Russell in Your Head-Roe (Re-Visited)

“Ordinary language is totally unsuited for expressing what physics really asserts, since the words of everyday life are not sufficiently abstract. Only mathematics and mathematical logic can say as little as the physicist means to say.” — Bertrand Russell, The Scientific Outlook (1931)


Previously pre-posted

Russell in Your Head-Roe — Bertrand Russell on mathematics
A Ladd Inane — Bertrand Russell on solipsism
Math Matters — Bertrand Russell on math and physics
Whip Poor Wilhelm — Bertrand Russell on Friedrich Nietzsche

Farnsicht

Photo of developing ferns by the German nature photographer Karl Blossfeldt (1866-1932)
(open in new window for full image)


Post-Performative Post-Scriptum

“Farnsicht” is a pun on German Farn, meaning “fern”, and Fernsicht, meaning “view” or “visibility” (literally fern, “far”, + Sicht, “visibility”).

Z-Fall

Do you want a haunting literary image? You’ll find one of the strangest and strongest in Borges’ “La Biblioteca de Babel” (1941), which is narrated by a librarian in an infinite library. The librarian anticipates the end of his life:

Muerto, no faltarán manos piadosas que me tiren por la baranda; mi sepultura será el aire insondable; mi cuerpo se hundirá largamente y se corromperá y disolverá en el viento engenerado por la caída, que es infinita. — “La Biblioteca de Babel

When I am dead, compassionate hands will throw me over the railing; my tomb will be the unfathomable air, my body will sink for ages, and will decay and dissolve in the wind engendered by my fall, which shall be infinite. — “The Library of Babel” (translation by Andrew Hurley)

The infinite fall is the haunting image. Falling is powerful; falling for ever is more powerful still. But it can’t happen in reality: soon or later a fall has to end. Objects crash to earth or splash into the ocean. Of course, you could call being in orbit a kind of infinite fall, but it doesn’t have the same power.

However, there’s more kinds of falling than one and I think the arithmophile Borges would have liked one of the other kinds a lot. Numbers can fall — you sum their digits, take the sum from the original number, and repeat. That is, n = n – digsum(n). Here are some examples:


10 → 9 → 0
100 → 99 → 81 → 72 → 63 → 54 → 45 → 36 → 27 → 18 → 9 → 0
1000 → 999 → 972 → 954 → 936 → 918 → 900 → 891 → 873 → 855 → 837 → 819 → 801 → 792 → 774 → 756 → 738 → 720 → 711 → 702 → 693 → 675 → 657 → 639 → 621 → 612 → 603 → 594 → 576 → 558 → 540 → 531 → 522 → 513 → 504 → 495 → 477 → 459 → 441 → 432 → 423 → 414 → 405 → 396 → 378 → 360 → 351 → 342 → 333 → 324 → 315 → 306 → 297 → 279 → 261 → 252 → 243 → 234 → 225 → 216 → 207 → 198 → 180 → 171 → 162 → 153 → 144 → 135 → 126 → 117 → 108 → 99 → 81 → 72 → 63 → 54 → 45 → 36 → 27 → 18 → 9 → 0

The details are different in other bases, like 2 or 16, but the destination is the same. The number falls to zero and the fall stops, because digsum(0) = 0:


102 → 1 → 0 (n=2)
100 → 11 → 1 → 0 (n=4)
1000 → 111 → 100 → 11 → 1 → 0 (n=8)
10000 → 1111 → 1011 → 1000 → 111 → 100 → 11 → 1 → 0 (n=16)
100000 → 11111 → 11010 → 10111 → 10011 → 10000 → 1111 → 1011 → 1000 → 111 → 100 → 11 → 1 → 0 (n=32)
1000000 → 111111 → 111001 → 110101 → 110001 → 101110 → 101010 → 100111 → 100011 → 100000 → 11111 → 11010 → 10111 → 10011 → 10000 → 1111 → 1011 → 1000 → 111 → 100 → 11 → 1 → 0 (n=64)


