Scaffscapes

A fractal is a shape that contains copies of itself on smaller and smaller scales. You can find fractals everywhere in nature. Part of a fern looks like the fern as a whole:

Fern as fractal (source)


Part of a tree looks like the tree as a whole:

Tree as fractal (source)


Part of a landscape looks like the landscape as a whole:

Landscape as fractal (source)


You can also create fractals for yourself. Here are three that I’ve constructed:

Fractal #1


Fractal #2


Fractal #3 — the T-square fractal


The three fractals look very different and, in one sense, that’s exactly what they are. But in another sense, they’re the same fractal. Each can morph into the other two:

Fractal #1 → fractal #2 → fractal #3 (animated)


Here are two more fractals taken en route from fractal #2 to fractal #3, as it were:

Fractal #4


Fractal #5


To understand how the fractals belong together, you have to see what might be called the scaffolding. The construction of fractal #3 is the easiest to understand. First you put up the scaffolding, then you take it away and leave the final fractal:

Fractal #3, scaffolding stage 1


Fractal #3, stage 2


Fractal #3, stage 3


Fractal #3, stage 4


Fractal #3, stage 5


Fractal #3, stage 6


Fractal #3, stage 7


Fractal #3, stage 8


Fractal #3, stage 9


Fractal #3, stage 10


Fractal #3 (scaffolding removed)


Construction of fractal #3 (animated)


Now here’s the construction of fractal #1:

Fractal #1, stage 1


Fractal #1, stage 2


Fractal #1, stage 3

↓

Construction of fractal #1 (animated)


Fractal #1 (static)


And the constructions of fractals #2, #4 and #5:

Fractal #2, stage 1


Fractal #2, stage 2


Fractal #2, stage 3

↓

Fractal #2 (animated)


Fractal #2 (static)


Fractal #4, stage 1


Fractal #4, stage 2


Fractal #4, stage 3

↓

Fractal #4 (animated)


Fractal #4 (static)


Fractal #5, stage 1


Fractal #5, stage 2


Fractal #5, stage 3

↓

Fractal #5 (animated)


Fractal #5


Root Pursuit

Roots are hard, powers are easy. For example, the square root of 2, or √2, is the mysterious and never-ending number that is equal to 2 when multiplied by itself:

• √2 = 1·414213562373095048801688724209698078569671875376948073...

It’s hard to calculate √2. But the powers of 2, or 2^p, are the straightforward numbers that you get by multiplying 2 repeatedly by itself. It’s easy to calculate 2^p:

• 2 = 2^1
• 4 = 2^2
• 8 = 2^3
• 16 = 2^4
• 32 = 2^5
• 64 = 2^6
• 128 = 2^7
• 256 = 2^8
• 512 = 2^9
• 1024 = 2^10
• 2048 = 2^11
• 4096 = 2^12
• 8192 = 2^13
• 16384 = 2^14
• 32768 = 2^15
• 65536 = 2^16
• 131072 = 2^17
• 262144 = 2^18
• 524288 = 2^19
• 1048576 = 2^20
[...]

But there is a way to find √2 by finding 2^p, as I discovered after I asked a simple question about 2^p and 3^p. What are the longest runs of matching digits at the beginning of each power?

• 131072 = 2^17
• 129140163 = 3^17
• 1255420347077336152767157884641... = 2^193
• 1214512980685298442335534165687... = 3^193
• 2175541218577478036232553294038... = 2^619
• 2177993962169082260270654106078... = 3^619
• 7524389324549354450012295667238... = 2^2016
• 7524012611682575322123383229826... = 3^2016

There’s no obvious pattern. Then I asked the same question about 2^p and 5^p. And an interesting pattern appeared:

• 32 = 2^5
• 3125 = 5^5
• 316912650057057350374175801344 = 2^98
• 3155443620884047221646914261131... = 5^98
• 3162535207926728411757739792483... = 2^1068
• 3162020133383977882730040274356... = 5^1068
• 3162266908803418110961625404267... = 2^127185
• 3162288411569894029343799063611... = 5^127185

The digits 31622 rang a bell. Isn’t that the start of √10? Yes, it is:

• √10 = 3·1622776601683793319988935444327185337195551393252168268575...

