Power Trap

Back in 2015, in an article called “Power Trip”, I looked at an unfamiliar sequence created by deleting zeroes from a familiar sequence. And I made a serious but fortunately-not-fatal error in my reasoning. The familiar sequence was powers of 2:

• 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 262144, 524288, 1048576…

This is what happens when you delete the zeroes from the powers of 2 (and carry on multiplying by two):

2 * 2 = 4
4 * 2 = 8
8 * 2 = 16
16 * 2 = 32
32 * 2 = 64
64 * 2 = 128
128 * 2 = 256
256 * 2 = 512
512 * 2 = 1024 → 124
124 * 2 = 248
248 * 2 = 496
496 * 2 = 992
992 * 2 = 1984
1984 * 2 = 3968
3968 * 2 = 7936
7936 * 2 = 15872
15872 * 2 = 31744
31744 * 2 = 63488
63488 * 2 = 126976
126976 * 2 = 253952
253952 * 2 = 507904 → 5794
5794 * 2 = 11588
11588 * 2 = 23176
23176 * 2 = 46352
46352 * 2 = 92704 → 9274…


I pointed out that this new sequence has to repeat, because deleting zeroes prevents n growing beyond a certain size. Eventually, then, a number will repeat and the sequence will fall into a loop: “This happens at step 526 with 366784, which matches 366784 at step 490.”

But that’s deleting every zero. What happens if you delete every second zero? That is, you start with a zero-count, zc, of 0. When you meet the first zero in the sequence, zc = zc + 1 = 1. When you meet the second zero in the sequence, zc = zc + 1 = 2. So you delete that second zero and reset zc to 0. The first zero occurs when 1024 = 2 * 512, so 1024 is left as it is. The second zero occurs when 2 * 1024 = 2048, so 2048 becomes 248. The sequence for zc=2 looks like this:

1 * 2 = 2
2 * 2 = 4
4 * 2 = 8
8 * 2 = 16
16 * 2 = 32
32 * 2 = 64
64 * 2 = 128
128 * 2 = 256
256 * 2 = 512
512 * 2 = 1024 → 1024
1024 * 2 = 2048 → 248
248 * 2 = 496
496 * 2 = 992
992 * 2 = 1984
1984 * 2 = 3968
3968 * 2 = 7936
7936 * 2 = 15872
15872 * 2 = 31744
31744 * 2 = 63488
63488 * 2 = 126976
126976 * 2 = 253952
253952 * 2 = 507904 → 50794
50794 * 2 = 101588 → 101588
101588 * 2 = 203176 → 23176
23176 * 2 = 46352
46352 * 2 = 92704 → 92704
92704 * 2 = 185408 → 18548

Again, the sequence has to repeat and I claimed that it did so “at step 9134 with 5458864, which matches 5458864 at step 4166”. I also said that I hadn’t found the loop for the delete-every-third-zero sequence, where zc=3. Coming back to this type of sequence in 2021, I wrote a much faster machine-code program to see if I could find the answer for zc=3. And I thought that I had. My program said that the sequence for zc=3 repeats at step 166369 with 6138486272, which matches 6138486272 at step 25429.

Or does it repeat? Does it match? In 2021 I suddenly realized that I had neglected to consider something vital back in 2015: whether the zero-count was the same when the sequence appeared to repeat. Take the zc=2 sequence. If zc=0 at at step 4166 and zc=1 at 9134 (or vice versa), the sequence isn’t in a loop, because it will be deleting a different set of zeroes after step 4166 than it is after step 9134.

I checked whether the zero-count for that sequence is the same when the sequence appears to repeat. Fortunately, it is the same and the zero-delete sequence for zc=2 does indeed begin looping “at step 9134 with 5458864, which matches 5458864 at step 4166”.

So my error wasn’t fatal for the zc=2 sequence. But what about the zc=3 sequence? Alas, the zero-count is different for 6138486272 at step 166369 than for 6138486272 at step 25429. The sequence doesn’t behave the same after those steps and hasn’t looped. I needed to find the n1 = n2 for steps s1 and s2 where zc1 = zc2. And even with the much faster machine-code program it took some time. But I can now say that 958718377984 at step 379046, with zc=0, matches 958718377984 at step 200906, with zc=0.