1013 → C → 0 (n=13)
100 → CC → B1 → A2 → 93 → 84 → 75 → 66 → 57 → 48 → 39 → 2A → 1B → C → 0 (n=169)
1000 → CCC → CA2 → C84 → C66 → C48 → C2A → C0C → BC1 → BA3 → B85 → B67 → B49 → B2B → B10 → B01 → AC2 → AA4 → A86 → A68 → A4A → A2C → A11 → A02 → 9C3 → 9A5 → 987 → 969 → 94B → 930 → 921 → 912 → 903 → 8C4 → 8A6 → 888 → 86A → 84C → 831 → 822 → 813 → 804 → 7C5 → 7A7 → 789 → 76B → 750 → 741 → 732 → 723 → 714 → 705 → 6C6 → 6A8 → 68A → 66C → 651 → 642 → 633 → 624 → 615 → 606 → 5C7 → 5A9 → 58B → 570 → 561 → 552 → 543 → 534 → 525 → 516 → 507 → 4C8 → 4AA → 48C → 471 → 462 → 453 → 444 → 435 → 426 → 417 → 408 → 3C9 → 3AB → 390 → 381 → 372 → 363 → 354 → 345 → 336 → 327 → 318 → 309 → 2CA → 2AC → 291 → 282 → 273 → 264 → 255 → 246 → 237 → 228 → 219 → 20A → 1CB → 1B0 → 1A1 → 192 → 183 → 174 → 165 → 156 → 147 → 138 → 129 → 11A → 10B → CC → B1 → A2 → 93 → 84 → 75 → 66 → 57 → 48 → 39 → 2A → 1B → C → 0 (n=2197)

But the fall to 0 made me think of another kind of number-fall. What if you count the 0s in a number, take that count away from the original number, and repeat? You could call this a z-fall (pronounced zee-fall). But unlike free-fall, z-fall doesn’t last long:


10 → 9
100 → 98
1000 → 997
10000 → 9996

And the number always comes to rest far above the ground, as it were. In a fall using digsum(n), the number descends to 0. In a fall using zerocount(n), the number never even reaches 1. At least, never in any base higher than 2. But in base-2, you get this:


10 → 1 (n=2)
100 → 10 → 1 (n=4)
1000 → 101 → 100 → 10 → 1 (n=8)
10000 → 1100 → 1010 → 1000 → 101 → 100 → 10 → 1 (n=16)
100000 → 11011 → 11010 → 11000 → 10101 → 10011 → 10001 → 1110 → 1101 → 1100 → 1010 → 1000 → 101 → 100 → 10 → 1 (n=32)
1000000 → 111010 → 111000 → 110101 → 110011 → 110001 → 101110 → 101100 → 101001 → 100110 → 100011 → 100000 → 11011 → 11010 → 11000 → 10101 → 10011 → 10001 → 1110 → 1101 → 1100 → 1010 → 1000 → 101 → 100 → 10 → 1 (n=64)

When I saw that, I had a wonderful vision of how even the biggest numbers in base 2 could z-fall all the way to 1. Almost all binary numbers contain 0, after all. So the z-falls would get longer and longer, paying tribute to la caída infinita, the infinite fall, of the librarian in Borges’ Library of Babel. Alas, binary numbers don’t behave like that. The highest number in base 2 that z-falls to 1 is this:


1010001 → 1001101 → 1001010 → 1000110 → 1000010 → 111101 → 111100 → 111010 → 111000 → 110101 → 110011 → 110001 → 101110 → 101100 → 101001 → 100110 → 100011 → 100000 → 11011 → 11010 → 11000 → 10101 → 10011 → 10001 → 1110 → 1101 → 1100 → 1010 → 1000 → 101 → 100 → 10 → 1 (n=81)

Above that, binary numbers land on what you might call a shelf:


1010010=82 → 1001110=78 → 1001011=75 → 1001000=72 → 1000011=67 → 111111=63 (n=82)

If binary numbers are an infinite tall mountain, 1 is at the foot of the mountain. 111111 = 63 is like a shelf a little way above the foot. But I conjecture that arbitrarily large binary numbers will z-fall to 63. For example, no matter how large the power of 2, I conjecture that it will z-fall to 63:


