 0·4 more min, Piglet walks 10 more min (m3 / m2 = 25, √25 * 0·4 = 5 * 0·4 = 2)

Therefore they walk for 2 min before they meet and Pooh walks 5x faster than Piglet.

And so on.

A Seed Indeed

Like plants, fractals grow from seeds. But plants start with a small seed that gets bigger. Fractals start with a big seed that gets smaller. For example, perhaps the most famous fractal of all is the Koch snowflake. The seed of the Koch snowflake is step #2 here:

Stages of the Koch snowflake (from Fractals and the coast of Great Britain)


To create the Koch snowflake, you replace each straight line in the initial triangle with the seed:

Creating the Koch snowflake (from Wikipedia)


Animated Koch snowflake (from Wikipedia)


Now here’s another seed for another fractal:

Fractal stage #1


The seed is like a capital “I”, consisting of a line of length l sitting between two lines of length l/2 at right angles. The rule this time is: Replace the center of the longer line and the two shorter lines with ½-sized versions of the seed:

Fractal stage #2


Try and guess what the final fractal looks like when this rule is applied again and again:

Fractal stage #3


Fractal stage #4


Fractal stage #5


Fractal stage #6


Fractal stage #7


Fractal stage #8


Fractal stage #9


Fractal stage #10


I call this fractal the hourglass. And there are a lot of ways to create it. Here’s an animated version of the way shown in this post:

Hourglass fractal (animated)


Hour Re-Re-Powered

In “Hour Power” I looked at my favorite fractal, the hourglass fractal:

The hourglass fractal


I showed three ways to create the fractal. Next, in “Hour Re-Powered”, I showed a fourth way. Now here’s a fifth (previously shown in “Tri Again”).

This is a rep-4 isosceles right triangle:

Rep-4 isosceles right triangle


If you divide and discard one of the four sub-triangles, then adjust one of the three remaining sub-triangles, then keep on dividing-and-discarding (and adjusting), you can create a certain fractal — the hourglass fractal:

Triangle to hourglass #1


Triangle to hourglass #2


Triangle to hourglass #3


Triangle to hourglass #4


Triangle to hourglass #5


Triangle to hourglass #6


Triangle to hourglass #7


Triangle to hourglass #8


Triangle to hourglass #9


Triangle to hourglass #10


Triangle to hourglass (anim) (open in new tab to see full-sized version)


And here is a zoomed version:

Triangle to hourglass (large)


Triangle to hourglass (large) (anim)


Hour Re-Powered

Pre-previously on Overlord in terms of the Über-Feral, I looked at my favorite member of the fractal community, the Hourglass Fractal:

The hourglass fractal


A real hourglass for comparison


As I described how I discovered the hourglass fractal indirectly and by accident, then showed how to create it directly, using two isosceles triangles set apex-to-apex in the form of an hourglass:

Triangles to hourglass #1


Triangles to hourglass #2


Triangles to hourglass #3


Triangles to hourglass #4


Triangles to hourglass #5


Triangles to hourglass #6

[…]

Triangles to hourglass #10


Triangles to hourglass #11


Triangles to hourglass #12


Triangles to hourglass (animated)


Now, here’s an even simpler way to create the hourglass fractal, starting with a single vertical line:

Line to hourglass #1


Line to hourglass #2


Line to hourglass #3


Line to hourglass #4


Line to hourglass #5


Line to hourglass #6


Line to hourglass #7


Line to hourglass #8


Line to hourglass #9


Line to hourglass #10


Line to hourglass #11


Line to hourglass (animated)


Hour Power

Would it be my favorite fractal if I hadn’t discovered it for myself? It might be, because I think it combines great simplicity with great beauty. I first came across it when I was looking at this rep-tile, that is, a shape that can be divided into smaller copies of itself:

Rep-4 L-Tromino


It’s called a L-tromino and is a rep-4 rep-tile, because it can be divided into four copies of itself. If you divide the L-tromino into four sub-copies and discard one particular sub-copy, then repeat again and again, you’ll get this fractal:

Tromino fractal #1


Tromino fractal #2


Tromino fractal #3


Tromino fractal #4


Tromino fractal #5


Tromino fractal #6


Tromino fractal #7


Tromino fractal #8


Tromino fractal #9


Tromino fractal #10


Tromino fractal #11


Hourglass fractal (animated)