I wrote a fast machine-code program to find even longer runs of matching initial digits. Sure enough, the pattern continued:

• 316227... = 2^2728361
• 316227... = 5^2728361
• 3162277... = 2^15917834
• 3162277... = 5^15917834
• 31622776... = 2^73482154
• 31622776... = 5^73482154
• 3162277660... = 2^961700165
• 3162277660... = 5^961700165

But why are powers of 2 and 5 generating the digits of √10? If you’re good at math, that’s a trivial question about a trivial discovery. Here’s the answer: We use base ten and 10 = 2 * 5, 10^2 = 100 = 2^2 * 5^2 = 4 * 25, 10^3 = 1000 = 2^3 * 5^3 = 8 * 125, and so on. When the initial digits of 2^p and 5^p match, those matching digits must come from the digits of √10. Otherwise the product of 2^p * 5^p would be too large or too small. Here are the records for matching initial digits multiplied by themselves:

• 32 = 2^5
• 3125 = 5^5
• 3^2 = 9

• 316912650057057350374175801344 = 2^98
• 3155443620884047221646914261131... = 5^98
• 31^2 = 961

• 3162535207926728411757739792483... = 2^1068
• 3162020133383977882730040274356... = 5^1068
• 3162^2 = 9998244

• 3162266908803418110961625404267... = 2^127185
• 3162288411569894029343799063611... = 5^127185
• 31622^2 = 999950884

• 316227... = 2^2728361
• 316227... = 5^2728361
• 316227^2 = 99999515529

• 3162277... = 2^15917834
• 3162277... = 5^15917834
• 3162277^2 = 9999995824729

• 31622776... = 2^73482154
• 31622776... = 5^73482154
• 31622776^2 = 999999961946176

• 3162277660... = 2^961700165
• 3162277660... = 5^961700165
• 3162277660^2 = 9999999998935075600

The square of each matching run falls short of 10^p. And so when the digits of 2^p and 5^p stop matching, one power must fall below √10, as it were, and one must rise above:

• 3 162266908803418110961625404267... = 2^127185
• 3·162277660168379331998893544432... = √10
• 3 162288411569894029343799063611... = 5^127185

In this way, 2^p * 5^p = 10^p. And that’s why matching initial digits of 2^p and 5^p generate the digits of √10. The same thing, mutatis mutandis, happens in base 6 with 2^p and 3^p, because 6 = 2 * 3:

• 2.24103122055214532500432040411... = √6 (in base 6)

• 24 = 2^4
• 213 = 3^4
• 225522024 = 2^34 in base 6 = 2^22 in base 10
• 22225525003213 = 3^34 (3^22)
• 2241525132535231233233555114533... = 2^1303 (2^327)
• 2240133444421105112410441102423... = 3^1303 (3^327)
• 2241055222343212030022044325420... = 2^153251 (2^15007)
• 2241003215453455515322105001310... = 3^153251 (3^15007)
• 2241032233315203525544525150530... = 2^233204 (2^20164)
• 2241030204225410320250422435321... = 3^233204 (3^20164)
• 2241031334114245140003252435303... = 2^2110415 (2^102539)
• 2241031103430053425141014505442... = 3^2110415 (3^102539)

And in base 30, where 30 = 2 * 3 * 5, you can find the digits of √30 in three different ways, because 30 = 2 * 15 = 3 * 10 = 5 * 6:

• 5·E9F2LE6BBPBF0F52B7385PE6E5CLN... = √30 (in base 30)