10 → 1 : 2 → 1 (count of steps=2)
100 ... → 1 : 4 ... → 1 (c=3)
1000 ... → 1 : 8 ... → 1 (c=5)
10000 ... → 1 : 16 ... → 1 (c=8)
100000 ... → 1 : 32 ... → 1 (c=16)
1000000 ... → 1 : 64 ... → 1 (c=27)
10000000 ... → 111111 : 128 ... → 63 (c=21)
100000000 ... → 111111 : 256 ... → 63 (c=60)
1000000000 ... → 111111 : 512 ... → 63 (c=130)
10000000000 ... → 111111 : 1024 ... → 63 (c=253)
100000000000 ... → 111111 : 2048 ... → 63 (c=473)
1000000000000 ... → 111111 : 4096 ... → 63 (c=869)
10000000000000 ... → 111111 : 8192 ... → 63 (c=1586)
100000000000000 ... → 111111 : 16384 ... → 63 (c=2899)
1000000000000000 ... → 111111 : 32768 ... → 63 (c=5327)
10000000000000000 ... → 111111 : 65536 ... → 63 (c=9851)
100000000000000000 ... → 111111 : 131072 ... → 63 (c=18340)
1000000000000000000 ... → 111111 : 262144 ... → 63 (c=34331)
10000000000000000000 ... → 111111 : 524288 ... → 63 (c=64559)
100000000000000000000 ... → 111111 : 1048576 ... → 63 (c=121831)
1000000000000000000000 ... → 111111 : 2097152 ... → 63 (c=230573)
10000000000000000000000 ... → 111111 : 4194304 ... → 63 (c=437435)
100000000000000000000000 ... → 111111 : 8388608 ... → 63 (c=831722)
1000000000000000000000000 ... → 111111 : 16777216 ... → 63 (c=1584701)
10000000000000000000000000 ... → 111111 : 33554432 ... → 63 (c=3025405)
100000000000000000000000000 ... → 111111 : 67108864 ... → 63 (c=5787008)
1000000000000000000000000000 ... → 111111 : 134217728 ... → 63 (c=11089958)
10000000000000000000000000000 ... → 111111 : 268435456 ... → 63 (c=21290279)
100000000000000000000000000000 ... → 111111 : 536870912 ... → 63 (c=40942711)
1000000000000000000000000000000 ... → 111111 : 1073741824 ... → 63 (c=78864154)

So the z-falls get longer and longer. But z-falling to 63 doesn’t have the power of z-falling to 1.

Period Panes

In The Penguin Dictionary of Curious and Interesting Numbers (1987), David Wells remarks that 142857 is “a number beloved of all recreational mathematicians”. He then explains that it’s “the decimal period of 1/7: 1/7 = 0·142857142857142…” and “the first decimal reciprocal to have maximum period, that is, the length of its period is only one less than the number itself.”

Why does this happen? Because when you’re calculating 1/n, the remainders can only be less than n. In the case of 1/7, you get remainders for all integers less than 7, i.e. there are 6 distinct remainders and 6 = 7-1:

(1*10) / 7 = 1 remainder 3, therefore 1/7 = 0·1...
(3*10) / 7 = 4 remainder 2, therefore 1/7 = 0·14...
(2*10) / 7 = 2 remainder 6, therefore 1/7 = 0·142...
(6*10) / 7 = 8 remainder 4, therefore 1/7 = 0·1428...
(4*10) / 7 = 5 remainder 5, therefore 1/7 = 0·14285...
(5*10) / 7 = 7 remainder 1, therefore 1/7 = 0·142857...
(1*10) / 7 = 1 remainder 3, therefore 1/7 = 0·1428571...
(3*10) / 7 = 4 remainder 2, therefore 1/7 = 0·14285714...
(2*10) / 7 = 2 remainder 6, therefore 1/7 = 0·142857142...

Mathematicians know that reciprocals with maximum period can only be prime reciprocals and with a little effort you can work out whether a prime will yield a maximum period in a particular base. For example, 1/7 has maximum period in bases 3, 5, 10, 12 and 17:

1/21 = 0·010212010212010212... in base 3
1/12 = 0·032412032412032412... in base 5
1/7 =  0·142857142857142857... in base 10
1/7 =  0·186A35186A35186A35... in base 12
1/7 =  0·274E9C274E9C274E9C... in base 17

To see where else 1/7 has maximum period, have a look at this graph:

Period pane for primes 3..251 and bases 2..39


I call it a “period pane”, because it’s a kind of window into the behavior of prime reciprocals. But what is it, exactly? It’s a graph where the x-axis represents primes from 3 upward and the y-axis represents bases from 2 upward. The red squares along the bottom aren’t part of the graph proper, but indicate primes that first occur after a power of two: 5 after 4=2^2; 11 after 8=2^3; 17 after 16=2^4; 37 after 32=2^5; 67 after 64=2^6; and so on.