I call it an hourglass fractal, because it reminds me of an hourglass:

A real hourglass


The hourglass fractal for comparison


I next came across the hourglass fractal when applying the same divide-and-discard process to a rep-4 square. The first fractal that appears is the Sierpiński triangle:

Square to Sierpiński triangle #1


Square to Sierpiński triangle #2


Square to Sierpiński triangle #3


[…]


Square to Sierpiński triangle #10


Square to Sierpiński triangle (animated)


However, you can rotate the sub-squares in various ways to create new fractals. Et voilà, the hourglass fractal appears again:

Square to hourglass #1


Square to hourglass #2


Square to hourglass #3


Square to hourglass #4


Square to hourglass #5


Square to hourglass #6


Square to hourglass #7


Square to hourglass #8


Square to hourglass #9


Square to hourglass #10


Square to hourglass #11


Square to hourglass (animated)


Finally, I was looking at variants of the so-called chaos game. In the standard chaos game, a point jumps half-way towards the randomly chosen vertices of a square or other polygon. In this variant of the game, I’ve added jump-towards-able mid-points to the sides of the square and restricted the point’s jumps: it can only jump towards the points that are first-nearest, seventh-nearest and eighth-nearest. And again the hourglass fractal appears:

Chaos game to hourglass #1


Chaos game to hourglass #2


Chaos game to hourglass #3


Chaos game to hourglass #4


Chaos game to hourglass #5


Chaos game to hourglass #6


Chaos game to hourglass (animated)


But what if you want to create the hourglass fractal directly? You can do it like this, using two isosceles triangles set apex-to-apex in the form of an hourglass:

Triangles to hourglass #1


Triangles to hourglass #2


Triangles to hourglass #3


Triangles to hourglass #4


Triangles to hourglass #5


Triangles to hourglass #6


Triangles to hourglass #7


Triangles to hourglass #8


Triangles to hourglass #9


Triangles to hourglass #10


Triangles to hourglass #11


Triangles to hourglass #12


Triangles to hourglass (animated)


Square Routes Re-Re-Re-Re-Re-Revisited

For a good example of how more can be less, try the chaos game. You trace a point jumping repeatedly 1/n of the way towards a randomly chosen vertex of a regular polygon. When the polygon is a triangle and 1/n = 1/2, this is what happens:

Chaos triangle #1


Chaos triangle #2


Chaos triangle #3


Chaos triangle #4


Chaos triangle #5


Chaos triangle #6


Chaos triangle #7


As you can see, this simple chaos game creates a fractal known as the Sierpiński triangle (or Sierpiński sieve). Now try more and discover that it’s less. When you play the chaos game with a square, this is what happens:

Chaos square #1


Chaos square #2


Chaos square #3


Chaos square #4


Chaos square #5


Chaos square #6


Chaos square #7


As you can see, more is less: the interior of the square simply fills with points and no attractive fractal appears. And because that was more is less, let’s see how less is more. What happens if you restrict the way in which the point inside the square can jump? Suppose it can’t jump twice towards the same vertex (i.e., the vertex v+0 is banned). This fractal appears:

Ban on choosing vertex [v+0]


And if the point can’t jump towards the vertex one place anti-clockwise of the currently chosen vertex, this fractal appears:

Ban on vertex [v+1] (or [v-1], depending on how you number the vertices)


And if the point can’t jump towards two places clockwise or anti-clockwise of the currently chosen vertex, this fractal appears:

Ban on vertex [v+2], i.e. the diagonally opposite vertex


At least, that is one possible route to those three particular fractals. You see another route, start with this simple fractal, where dividing and discarding parts of a square creates a Sierpiński triangle:

Square to Sierpiński triangle #1


Square to Sierpiński triangle #2


Square to Sierpiński triangle #3


Square to Sierpiński triangle #4


[…]


Square to Sierpiński triangle #10


Square to Sierpiński triangle (animated)


By taking four of these square-to-Sierpiński-triangle fractals and rotating them in the right way, you can re-create the three chaos-game fractals shown above. Here’s the [v+0]-ban fractal:

[v+0]-ban fractal #1


[v+0]-ban #2


[v+0]-ban #3


[v+0]-ban #4


[v+0]-ban #5


[v+0]-ban #6


[v+0]-ban #7


[v+0]-ban #8


[v+0]-ban #9


[v+0]-ban (animated)


