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.

RevNumSum

If you take an integer, n, and reverse its digits to get the integer r, there are three possibilities:


n > r (e.g. 85236 > 63258)
n < r (e.g. 17783 < 38771)
n = r (e.g. 45154 = 45154)

If n = r, n is a palindrome. If n > r, I call n a major number. If n < r, I call n a minor number. And here are the minor and major numbers represented as white squares on an Ulam-like spiral (the negative of a minor spiral is a major spiral, and vice versa — sometimes one looks better than the other):

b=2 (minor numbers)


b=3


b=4


b=5


b=6


b=7 (major numbers)


b=8 (minor numbers)


b=9 (mjn)


b=10 (mjn)


b=11 (mjn)


b=12 (mjn)


b=13 (mjn)


b=14 (mjn)


b=15 (mjn)


b=16 (mjn)


b=17 (mjn)


b=18 (mjn)


b=19 (mjn)


b=20 (mjn)


Minor numbers, b=2..20 (animated)


Now let’s look at a sequence formed by summing the reversed numbers, minor ones, major ones and palindromes. Here are the standard integers:


1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17...

If you sum the integers, you get what are called the triangular numbers:


1 = 1
3 = 1 + 2
6 = 1 + 2 + 3
10 = 1 + 2 + 3 + 4
15 = 1 + 2 + 3 + 4 + 5
21 = 1 + 2 + 3 + 4 + 5 + 6
28 = 1 + 2 + 3 + 4 + 5 + 6 + 7
36 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8
45 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9
55 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10
66 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11
78 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12
91 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13
105 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14
120 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15
136 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16
153 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17
171 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18
190 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19
210 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20

But what happens if you reverse the integers before summing them? Here side-by-side are the triangular numbers and the underlined revnumsums (as they might be called):


45 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9
45 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9
55 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10
46 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 1
66 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11
57 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 1 + 11
78 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12
78 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 1 + 11 + 21
91 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13
109 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 1 + 11 + 21 + 31
105 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14
150 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 1 + 11 + 21 + 31 + 41
120 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15
201 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 1 + 11 + 21 + 31 + 41 + 51
136 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16
262 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 1 + 11 + 21 + 31 + 41 + 51 + 61
153 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17
333 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 1 + 11 + 21 + 31 + 41 + 51 + 61 + 71
171 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18
414 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 1 + 11 + 21 + 31 + 41 + 51 + 61 + 71 + 81
190 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19
505 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 1 + 11 + 21 + 31 + 41 + 51 + 61 + 71 + 81 + 91
210 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20
507 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 1 + 11 + 21 + 31 + 41 + 51 + 61 + 71 + 81 + 91
+ 2

Unlike triangular numbers, revnumsums are dependent on the base they’re calculated in. In base 2, the revnumsum is always smaller than the triangular number, except at step 1. In base 3, the revnumsum is equal to the triangular number at steps 1, 2 and 15 (= 120 in base 3). Otherwise it’s smaller than the triangular number.

And in higher bases? In bases > 3, the revnumsum rises and falls above the equivalent triangular number. When it’s higher, it tends towards a maximum height of (base+1)/4 * triangular number.

Palindrought

The alchemists dreamed of turning dross into gold. In mathematics, you can actually do that, metaphorically speaking. If palindromes are gold and non-palindromes are dross, here is dross turning into gold:


22 = 10 + 12
222 = 10 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 + 23 + 24
484 = 10 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 + 21 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 30 + 31 + 32 + 34
555 = 10 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 + 21 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 30 + 31 + 32 + 34 + 35 + 36
2002 = nonpalsum(10,67)
36863 = nonpalsum(10,286)
45954 = nonpalsum(10,319)
80908 = nonpalsum(10,423)
113311 = nonpalsum(10,501)
161161 = nonpalsum(10,598)
949949 = nonpalsum(10,1417)
8422248 = nonpalsum(10,4136)
13022031 = nonpalsum(10,5138)
14166141 = nonpalsum(10,5358)
16644661 = nonpalsum(10,5806)
49900994 = nonpalsum(10,10045)
464939464 = nonpalsum(10,30649)
523434325 = nonpalsum(10,32519)
576656675 = nonpalsum(10,34132)
602959206 = nonpalsum(10,34902)
[...]