• 55AA4 = 2^M in base 30 = 2^22 in base 10
• 5NO6CQN69C3Q0E1Q7F = F^M = 15^22
• 5E63NMOAO4JPQD6996F3HPLIMLIRL6F... = 2^K6 (2^606)
• 5ECQDMIOCIAIR0DGJ4O4H8EN10AQ2GR... = F^K6 (15^606)
• 5E9DTE7BO41HIQDDO0NB1MFNEE4QJRF... = 2^B14 (2^9934)
• 5E9G5SL7KBNKFLKSG89J9J9NT17KHHO... = F^B14 (15^9934)
[...]
• 5R4C9 = 3^E in base 30 = 3^14 in base 10
• 52CE6A3L3A = A^E = 10^14
• 5E6SOQE5II5A8IRCH9HFBGO7835KL8A = 3^3N (3^113)
• 5EC1BLQHNJLTGD00SLBEDQ73AH465E3... = A^3N (10^113)
• 5E9FI455MQI4KOJM0HSBP3GG6OL9T8P... = 3^EJH (3^13187)
• 5E9EH8N8D9TR1AH48MT7OR3MHAGFNFQ... = A^EJH (10^13187)
[...]
• 5OCNCNRAP = 5^I in base 30 = 5^18 in base 10
• 54NO22GI76 = 6^I (6^18)
• 5EG4RAMD1IGGHQ8QS2QR0S0EH09DK16... = 5^1M7 (5^1567)
• 5E2PG4Q2G63DOBIJ54E4O035Q9TEJGH... = 6^1M7 (6^1567)
• 5E96DB9T6TBIM1FCCK8A8J7IDRCTM71... = 5^F9G (5^13786)
• 5E9NM222PN9Q9TEFTJ94261NRBB8FCH... = 6^F9G (6^13786)
[...]

So that’s √10, √6 and √30. But I said at the beginning that you can find √2 by finding 2^p. How do you do that? By offsetting the powers, as it were. With 2^p and 5^p, you can find the digits of √10. With 2^(p+1) and 5^p, you can find the digits of √2 and √20, because 2^(p+1) * 5^p = 2 * 2^p * 5^p = 2 * 10^p:

•  √2 = 1·414213562373095048801688724209698078569671875376948073...
• √20 = 4·472135954999579392818347337462552470881236719223051448...

• 16 = 2^4
• 125 = 5^3
• 140737488355328 = 2^47
• 142108547152020037174224853515625 = 5^46
• 1413... = 2^243
• 1414... = 5^242
• 14141... = 2^6651
• 14142... = 5^6650
• 141421... = 2^35389
• 141420... = 5^35388
• 4472136... = 2^162574
• 4472135... = 5^162573
• 141421359... = 2^3216082
• 141421352... = 5^3216081
• 447213595... = 2^172530387
• 447213595... = 5^172530386
[...]

A Vulcar Display of Power


Post Performative Post-Scriptum

The title of this incendiary intervention is a paronomasia on Pantera’s Vulgar Display of Power (1992). I don’t like Pantera, but that’s a good title.

Performativizing Papyrocentricity #72

Papyrocentric Performativity Presents…

• Homing in the Gloaming – Homing: On Pigeons, Dwellings and Why We Return, Jon Day (John Murray 2019)

• Niceberg – The Storyteller: Tales of Life and Music, Dave Grohl (Simon & Schuster 2021)

• Nasty Lastly – Nasty Endings 1, compiled by Dennis Pepper (Oxford University Press 2001)

• Daysed and Confused – Hawkwind: Days of the Underground: Radical Escapism in the Age of Paranoia, Joe Banks (Strange Attractor 2020)

• World-Wide Wipe-Out – Empty World, John Christopher (1977)

• Chuck Off – Post Office, Charles Bukowski (1971)

• #AllDayDong – Dong, Peter Sotos and Sam Salatta (TransVisceral Books 2022)

• Meet the Maverick Munch-Bunch… – Naked Krunch: The Sinister, Sordid and Strangely Scrumptious Story of SavSnaq, Dr David M. Mitchell (Savoy Books 2022)


Or Read a Review at Random: RaRaR

Les Mains de Montesquiou

Comte Robert de Montesquiou-Fézensac, Henri Lucien Doucet (1879)


Passionately Pre-Posted…

• Portrait of a Peacock — Cornelia Otis Skinner’s biographical sketch of Montesquiou
• Le Paon dans les Pyrénées — review of Julian Barnes’ The Man in the Red Coat (2019)