And here’s the [v+1]-ban fractal:

[v+1]-ban fractal #1


[v+1]-ban #2


[v+1]-ban #3


[v+1]-ban #4


[v+1]-ban #5


[v+1]-ban #6


[v+1]-ban #7


[v+1]-ban #8


[v+1]-ban #9


[v+1]-ban (animated)


And here’s the [v+2]-ban fractal:

[v+2]-ban fractal #1


[v+2]-ban #2


[v+2]-ban #3


[v+2]-ban #4


[v+2]-ban #5


[v+2]-ban #6


[v+2]-ban #7


[v+2]-ban #8


[v+2]-ban #9


[v+2]-ban (animated)

And taking a different route means that you can find more fractals — as I will demonstrate.


Previously pre-posted (please peruse):

Square Routes
Square Routes Revisited
Square Routes Re-Revisited
Square Routes Re-Re-Revisited
Square Routes Re-Re-Re-Revisited
Square Routes Re-Re-Re-Re-Revisited

Koch Rock

The Koch snowflake, named after the Swedish mathematician Helge von Koch, is a famous fractal that encloses a finite area within an infinitely long boundary. To make a ’flake, you start with an equilateral triangle:

Koch snowflake stage #1 (with room for manœuvre)


Next, you divide each side in three and erect a smaller equilateral triangle on the middle third, like this:

Koch snowflake #2


Each original straight side of the triangle is now 1/3 longer, so the full perimeter has also increased by 1/3. In other words, perimeter = perimeter * 1⅓. If the perimeter of the equilateral triangle was 3, the perimeter of the nascent Koch snowflake is 4 = 3 * 1⅓. The area of the original triangle also increases by 1/3, because each new equalitarian triangle is 1/9 the size of the original and there are three of them: 1/9 * 3 = 1/3.

Now here’s stage 3 of the snowflake:

Koch snowflake #3, perimeter = 4 * 1⅓ = 5⅓


Again, each straight line on the perimeter has been divided in three and capped with a smaller equilateral triangle. This increases the length of each line by 1/3 and so increases the full perimeter by a third. 4 * 1⅓ = 5⅓. However, the area does not increase by 1/3. There are twelve straight lines in the new perimeter, so twelve new equilateral triangles are erected. However, because their sides are 1/9 as long as the original side of the triangle, they have 1/(9^2) = 1/81 the area of the original triangle. 1/81 * 12 = 4/27 = 0.148…

Koch snowflake #4, perimeter = 7.11


Koch snowflake #5, p = 9.48


Koch snowflake #6, p = 12.64


Koch snowflake #7, p = 16.85


Koch snowflake (animated)


The perimeter of the triangle increases by 1⅓ each time, while the area reaches a fixed limit. And that’s how the Koch snowflake contains a finite area within an infinite boundary. But the Koch snowflake isn’t confined to itself, as it were. In “Dissecting the Diamond”, I described how dissecting and discarding parts of a certain kind of diamond could generate one side of a Koch snowflake. But now I realize that Koch snowflakes are everywhere in the diamond — it’s a Koch rock. To see how, let’s start with the full diamond. It can be divided, or dissected, into five smaller versions of itself:

Dissectable diamond


When the diamond is dissected and three of the sub-diamonds are discarded, two sub-diamonds remain. Let’s call them sub-diamonds 1 and 2. When this dissection-and-discarding is repeated again and again, a familiar shape begins to appear:

Koch rock stage 1


Koch rock #2


Koch rock #3


Koch rock #4


Koch rock #5


Koch rock #6


Koch rock #7


Koch rock #8


Koch rock #9


Koch rock #10


Koch rock #11


Koch rock #12


Koch rock #13


Koch rock (animated)


Dissecting and discarding the diamond creates one side of a Koch triangle. Now see what happens when discarding is delayed and sub-diamonds 1 and 2 are allowed to appear in other parts of the diamond. Here again is the dissectable diamond:

Dia-flake stage 1


If no sub-diamonds are discarded after dissection, the full diamond looks like this when each sub-diamond is dissected in its turn:

Dia-flake #2


Now let’s start discarding sub-diamonds:

Dia-flake #3


And now discard everything but sub-diamonds 1 and 2:

Dia-flake #4


Dia-flake #5


Dia-flake #6


Dia-flake #7


Dia-flake #8


Dia-flake #9


Dia-flake #10


Now full Koch snowflakes have appeared inside the diamond — count ’em! I see seven full ’flakes:

Dia-flake #11


Dia-flake (animated)


But that isn’t the limit. In fact, an infinite number of full ’flakes appear inside the diamond — it truly is a Koch rock. Here are examples of how to find more full ’flakes:

Dia-flake 2 (static)


Dia-flake 2 (animated)


Dia-flake 3 (static)


Dia-flake 3 (animated)


Previously pre-posted:

Dissecting the Diamond — other fractals in the dissectable diamond

Dissecting the Diamond

Pre-previously on O.o.t.Ü.-F., I dilated the delta. Now I want to dissect the diamond. In geometry, a shape is dissected when it is completely divided into smaller shapes of some kind. If the smaller shapes are identical (except for size) to the original, the original shape is called a rep-tile (because it can be tiled with repeating versions of itself). If the smaller identical shapes are equal in size to each other, the rep-tile is regular; if the smaller shapes are not equal, the rep-tile is irregular. This diamond is an irregular rep-tile or irrep-tile:

Dissectable diamond

Dissected diamond


As you can see, the diamond can be dissected into five smaller versions of itself, two larger ones and three smaller ones. This makes it a rep-5 irrep-tile. And the smaller versions, or sub-diamonds, can themselves be dissected ad infinitum, like this:

Dissected diamond stage #1


Dissected diamond #2


Dissected diamond #3


Dissected diamond #4


Dissected diamond #5


Dissected diamond #6


Dissected diamond #7


Dissected diamond #8


Dissected diamond #9


Dissected diamond (animated)


The full dissected diamond is a fractal, or shape that is similar to itself at varying scales. However, the fractality of the diamond becomes most obvious when you dissect-and-discard. That is, first you dissect the diamond, then you discard one (or more) of the sub-diamonds, like this:

Diamond fractal (retaining sub-diamonds 1,2,3,4) stage #1


1234-Diamond #2


1234-Diamond #3


1234-Diamond #4


1234-Diamond #5


1234-Diamond #6


1234-Diamond #7


1234-Diamond #8


1234-Diamond #9


1234-Diamond (animated)


Here are some more fractals created by dissecting and discarding one sub-diamond:

Diamond fractal (retaining sub-diamonds 1,2,4,5)


1245-Diamond (anim)


2345-Diamond


2345-Diamond (anim)


The 2345-diamond fractal has variants created by mirroring one or more sub-diamonds, so that the orientation of the sub-dissections changes. Here is one of the variants:

2345-Diamond (variation)


2345-Diamond (variant) (anim)


And here is a fractal created by dissecting and discarding two sub-diamonds:

Diamond fractal (retaining sub-diamonds 1,2,3)


123-Diamond (anim)


Again, the fractal has variants created by mirroring one or more of the sub-diamonds:

123-Diamond (variant #1)


123-Diamond (variant #2)


123-Diamond (variant #3)


123-Diamond (variant #4)


Some more fractals created by dissecting and discarding two sub-diamonds:

125-Diamond


125-Diamond (anim)


134-Diamond


134-Diamond (anim)


235-Diamond


235-Diamond (anim)


135-Diamond


135-Diamond (anim)


A variant of the 135-Diamond fractal looks like one side of a Koch snowflake:

135-Diamond (variant #1) — like Koch snowflake


135-Diamond (variant #2)


Finally, here are some colour variants of the full dissected diamond:






Full diamond colour variants (anim)


Elsewhere other-engageable:

Dilating the Delta

Total Score

The number 23 is always (and trivially) equal to some running total of the digits of its roots in base 2. In other bases, that’s not always true (n.b. numbers inside square brackets represent single digits in that base):

√23 = 23^(1/2) = 100.1100101110111011100111010101110111000001000... in base 2
23 = digsum(100.110010111011101110011101010111011)
23^(1/2) = 11.21011101110011111122022101121121... in base 3
23 = digsum(11.2101110111001111112202)
23^(1/2) = 4.8832850[10]89028... in base 11
23 = digsum(4.883)
23^(1/2) = 4.[14]5[15]53[14]0[12]0[14]5[13]... in base 18
23 = digsum(4.[14]5)
23^(1/2) = 4.[19]29[13][19]4[11][23][19][11][20]... in base 24
23 = digsum(4.[19])
23^(1/2) = 4.[19][22]9[21][17]5[12][10]456... in base 25
23 = digsum(4.[19])