The palindromes don’t seem to stop arriving. But something unexpected happens when you try to turn gold into gold. If you sum palindromes to get palindromes, you’re soon hit by what you might call a palindrought, where no palindromes appear:


1 = 1
3 = 1 + 2
6 = 1 + 2 + 3
111 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 11 + 22 + 33
353 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 11 + 22 + 33 + 44 + 55 + 66 + 77
7557 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 11 + 22 + 33 + 44 + 55 + 66 + 77 + 88 + 99 + 101 + 111 + 121 + 131 + 141 + 151 + 161 + 171 + 181 + 191 + 202 + 212 + 222 + 232 + 242 + 252 + 262 + 272 + 282 + 292 + 303 + 313 + 323 + 333 + 343 + 353 + 363 + 373 + 383
2376732 = palsum(1,21512)

That’s sequence A046488 at the OEIS. And I suspect that the sequence is complete and that the palindrought never ends. For some evidence of that, here’s an interesting pattern that emerges if you look at palsums of 1 to repdigits 9[…]9:


50045040 = palsum(1,99999)
50045045040 = palsum(1,9999999)
50045045045040 = palsum(1,999999999)
50045045045045040 = palsum(1,99999999999)
50045045045045045040 = palsum(1,9999999999999)
50045045045045045045040 = palsum(1,999999999999999)
50045045045045045045045040 = palsum(1,99999999999999999)
50045045045045045045045045040 = palsum(1,9999999999999999999)
50045045045045045045045045045040 = palsum(1,999999999999999999999)

As the sums get bigger, the carries will stop sweeping long enough and the sums may fall into semi-regular patterns of non-palindromic numbers like 50045040. If you try higher bases like base 909, you get more palindromes by summing palindromes, but a palindrought arrives in the end there too:


1 = palsum(1)
3 = palsum(1,2)
6 = palsum(1,3)
A = palsum(1,4)
[...]
66 = palsum(1,[104]) (palindromes = 43)
LL = palsum(1,[195]) (44)
[37][37] = palsum(1,[259]) (45)
[73][73] = palsum(1,[364]) (46)
[114][114] = palsum(1,[455]) (47)
[172][172] = palsum(1,[559]) (48)
[369][369] = palsum(1,[819]) (49)
6[466]6 = palsum(1,[104][104]) (50)
L[496]L = palsum(1,[195][195]) (51)
[37][528][37] = palsum(1,[259][259]) (52)
[73][600][73] = palsum(1,[364][364]) (53)
[114][682][114] = palsum(1,[455][455]) (54)
[172][798][172] = palsum(1,[559][559]) (55)
[291][126][291] = palsum(1,[726][726]) (56)
[334][212][334] = palsum(1,[778][778]) (57)
[201][774][830][774][201] = palsum(1,[605][707][605]) (58)
[206][708][568][708][206] = palsum(1,[613][115][613]) (59)
[456][456][569][569][456][456] = palsum(1,11[455]11) (60)
22[456][454][456]22 = palsum(1,21012) (61)

Note the palindrome for palsum(1,21012). All odd bases higher than 3 seem to produce a palindrome for 1 to 21012 in that base (21012 in base 5 = 1382 in base 10, 2012 in base 7 = 5154 in base 10, and so on):


2242422 = palsum(1,21012) (base=5)
2253522 = palsum(1,21012) (b=7)
2275722 = palsum(1,21012) (b=11)
2286822 = palsum(1,21012) (b=13)
2297922 = palsum(1,21012) (b=15)
22A8A22 = palsum(1,21012) (b=17)
22B9B22 = palsum(1,21012) (b=19)
22CAC22 = palsum(1,21012) (b=21)
22DBD22 = palsum(1,21012) (b=23)

And here’s another interesting pattern created by summing squares in base 9 (where 17 = 16 in base 10, 40 = 36 in base 10, and so on):


1 = squaresum(1)
5 = squaresum(1,4)
33 = squaresum(1,17)
111 = squaresum(1,40)
122221 = squaresum(1,4840)
123333321 = squaresum(1,503840)
123444444321 = squaresum(1,50483840)
123455555554321 = squaresum(1,5050383840)
123456666666654321 = squaresum(1,505048383840)
123456777777777654321 = squaresum(1,50505038383840)
123456788888888887654321 = squaresum(1,5050504838383840)

Then a palindrought strikes again. But you don’t get a palindrought in the triangular numbers, or numbers created by summing the integers, palindromic and non-palindromic alike:


1 = 1
3 = 1 + 2
6 = 1 + 2 + 3
55 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10
66 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11
171 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18
595 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 + 21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 30 + 31 + 32 + 33 + 34
666 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 + 21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 30 + 31 + 32 + 33 + 34 + 35 + 36
3003 = palsum(1,77)
5995 = palsum(1,109)
8778 = palsum(1,132)
15051 = palsum(1,173)
66066 = palsum(1,363)
617716 = palsum(1,1111)
828828 = palsum(1,1287)
1269621 = palsum(1,1593)
1680861 = palsum(1,1833)
3544453 = palsum(1,2662)
5073705 = palsum(1,3185)
5676765 = palsum(1,3369)
6295926 = palsum(1,3548)
35133153 = palsum(1,8382)
61477416 = palsum(1,11088)
178727871 = palsum(1,18906)
1264114621 = palsum(1,50281)
1634004361 = palsum(1,57166)
5289009825 = palsum(1,102849)
6172882716 = palsum(1,111111)
13953435931 = palsum(1,167053)
16048884061 = palsum(1,179158)
30416261403 = palsum(1,246642)
57003930075 = palsum(1,337650)
58574547585 = palsum(1,342270)
66771917766 = palsum(1,365436)
87350505378 = palsum(1,417972)
[...]

If 617716 = palsum(1,1111) and 6172882716 = palsum(1,111111), what is palsum(1,11111111)? Try it for yourself — there’s an easy formula for the triangular numbers.

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.

Digital Dissection

As I never tire of pointing out, the three most powerful drugs in the universe are water, maths and language. And I never tire of snorting the fact that numbers can come in many different guises. You can take a trivial, everyday number like a hundred and see it transform like this:


100 = 1100100 in base 2; 10201 in base 3; 1210 in base 4; 400 in base 5; 244 in base 6; 202 in base 7; 144 in base 8; 121 in base 9; 100 in b10; 91 in b11; 84 in b12; 79 in b13; 72 in b14; 6A in b15; 64 in b16; 5F in b17; 5A in b18; 55 in b19; 50 in b20; 4G in b21; 4C in b22; 48 in b23; 44 in b24; 40 in b25; 3M in b26; 3J in b27; 3G in b28; 3D in b29; 3A in b30; 37 in b31; 34 in b32; 31 in b33; 2W in b34; 2U in b35; 2S in b36; 2Q in b37; 2O in b38; 2M in b39; 2K in b40; 2I in b41; 2G in b42; 2E in b43; 2C in b44; 2A in b45; 28 in b46; 26 in b47; 24 in b48; 22 in b49; 20 in b50; 1[49] in b51; 1[48] in b52; 1[47] in b53; 1[46] in b54; 1[45] in b55; 1[44] in b56; 1[43] in b57; 1[42] in b58; 1[41] in b59; 1[40] in b60; 1[39] in b61; 1[38] in b62; 1[37] in b63; 1[36] in b64; 1Z in b65; 1Y in b66; 1X in b67; 1W in b68; 1V in b69; 1U in b70; 1T in b71; 1S in b72; 1R in b73; 1Q in b74; 1P in b75; 1O in b76; 1N in b77; 1M in b78; 1L in b79; 1K in b80; 1J in b81; 1I in b82; 1H in b83; 1G in b84; 1F in b85; 1E in b86; 1D in b87; 1C in b88; 1B in b89; 1A in b90; 19 in b91; 18 in b92; 17 in b93; 16 in b94; 15 in b95; 14 in b96; 13 in b97; 12 in b98; 11 in b99

I like the shifts from 1100100 to 10201 to 1210 to 400 to 244 to 202 to 144 to 121. How can 1100100 and 244 be the same number? Well, they are — or they’re not, as you please. In base 2, 1100100 = 244 in base 6 = 100 in base 10. But if all those numbers are in the same base, they’re completely different and 1100100 dwarfs the other two.