El Sabor de Salvador

“Más es menos” is Spanish for “more is less”. And you can certainly see “más es menos” at work in the paintings of that greatest of Spaniards, Salvador Dalí. The more technically skilled and detailed his art became, the less powerful and interesting it was. Compare Sleep, from 1937…

Le Sommeil (Sleep), 1937

…with Still Life — Fast Moving from 1956:

Still Life — Fast Moving,1956

Toxic Turntable #23

Currently listening…

• Transylv Nexus, Vamplifier (1996)
• Jotmu Bkhu, We Stay Zipped (Songs for the Carnival) (1999)
• Ranfha, Deep to Deep (1979)
• Nade Famborne, Odū Pkeem x’Siqa (1985)
• Adrienne Prunier, Pour la Déesse (1982)
• Yoagoįh, Rhythmic Jellifications (1993)
• Caedicore, As Weird Is Null (1999)
• XS-Doz, Texanized (1985)
• Epics in the Underworld, Khviu (2012)
• Todt-89, Numina (LXVII) (2014)
• Ussia, My Kayak (Live Mixes) (1992)
• Ekkokoz, Qualis Tu Es (1997)
• Yoke of Cud, Red Leap (Led Reap) (1990)
• Fixenhoff, Swedish Amiff (1994)
• Aiyhor, Ihqxelyy-043478 (2006)
• Caiunic, HYH (1988)
• Uz R Under, Deborah the Henge (1983)
• Loftmaft, Horse for the Silent Shore (1996)
• Futility in Mexborough, Axolotl Dreams (2003)
• Sleek Boutique, Canopy Collapse (2020)
• Mmjojg Siki, A Height to Savor (1983)
• Franz Anton Hoffmeister, Viola Concerto in D major (1949)
• Froschkönig Gabriel, Aros Dillidia (1995)


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

Un Paon Papyrocentrique

Le Paon dans les Pyrénées — a review at Papyrocentric Performativity of Julian Barnes’ The Man in the Red Coat (2019), which contains a lot about Robert de Montesquiou


Elsewhere other-accessible

Portrait of a Peacock — Cornelia Otis Skinner’s biographical sketch of Montesquiou

Ciss Bliss

Si hortum in bibliotheca habes, deerit nihil. – Cicero (106-43 BC), Epistulae ad Familiares, Liber IX, Epistula IV

• “If you have a garden and a library, you lack for nothing.” — Cicero, Letters to Friends, Book 9, Letter 4

Spiral Artefact

What’s the next number in this sequence of integers?


5, 14, 19, 23, 28, 32, 37, 41, 46, 50, 55... (A227793 at the OEIS)

It shouldn’t be hard to work out that it’s 64 — the sum-of-digits of n is divisible by 5, i.e., digsum(n) mod 5 = 0. Now try summing the numbers in that sequence:


5 + 14 = 19
19 + 19 = 38
38 + 23 = 61
61 + 28 = 89
89 + 32 = 121
121 + 37 = 158
158 + 41 = 199
199 + 46 = 245
[...]

Here are the cumulative sums as another sequence:


5, 19, 38, 61, 89, 121, 158, 199, 245, 295, 350, 414, 483, 556, 634, 716, 803, 894, 990, 1094, 1203, 1316, 1434, 1556, 1683, 1814, 1950, 2090, 2235, 2389, 2548, 2711, 2879, 3051, 3228, 3409, 3595, 3785, 3980, 4183, 4391, 4603, 4820, 5041, 5267, 5497, 5732, 5976, 6225...

And there’s that cumulative-sum sequence represented as a spiral:

Spiral for cumulative sum of n where digsum(n) mod 5 = 0


You can see how the spiral is created by following 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E… from the center:


ZYXWVU
GFEDCT
H432BS
I501AR
J6789Q
KLMNOP

What about other values for the cumulative sums of digsum(n) mod m = 0? Here’s m = 2,3,4,5,6,7:

Spiral for cumulative sum of n where digsum(n) mod 2 = 0
s1 = 2, 4, 6, 8, 11, 13, 15, 17, 19, 20, 22…
s2 = 2, 6, 12, 20, 31, 44, 59, 76, 95, 115… (cumulative sum of s1)


sum of digsum(n) mod 3 = 0
s1 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33…
s2 = 3, 9, 18, 30, 45, 63, 84, 108, 135, 165…


sum of digsum(n) mod 4 = 0
s1 = 4, 8, 13, 17, 22, 26, 31, 35, 39, 40, 44…
s2 = 4, 12, 25, 42, 64, 90, 121, 156, 195, 235…


sum of digsum(n) mod 5 = 0
s1 = 5, 14, 19, 23, 28, 32, 37, 41, 46, 50, 55…
s2 = 5, 19, 38, 61, 89, 121, 158, 199, 245, 295…


sum of digsum(n) mod 6 = 0
s1 = 6, 15, 24, 33, 39, 42, 48, 51, 57, 60, 66…
s2 = 6, 21, 45, 78, 117, 159, 207, 258, 315, 375…


sum of digsum(n) mod 7 = 0
s1 = 7, 16, 25, 34, 43, 52, 59, 61, 68, 70, 77…
s2 = 7, 23, 48, 82, 125, 177, 236, 297, 365, 435…


The spiral for m = 2 is strange, but the spirals are similar after that. Until m = 8, when something strange happens again:

sum of digsum(n) mod 8 = 0
s1 = 8, 17, 26, 35, 44, 53, 62, 71, 79, 80, 88…
s2 = 8, 25, 51, 86, 130, 183, 245, 316, 395, 475…


Then the spirals return to normal for m = 9, 10:

sum of digsum(n) mod 9 = 0
s1 = 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99…
s2 = 9, 27, 54, 90, 135, 189, 252, 324, 405, 495…


sum of digsum(n) mod 10 = 0
s1 = 19, 28, 37, 46, 55, 64, 73, 82, 91, 109, 118…
s2 = 19, 47, 84, 130, 185, 249, 322, 404, 495, 604…


Here’s an animated gif of m = 8 at higher and higher resolution:

sum of digsum(n) mod 8 = 0 (animated gif)


You might think this strange behavior is dependant on the base in which the dig-sum is calculated. It isn’t. Here’s an animated gif for other bases in which the mod-8 spiral behaves strangely:

sum of digsum(n) mod 8 = 0 in base b = 5, 6, 7, 9, 11, 12, 13 (animated gif)


But the mod-8 spiral stops behaving strangely when the spiral is like this, as a diamond:


   W
  XIV
 YJ8HU
ZK927GT
LA3016FS
 MB45ER
  NCDQ
   OP

Now the mod-8 spiral looks like this:

sum of digsum(n) mod 8 = 0 (diamond spiral)


But the mod-4 and mod-9 spirals look like this:

sum of digsum(n) mod 4 = 0 (diamond spiral)


sum of digsum(n) mod 9 = 0 (diamond spiral)


You can also construct the spirals as a triangle, like this:


     U
    VCT
   WD2CS
  XE301AR
 YF456789Q
ZGHIJKLMNOP

Here’s the beginning of the mod-5 triangular spiral:

sum of digsum(n) mod 5 = 0 (triangular spiral) (open in new window for full size)


And the beginning of the mod-8 triangular spiral:

sum of digsum(n) mod 8 = 0 (triangular spiral) (open in new window for full size)


The mod-8 spiral is behaving strangely again. So the strangeness is partly an artefact of the way the spirals are constructed.


Post-Performative Post-Scriptum

“Spiral Artefact”, the title of this incendiary intervention, is of course a tip-of-the-hat to core Black-Sabbath track “Spiral Architect”, off core Black-Sabbath album Sabbath Bloody Sabbath, issued in core Black-Sabbath success-period of 1973.

Terminal Transgressivity

“If this work is about hell,” he says, “it’s not only about hell in terms of content. It’s also about hell in terms of its hellishness in terms of production.” — maximally maverick artist Jake Chapman describes how he and his brother Dinos made the transgressive sculpture Hell (2000), as quoted in Simon Garfield’s In Miniature: How Small Things Illuminate the World (2018)


Elsewhere Other-Accessible

Ex-term-in-nate! — incendiarily interrogating issues around “in terms of”…
All O.o.t.Ü.-F. posts interrogating issues around “in terms of”…


Peri-Performative Post-Scriptum…

Yes, this was an über-ideal quote for posting on the 23rd in terms of the month… But I was so taken with it that I couldn’t delay any longer. And anyway: it is the 23rd of the months in base 11. (I.e., 2111 = 2 * 11 + 1 = 22 + 1 = 23.)