23^(1/3) = 10.11011000000001111010101010011000101000110000001100000010010000101011... in base 2
23 = digsum(10.1101100000000111101010101001100010100011000000110000001001)
23^(1/3) = 2.21121001121111121022212100220... in base 3
23 = digsum(2.2112100112111112102)
23^(1/3) = 2.312000132222212022030003... in base 4
23 = digsum(2.31200013222221)
23^(1/3) = 2.6600365246121403... in base 8
23 = digsum(2.660036)
23^(1/3) = 2.753154453877080... in base 9
23 = digsum(2.75315)
23^(1/3) = 2.93120691571[10]001[10]... in base 11
23 = digsum(2.931206)
23^(1/3) = 2.[12]9[13]0[11]74[11]61[14]2... in base 15
23 = digsum(2.[12]9)
23^(1/3) = 2.[13]807[10][10]98[10]303... in base 16
23 = digsum(2.[13]8)
23^(1/3) = 2.[21]2[10][10][13][11][21][23][15][24][21]... in base 25
23 = digsum(2.[21])
23^(1/3) = 2.[21][24][11][20][24][22][23][25]0[11][11]... in base 26
23 = digsum(2.[21])

23^(1/4) = 10.0011000010011111110100101010011000001001011110001110101... in base 2
23 = digsum(10.001100001001111111010010101001100000100101111)
23^(1/4) = 2.1411772251404570... in base 8
23 = digsum(2.141177)
23^(1/4) = 2.1634161832077814... in base 9
23 = digsum(2.163416)
23^(1/4) = 2.33[15]2[14][13]967[10]6[12]5... in base 17
23 = digsum(2.33[15])
23^(1/4) = 2.6[15][19][11][31][17][10][18][21]30[27]... in base 34
23 = digsum(2.6[15])
23^(1/4) = 2.[12]9[63][18][41][32][37][56][58][60]1[17]... in base 64
23 = digsum(2.[12]9)
23^(1/4) = 2.[21]9[26]6[54][21][20]3[64][86][110]... in base 111
23 = digsum(2.[21])
23^(1/4) = 2.[21][30][66][22][73][19]3[15][51][24]8... in base 112
23 = digsum(2.[21])
23^(1/4) = 2.[21][52][36][111][32][104][66][40][95][33]5... in base 113
23 = digsum(2.[21])
23^(1/4) = 2.[21][74][50][62][27]19[100][70][48][89]... in base 114
23 = digsum(2.[21])
23^(1/4) = 2.[21][96][108]2[101][62][43][18][71][113][37]... in base 115
23 = digsum(2.[21])

23^(1/5) = 1.110111110100011010011101000111111011111011000... in base 2
23 = digsum(1.11011111010001101001110100011111101)
23^(1/5) = 1.313310122131013323323010... in base 4
23 = digsum(1.31331012213101)
23^(1/5) = 1.[10]5714140[10][11][11]61... in base 12
23 = digsum(1.[10]57)
23^(1/5) = 1.[11]45210[12]3974[12]0[11]... in base 13
23 = digsum(1.[11]452)
23^(1/5) = 1.[22][17][15]788[12][20][10][16]5... in base 26
23 = digsum(1.[22])

And in base 10:

23^(1/7) = 1.565065607960239...
23 = digsum(1.56506)

23^(1/11) = 1.32982177397055...
23 = digsum(1.3298)

23^(1/25) = 1.133624213096260543...
23 = digsum(1.13362421)

23^(1/43) = 1.075642836327515...
23 = digsum(1.07564)

23^(1/51) = 1.0634095245502272...
23 = digsum(1.063409)

23^(1/59) = 1.054581462032154...
23 = digsum(1.05458)

23^(1/74) = 1.043282031364111825...
23 = digsum(1.04328203)

23^(1/78) = 1.041017545329593513...
23 = digsum(1.04101754)

23^(1/81) = 1.039468791371841...
23 = digsum(1.03946)

23^(1/85) = 1.037576979258809...
23 = digsum(1.03757)

23^(1/86) = 1.0371320245405187874...
23 = digsum(1.037132024)

23^(1/101) = 1.031531403111493041428...
23 = digsum(1.03153140311)