Elsewhere Other-Accessible…

• The hands of Robert de Montesquiou at Strange Flowers

Toxic Turntable #25

Currently listening…

• Lupine Katie, Dark Calico (1976)
• Ô Grecq, Mouse Makes Cop (2005)
• Julia Sedburgh, Vacant Reef (1975)
• Fox in Quake, Drifting on Sonic Seas (2002)
• الاصداف, ميوريكس ﮪاوستلم (1995)
• Pidita Kick, I Leap for Hawaii (1998)
• Yuperfen Lagekim, Jeptic (2013)
• Quiggy and the Grin, Soda Family (1983)
• Oxton Bvulpsi, Dealt a Devil’s Hand (2017)
• მარმარილი მანქანა, აბრეშუმის ჭია (1997)
• Dogmatic Cajuns, Little Bit (1996)
• Pamela Kelpstan, Episodic Outtakes (1970)
• Okeäp V, Räucherfisch (1994)
• Bells in the Battery, Hue You (1990)
• Showt Nowt, Our Wild Folkloric Dreams (2012)
• ¡XtrmDrm!, LIVE in ¡Thunderland! (2004)
• Jumjin, Mqoag i Hio (1979)
• Ѫй Ѯяоѱ Ѩѽ, Koxiio (1997)
• Edgar Roslin Quintet, Zeppelin Zoo (2017)
• Uh Bpuerw, Sqiaamn (1983)
• Stellae Vulpis, Umbrian Folk Revival (2006)
• Milirad im Untergang, Ab’xu Mim Myaceleg (1973)


Previously pre-posted:

Toxic Turntable #1 • #2 • #3 • #4 • #5 • #6 • #7 • #8 • #9 • #10 • #11 • #12 • #13 • #14 • #15 • #16 • #17 • #18 • #19 • #20 • #21 • #22 • #23 • #24 •

God Give Me Benf’

In “Wake the Snake”, I looked at the digits of powers of 2 and mentioned a fascinating mathematical phenomenon known as Benford’s law, which governs — in a not-yet-fully-explained way — the leading digits of a wide variety of natural and human statistics, from the lengths of rivers to the votes cast in elections. Benford’s law also governs a lot of mathematical data. It states, for example, that the first digit, d, of a power of 2 in base b (except b = 2, 4, 8, 16…) will occur with the frequency logb(1 + 1/d). In base 10, therefore, Benford’s law states that the digits 1..9 will occur with the following frequencies at the beginning of 2^p:

1: 30.102999%
2: 17.609125%
3: 12.493873%
4: 09.691001%
5: 07.918124%
6: 06.694678%
7: 05.799194%
8: 05.115252%
9: 04.575749%

Here’s a graph of the actual relative frequencies of 1..9 as the leading digit of 2^p (open images in a new window if they appear distorted):


And here’s a graph for the predicted frequencies of 1..9 as the leading digit of 2^p, as calculated by the log(1+1/d) of Benford’s law:


The two graphs agree very well. But Benford’s law applies to more than one leading digit. Here are actual and predicted graphs for the first two leading digits of 2^p, 10..99:



And actual and predicted graphs for the first three leading digits of 2^p, 100..999:



But you can represent the leading digit of 2^p in another way: using an adaptation of the famous Ulam spiral. Suppose powers of 2 are represented as a spiral of squares that begins like this, with 2^0 in the center, 2^1 to the right of center, 2^2 above 2^1, and so on:

←←←⮲
432↑
501↑
6789

If the digits of 2^p start with 1, fill the square in question; if the digits of 2^p don’t start with 1, leave the square empty. When you do this, you get this interesting pattern (the purple square at the very center represents 2^0):

Ulam-like power-spiral for 2^p where 1 is the leading digit


Here’s a higher-resolution power-spiral for 1 as the leading digit:

Power-spiral for 2^p, leading-digit = 1 (higher resolution)


And here, at higher resolution still, are power-spirals for all the possible leading digits of 2^p, 1..9 (some spirals look very similar, so you have to compare those ones carefully):