For Flake’s Sake

It caught my eye, it caught my eye,
That fluttering flake of fallen sky.

It rode the wind as cars bored by
And did not die:

And shall not die,
That fluttering flake of fallen sky.


Post-Performative Post-Scriptum

A poem written months ago about a briefly glimpsed blue butterfly flying along — and over — a busy road. I don’t know the species, but Polyommatus icarus seems a reasonable guess.

Octobyss

The deep-sea octopus Vulcanoctopus hydrothermalis, which lives around hydrothermal vents on the floor of the Pacific (image from Octolab)


Elsewhere Other-Engageable

Guise and Molls — review of Octopus: The Ocean’s Intelligent Invertebrate: A Natural History (2010)
Magna Mater Marina — review of The Illustrated World Encyclopedia of Marine Fish and Sea Creatures (2007)

There are 719 errors in this sentence

Here’s a famous paradox (or a variant of it at least):

• There are two errers in this sentence.

The only visible error is the misspelt “errers”. But if the sentence claims to have two errors while having only one, that is another error and there are two errors after all.

Now for another variant. I’m not sure if I’ve thought this up for myself, but try this sentence:

• There are three errors in this sentence.

There are no visible errors in the sentence. Therefore it has one error: the claim that it has three errors when there is in fact no error. But if it has one error, it’s in error to claim that it has three errors. Therefore the sentence has two errors. And if it has two errors, again it’s in error, because it claims to have three errors while having only two. Therefore it has three errors after all.

The same reasoning can be applied to any integral number of errors:

• There are five errors in this sentence.
• There are 719 errors in this sentence.
• There are 1,000,000 errors in this sentence.
• There are 1,000,000,000 errors in this sentence.

No matter how large the number of errors, the sentence becomes true instantly, because each time the sentence makes a false claim, it makes another error. But those “times of error” don’t take place in time, any more than this equation does:

• 2 = 1 + 1/2 + 1/4 + 1/8 + 1/16…

So I think these sentences are instantly true:

• There are infinitely many errors in this sentence.
• There are ∞ errors in this sentence.

But there are infinitely many infinities. Ordinary infinity, the infinity of 1,2,3…, is called ℵ0 or aleph-zero. It’s a countable infinity. Above that comes ℵ1, an uncountable infinity. So does this sentence instantly become true?

• There are ℵ1 errors in this sentence.

I’m not sure. But I think I can argue for the validity of sentences claiming fractional or irrational number of errors:

• There is 1.5 errors in this sentence.
• There are π errors in this sentence.

Let’s have a look at “There is 1.5 errors in this sentence”. There are no visible errors, so there’s one error: the claim that sentence contains 1.5 errors. So now there seems to be another error: the sentence has one error but claims to have 1.5 errors. But does it therefore have two errors? No, because if it has two errors, it’s still in error and has three errors. And that generates another error and another and another, and so on for ever. The sentence becomes unstoppably and infinitely false.

So let’s go back to the point at which the sentence contains one error. Now, the difference between 1 error and 1.5 errors is small — less than a full error. So how big is the error of claiming to have 1.5 errors when having 1 error? Well, it’s obviously 0.5 of an error. So the sentence contains 1.5 errors after all.

Now for “There are π errors in this sentence”. There are no visible errors, so there’s one error: the claim that the sentence contains π errors. Therefore it contains one error. But it claims to have π errors, so it has another error. And if it has 2 errors and claims to have π errors, it has another and third error. But if it has three errors and claims to have π error, it’s still in error. But only slightly — it’s now committing a small amount of an error. How much? It can only be 0.14159265… of an error. Therefore it’s committing 3.14159265… = π errors and is a true sentence.

Now try:

• There is -1 error in this sentence.

What is a negative error? A truth. So I think that sentence is valid too. But I can’t think of how to use i, or the square root of -1, in a sentence like that.