If a prime reciprocal has maximum period in a particular base, the graph has a solid colored square. Accordingly, the purple square at the bottom left represents 1/7 in base 10. And as though to signal the approval of the goddess of mathematics, the graph contains a lower-case b-for-base, which I’ve marked in green. Here are more period panes in higher resolution (open the images in a new window to see them more clearly):

Period pane for primes 3..587 and bases 2..77


Period pane for primes 3..1303 and bases 2..152


An interesting pattern has begun to appear: note the empty lanes, free of reciprocals with maximum period, that stretch horizontally across the period panes. These lanes are empty because there are no prime reciprocals with maximum period in square bases, that is, bases like 4, 9, 25 and 36, where 4 = 2*2, 9 = 3*3, 25 = 5*5 and 36 = 6*6. I don’t know why square bases don’t have max-period prime reciprocals, but it’s probably obvious to anyone with more mathematical nous than me.

Period pane for primes 3..2939 and bases 2..302


Period pane for primes 3..6553 and bases 2..602


Like the Ulam spiral, other and more mysterious patterns appear in the period panes, hinting at the hidden regularities in the primes.

Bent Pent

This is a beautiful and interesting shape, reminiscent of a piece of jewellery:

Pentagons in a ring


I came across it in this tricky little word-puzzle:

Word puzzle using pentagon-ring


Here’s a printable version of the puzzle:

Printable puzzle


Let’s try placing some other regular polygons with s sides around regular polygons with s*2 sides:

Hexagonal ring of triangles


Octagonal ring of squares


Decagonal ring of pentagons


Dodecagonal ring of hexagons


Only regular pentagons fit perfectly, edge-to-edge, around a regular decagon. But all these polygonal-rings can be used to create interesting and beautiful fractals, as I hope to show in a future post.

Performativizing Polyhedra

Τα Στοιχεία του Ευκλείδου, ια΄

κεʹ. Κύβος ἐστὶ σχῆμα στερεὸν ὑπὸ ἓξ τετραγώνων ἴσων περιεχόμενον.
κϛʹ. ᾿Οκτάεδρόν ἐστὶ σχῆμα στερεὸν ὑπὸ ὀκτὼ τριγώνων ἴσων καὶ ἰσοπλεύρων περιεχόμενον.
κζʹ. Εἰκοσάεδρόν ἐστι σχῆμα στερεὸν ὑπὸ εἴκοσι τριγώνων ἴσων καὶ ἰσοπλεύρων περιεχόμενον.
κηʹ. Δωδεκάεδρόν ἐστι σχῆμα στερεὸν ὑπὸ δώδεκα πενταγώνων ἴσων καὶ ἰσοπλεύρων καὶ ἰσογωνίων περιεχόμενον.

Euclid’s Elements, Book 11

25. A cube is a solid figure contained by six equal squares.
26. An octahedron is a solid figure contained by eight equal and equilateral triangles.
27. An icosahedron is a solid figure contained by twenty equal and equilateral triangles.
28. A dodecahedron is a solid figure contained by twelve equal, equilateral, and equiangular pentagons.

Sprime Time

All fans of recreational math love palindromic numbers. It’s mandatory, man. 101, 727, 532235, 8810188, 1367755971795577631 — I love ’em! But where can you go after palindromes? Well, you can go to palindromes in a higher dimension. Numbers like 101, 727, 532235 and 8810188 are 1-d palindromes. That is, they’re palindromic in one dimension: backwards and forwards. But numbers like 181818189 and 646464640 aren’t palindromic in one dimension. They’re palindromic in two dimensions:


1 8 1
8 9 8
1 8 1

n=181818189


6 4 6
4 0 4
6 4 6

n=646464640



They’re 2-d palindromes or spiral numbers, that is, numbers that are symmetrical when written as a spiral. You start with the first digit on the top left, then spiral inwards to the center, like this for a 9-digit spiral (9 = 3×3):


And this for a 36-digit spiral (36 = 6×6):


Spiral numbers are easy to construct, because you can reflect and rotate the numbers in one triangular slice of the spiral to find all the others:


You could say that the seed for the spiral number above is 7591310652, because you can write that number in descending lines, left-to-right, as a triangle.

Here are some palindromic numbers with nine digits in base 3 — as you can see, some are both palindromic numbers and spiral numbers. That is, some are palindromic in both one and two dimensions:

1  0  1

0  1  0

1  0  1

n=101010101


1  0  1

0  2  0

1  0  1

n=101010102


1  1  1

1  0  1

1  1  1

n=111111110


1  1  1

1  1  1

1  1  1

n=111111111


2  0  2

0  1  0

2  0  2

n=202020201


2  0  2

0  2  0

2  0  2

n=202020202


2  2  2

2  1  2

2  2  2

n=222222221


2  2  2

2  2  2

2  2  2

n=222222222


But palindromic primes are even better than ordinary palindromes. Here are a few 1-d palindromic primes in base 10:

101
151
73037
7935397
97356765379
1091544334334451901
1367755971795577631
70707270707
39859395893
9212129
7436347
166000661
313
929


And after 1-d palindromic primes, you can go to 2-d palindromic primes. That is, to spiral primes or sprimes — primes that are symmetrical when written as a spiral:

3 6 3
6 7 6
3 6 3

n=363636367 (prime)
seed=367 (see definition above)


9 1 9
1 3 1
9 1 9

n=919191913 (prime)
seed=913


3 7 8 6 3 6 8 7 3
7 9 1 8 9 8 1 9 7
8 1 9 0 9 0 9 1 8
6 8 0 5 5 5 0 8 6
3 9 9 5 7 5 9 9 3
6 8 0 5 5 5 0 8 6
8 1 9 0 9 0 9 1 8
7 9 1 8 9 8 1 9 7
3 7 8 6 3 6 8 7 3

n=378636873786368737863687378636879189819189819189819189819090909090909090555555557 (prime)
seed=378639189909557 (l=15)


And why stop with spiral numbers — and sprimes — in two dimensions? 363636367 is a 2-sprime, being palindromic in two dimensions. But the digits of a number could be written to form a symmetrical cube in three, four, five and more dimensions. So I assume that there are 3-sprimes, 4-sprimes, 5-sprimes and more out there. Watch this space.

The Power of Powder

• Racine carrée de 2, c’est 1,414 et des poussières… Et quelles poussières ! Des grains de sable qui empêchent d’écrire racine de 2 comme une fraction. Autrement dit, cette racine n’est pas dans Q. — Rationnel mon Q: 65 exercices de styles, Ludmilla Duchêne et Agnès Leblanc (2010)

• The square root of 2 is 1·414 and dust… And what dust! Grains of sand that stop you writing the root of 2 as a fraction. Put another way, this root isn’t in Q [the set of rational numbers].

Thrice Dice Twice

A once very difficult but now very simple problem in probability from Ian Stewart’s Do Dice Play God? (2019):

For three dice [Girolamo] Cardano solved a long-standing conundrum [in the sixteenth century]. Gamblers had long known from experience that when throwing three dice, a total of 10 is more likely than 9. This puzzled them, however, because there are six ways to get a total of 10:

1+4+5; 1+3+6; 2+4+4; 2+2+6; 2+3+5; 3+3+4

But also six ways to get a total of 9:

1+2+6; 1+3+5; 1+4+4; 2+2+5; 2+3+4; 3+3+3

So why does 10 occur more often?

To see the answer, imagine throwing three dice of different colors: red, blue and yellow. How many ways can you get 9 and how many ways can you get 10?

Roll Total=9 Dice #1 (Red) Dice #2 (Blue) Dice #3 (Yellow)
01 9 = 1 2 6
02 9 = 1 3 5
03 9 = 1 4 4
04 9 = 1 5 3
05 9 = 1 6 2
06 9 = 2 1 6
07 9 = 2 2 5
08 9 = 2 3 4
09 9 = 2 4 3
10 9 = 2 5 2
11 9 = 2 6 1
12 9 = 3 1 5
13 9 = 3 2 4
14 9 = 3 3 3
15 9 = 3 4 2
16 9 = 3 5 1
17 9 = 4 1 4
18 9 = 4 2 3
19 9 = 4 3 2
20 9 = 4 4 1
21 9 = 5 1 3
22 9 = 5 2 2
23 9 = 5 3 1
24 9 = 6 1 2
25 9 = 6 2 1
Roll Total=10 Dice #1 (Red) Dice #2 (Blue) Dice #3 (Yellow)
01 10 = 1 3 6
02 10 = 1 4 5
03 10 = 1 5 4
04 10 = 1 6 3
05 10 = 2 2 6
06 10 = 2 3 5
07 10 = 2 4 4
08 10 = 2 5 3
09 10 = 2 6 2
10 10 = 3 1 6
11 10 = 3 2 5
12 10 = 3 3 4
13 10 = 3 4 3
14 10 = 3 5 2
15 10 = 3 6 1
16 10 = 4 1 5
17 10 = 4 2 4
18 10 = 4 3 3
19 10 = 4 4 2
20 10 = 4 5 1
21 10 = 5 1 4
22 10 = 5 2 3
23 10 = 5 3 2
24 10 = 5 4 1
25 10 = 6 1 3
26 10 = 6 2 2
27 10 = 6 3 1