Power-spiral for 2^p, leading-digit = 1 (very high resolution)


Power-spiral for 2^p, leading-digit = 2


Power-spiral for 2^p, ld = 3


Power-spiral for 2^p, ld = 4


Power-spiral for 2^p, ld = 5


Power-spiral for 2^p, ld = 6


Power-spiral for 2^p, ld = 7


Power-spiral for 2^p, ld = 8


Power-spiral for 2^p, ld = 9


Power-spiral for 2^p, ld = 1..9 (animated)


Now try the power-spiral of 2^p, ld = 1, in some other bases:

Power-spiral for 2^p, leading-digit = 1, base = 9


Power-spiral for 2^p, ld = 1, b = 15


You can also try power-spirals for other n^p. Here’s 3^p:

Power-spiral for 3^p, ld = 1, b = 10


Power-spiral for 3^p, ld = 2, b = 10


Power-spiral for 3^p, ld = 1, b = 4


Power-spiral for 3^p, ld = 1, b = 7


Power-spiral for 3^p, ld = 1, b = 18


Elsewhere Other-Accessible…

• Wake the Snake — an earlier look at the digits of 2^p

Wake the Snake

In my story “Kopfwurmkundalini”, I imagined the square root of 2 as an infinitely long worm or snake whose endlessly varying digit-segments contained all stories ever (and never) written:

• √2 = 1·414213562373095048801688724209698078569671875376948073…

But there’s another way to get all stories ever written from the number 2. You don’t look at the root(s) of 2, but at the powers of 2:

• 2 = 2^1 = 2
• 4 = 2^2 = 2*2
• 8 = 2^3 = 2*2*2
• 16 = 2^4 = 2*2*2*2
• 32 = 2^5 = 2*2*2*2*2
• 64 = 2^6 = 2*2*2*2*2*2
• 128 = 2^7 = 2*2*2*2*2*2*2
• 256 = 2^8 = 2*2*2*2*2*2*2*2
• 512 = 2^9 = 2*2*2*2*2*2*2*2*2
• 1024 = 2^10
• 2048 = 2^11
• 4096 = 2^12
• 8192 = 2^13
• 16384 = 2^14
• 32768 = 2^15
• 65536 = 2^16
• 131072 = 2^17
• 262144 = 2^18
• 524288 = 2^19
• 1048576 = 2^20
• 2097152 = 2^21
• 4194304 = 2^22
• 8388608 = 2^23
• 16777216 = 2^24
• 33554432 = 2^25
• 67108864 = 2^26
• 134217728 = 2^27
• 268435456 = 2^28
• 536870912 = 2^29
• 1073741824 = 2^30
[...]

The powers of 2 are like an ever-lengthening snake swimming across a pool. The snake has an endlessly mutating head and a rhythmically waving tail with a regular but ever-more complex wake. That is, the leading digits of 2^p don’t repeat but the trailing digits do. Look at the single final digit of 2^p, for example:

• 02 = 2^1
• 04 = 2^2
• 08 = 2^3
• 16 = 2^4
• 32 = 2^5
• 64 = 2^6
• 128 = 2^7
• 256 = 2^8
• 512 = 2^9
• 1024 = 2^10
• 2048 = 2^11
• 4096 = 2^12
• 8192 = 2^13
• 16384 = 2^14
• 32768 = 2^15
• 65536 = 2^16
• 131072 = 2^17
• 262144 = 2^18
• 524288 = 2^19
• 1048576 = 2^20
• 2097152 = 2^21
• 4194304 = 2^22
[...]

The final digit of 2^p falls into a loop: 2 → 4 → 8 → 6 → 2 → 4→ 8…

Now try the final two digits of 2^p:

• 02 = 2^1
• 04 = 2^2
• 08 = 2^3
• 16 = 2^4
• 32 = 2^5
• 64 = 2^6
• 128 = 2^7
• 256 = 2^8
• 512 = 2^9
• 1024 = 2^10
• 2048 = 2^11
• 4096 = 2^12
• 8192 = 2^13
• 16384 = 2^14
• 32768 = 2^15
• 65536 = 2^16
• 131072 = 2^17
• 262144 = 2^18
• 524288 = 2^19
• 1048576 = 2^20
• 2097152 = 2^21
• 4194304 = 2^22
• 8388608 = 2^23
• 16777216 = 2^24
• 33554432 = 2^25
• 67108864 = 2^26
• 134217728 = 2^27
• 268435456 = 2^28
• 536870912 = 2^29
• 1073741824 = 2^30
[...]

Now there’s a longer loop: 02 → 04 → 08 → 16 → 32 → 64 → 28 → 56 → 12 → 24 → 48 → 96 → 92 → 84 → 68 → 36 → 72 → 44 → 88 → 76 → 52 → 04 → 08 → 16 → 32 → 64 → 28… Any number of trailing digits, 1 or 2 or one trillion, falls into a loop. It just takes longer as the number of trailing digits increases.

That’s the tail of the snake. At the other end, the head of the snake, the digits don’t fall into a loop (because of the carries from the lower digits). So, while you can get only 2, 4, 8 and 6 as the final digits of 2^p, you can get any digit but 0 as the first digit of 2^p. Indeed, I conjecture (but can’t prove) that not only will all integers eventually appear as the leading digits of 2^p, but they will do so infinitely often. Think of a number and it will appear as the leading digits of 2^p. Let’s try the numbers 1, 12, 123, 1234, 12345…:

• 16 = 2^4
• 128 = 2^7
• 12379400392853802748... = 2^90
• 12340799625835686853... = 2^1545
• 12345257952011458590... = 2^34555
• 12345695478410965346... = 2^63293
• 12345673811591269861... = 2^4869721
• 12345678260232358911... = 2^5194868
• 12345678999199154389... = 2^62759188

But what about the numbers 9, 98, 987, 986, 98765… as leading digits of 2^p? They don’t appear as quickly:

• 9007199254740992 = 2^53
• 98079714615416886934... = 2^186
• 98726397006685494828... = 2^1548
• 98768356967522174395... = 2^21257
• 98765563827287722773... = 2^63296
• 98765426081858871289... = 2^5194871
• 98765430693066680199... = 2^11627034
• 98765432584491513519... = 2^260855656
• 98765432109571471006... = 2^1641098748

Why do fragments of 123456789 appear much sooner than fragments of 987654321? Well, even though all integers occur infinitely often as leading digits of 2^p, some integers occur more often than others, as it were. The leading digits of 2^p are actually governed by a fascinating mathematical phenomenon known as Benford’s law, which states, for example, that the single first digit, d, will occur with the frequency log10(1 + 1/d). Here are the actual frequencies of 1..9 for all powers of 2 up to 2^101000, compared with the estimate by Benford’s law:

1: 30% of leading digits ↔ 30.1% estimated
2: 17.55% ↔ 17.6%
3: 12.45% ↔ 12.49%
4: 09.65% ↔ 9.69%
5: 07.89% ↔ 7.92%
6: 06.67% ↔ 6.69%
7: 05.77% ↔ 5.79%
8: 05.09% ↔ 5.11%
9: 04.56% ↔ 4.57%

Because (inter alia) 1 appears as the first digit of 2^p far more often than 9 does, the fragments of 123456789 appear faster than the fragments of 987654321. Mutatis mutandis, the same applies in all other bases (apart from bases that are powers of 2, where there’s a single leading digit, 1, 2, 4, 8…, followed by 0s). But although a number like 123456789 occurs much frequently than 987654321 in 2^p expressed in base 10 (and higher), both integers occur infinitely often.

As do all other integers. And because stories can be expressed as numbers, all stories ever (and never) written appear in the powers of 2. Infinitely often. You’ll just have to trim the tail of the story-snake.

I Say, I Sigh, I Sow #14

“In a very real sense, the Holocaust, as the ultimate moral and aesthetic obscenity, was also the ultimate drum-solo.” — Simon Whitechapel, 31i18