But some things you can’t please yourself about. Suppose you take the different representations of 6561 in bases 2..6560 and add up the 1s, the 2s, the 3s and so on, like this:


n=6561

digsum(1,6561,b=2..6560) = 3343 (50.95% of 6561)
digsum(2,6561,b=2..6560) = 2246 (34.23% of 6561)
digsum(3,6561,b=2..6560) = 1680 (25.61% of 6561)
digsum(4,6561,b=2..6560) = 1368 (20.85% of 6561)
digsum(5,6561,b=2..6560) = 1185 (18.06% of 6561)
digsum(6,6561,b=2..6560) = 1074 (16.37% of 6561)
digsum(7,6561,b=2..6560) = 875 (13.34% of 6561)
digsum(8,6561,b=2..6560) = 768 (11.71% of 6561)
digsum(9,6561,b=2..6560) = 1080 (16.46% of 6561)
[...]
digcount(0,6561,b=2..6560) = 31

Is there a pattern in the percentages? Let’s apply the same process to some bigger numbers (and note that 0 does not behave like the other digits):


n=59049

digsum(1,59049) = 29648 (50.21%)
digsum(2,59049) = 19790 (33.51%)
digsum(3,59049) = 14901 (25.23%)
digsum(4,59049) = 11956 (20.25%)
digsum(5,59049) = 9970 (16.88%)
digsum(6,59049) = 8550 (14.48%)
digsum(7,59049) = 7539 (12.77%)
digsum(8,59049) = 6672 (11.30%)
digsum(9,59049) = 6579 (11.14%)
digcount(0,59049) = 41


n=531441

digsum(1,531441) = 266065 (50.06%)
digsum(2,531441) = 177394 (33.38%)
digsum(3,531441) = 133128 (25.05%)
digsum(4,531441) = 106532 (20.05%)
digsum(5,531441) = 88815 (16.71%)
digsum(6,531441) = 76224 (14.34%)
digsum(7,531441) = 66661 (12.54%)
digsum(8,531441) = 59320 (11.16%)
digsum(9,531441) = 53928 (10.15%)
digcount(0,531441) = 62


n=4782969

digsum(1,4782969) = 2392219 (50.02%)
digsum(2,4782969) = 1595000 (33.35%)
digsum(3,4782969) = 1196370 (25.01%)
digsum(4,4782969) = 957300 (20.01%)
digsum(5,4782969) = 797700 (16.68%)
digsum(6,4782969) = 683850 (14.30%)
digsum(7,4782969) = 598444 (12.51%)
digsum(8,4782969) = 531944 (11.12%)
digsum(9,4782969) = 480870 (10.05%)
digcount(0,4782969) = 66

Yes, the pattern’s getting stronger. Let’s try even bigger numbers:


n=43046721

digsum(1,43046721) = 21525521 (50.01%)
digsum(2,43046721) = 14350754 (33.34%)
digsum(3,43046721) = 10763496 (25.00%)
digsum(4,43046721) = 8610980 (20.00%)
digsum(5,43046721) = 7175955 (16.67%)
digsum(6,43046721) = 6150924 (14.29%)
digsum(7,43046721) = 5382167 (12.50%)
digsum(8,43046721) = 4784232 (11.11%)
digsum(9,43046721) = 4306257 (10.00%)
digcount(0,43046721) = 86


n=387420489

digsum(1,387420489) = 193716365 (50.00%)
digsum(2,387420489) = 129145522 (33.33%)
digsum(3,387420489) = 96859980 (25.00%)
digsum(4,387420489) = 77488588 (20.00%)
digsum(5,387420489) = 64574220 (16.67%)
digsum(6,387420489) = 55349742 (14.29%)
digsum(7,387420489) = 48431250 (12.50%)
digsum(8,387420489) = 43050264 (11.11%)
digsum(9,387420489) = 38748357 (10.00%)
digcount(0,387420489) = 95

To the given precision, the sum of 1s is 1/2 of n; the sum of 2s is 1/3; the sum of 3 is 1/4; and the sum of 4s is 1/5. It looks as though the sum of a given digit d → 1/(d+1) of n as n → ∞. But why? My mathematical intuition is bad, so it took me a while to see what some people will see in a flash. To see what’s going on, let’s go back to the all-base representations of 100:


100 = 1100100 in base 2; 10201 in base 3; 1210 in base 4; 400 in base 5; 244 in base 6; 202 in base 7; 144 in base 8; 121 in base 9; 100 in b10; 91 in b11; 84 in b12; 79 in b13; 72 in b14; 6A in b15; 64 in b16; 5F in b17; 5A in b18; 55 in b19; 50 in b20; 4G in b21; 4C in b22; 48 in b23; 44 in b24; 40 in b25; 3M in b26; 3J in b27; 3G in b28; 3D in b29; 3A in b30; 37 in b31; 34 in b32; 31 in b33; 2W in b34; 2U in b35; 2S in b36; 2Q in b37; 2O in b38; 2M in b39; 2K in b40; 2I in b41;
2G in b42; 2E in b43; 2C in b44; 2A in b45; 28 in b46; 26 in b47; 24 in b48; 22 in b49; 20 in b50; 1[49] in b51; 1[48] in b52; 1[47] in b53; 1[46] in b54; 1[45] in b55; 1[44] in b56; 1[43] in b57; 1[42] in b58; 1[41] in b59; 1[40] in b60; 1[39] in b61; 1[38] in b62; 1[37] in b63; 1[36] in b64; 1Z in b65; 1Y in b66; 1X in b67; 1W in b68; 1V in b69; 1U in b70; 1T in b71; 1S in b72; 1R in b73; 1Q in b74; 1P in b75; 1O in b76; 1N in b77; 1M in b78; 1L in b79; 1K in b80; 1J in b81
; 1I in b82; 1H in b83; 1G in b84; 1F in b85; 1E in b86; 1D in b87; 1C in b88; 1B in b89; 1A in b90; 19 in b91; 18 in b92; 17 in b93; 16 in b94; 15 in b95; 14 in b96; 13 in b97; 12 in b98; 11 in b99

When the base b is higher than half of 100, the representations of 100 consist of a digit 1 followed by another digit. Half of a hundred = 50, therefore 100 in base 10 = 1[49] in b51, 1[48] in b52, 1[47] in b53, 1[46] in b54, 1[45] in b55, 1[44] in b56, 1[43] in b57, 1[42] in b58, 1[41] in b59… If you take binary and so on into account, 1 is the first digit of slightly over half the representations of 100. And 1 also occurs in other positions. Therefore digsum(1,100,b=2..99) > 50. As the number n gets larger and larger, the contribution of leading 1s in bases b > n/2 begins to swamp the contributions of 1s in other positions, therefore digsum(1,n) → 1/2 of n as n → ∞.

And what about 2s and 3s? Similar reasoning applies. One hundred has a leading digit of 2 in bases b where b > 1/3 of 100 and b <= 1/2 of 100. So 100 = 2W in b34, 2U in b35, 2S in b36, 2Q in b37, 2O in b38… In other words, roughly 1/2 – 1/3 of the representations of 100 have a leading 2. Now, 1/2 – 1/3 = 3/6 – 2/6 = 1/6 and 1/6 * 2 = 1/3 (i.e., 1/6 of the representations contribute a leading 2 to the sum of 2s). Therefore the all-base digsum(2,n) → 1/3 of n as n → ∞. Next, one hundred has a leading digit of 3 in bases b where b > 1/4 of 100 and b <= 1/3. So 100 = 3M in b26, 3J in b27, 3G in b28, 3D in b29, 3A in b30… Now, 1/3 – 1/4 = 4/12 – 3/12 = 1/12 and 1/12 * 3 = 1/4. Therefore the all-base digsum(3,n) → 1/4 of n as n → ∞.

And so on.

Count Amounts

One of my favourite integer sequences is what I call the digit-line. You create it by taking this very familiar integer sequence:

• 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20…

And turning it into this one:

• 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 0, 1, 1, 1, 2, 1, 3, 1, 4, 1, 5, 1, 6, 1, 7, 1, 8, 1, 9, 2, 0… (A033307 in the Online Encyclopedia of Integer Sequences)

You simply chop all numbers into single digits. What could be simpler? Well, creating the digit-line couldn’t be simpler, but it is in fact a very complex object. There are hidden depths in its patterns, as even a brief look will uncover. For example, you can try counting the digits as they appear one-by-one in the line and seeing whether the digit-counts compare. Do the 1s of the digit-line always outnumber the 0s, as you might expect? Yes, they do (unless you start the digit-line 0, 1, 2, 3…). But do the 2s always outnumber the 0s? No: at position 2, there’s a 2, and at position 11 there’s a 0. So that’s one 2 and one 0. Does it happen again? Yes, it happens again at the 222nd digit of the digit-line, as below:

1, 2count=1, 3, 4, 5, 6, 7, 8, 9, 1, 0count=1, 1, 1, 1, 22, 1, 3, 1, 4, 1, 5, 1, 6, 1, 7, 1, 8, 1, 9, 23, 02, 24, 1, 25, 26, 27, 3, 28, 4, 29, 5, 210, 6, 211, 7, 212, 8, 213, 9, 3, 03, 3, 1, 3, 214, 3, 3, 3, 4, 3, 5, 3, 6, 3, 7, 3, 8, 3, 9, 4, 04, 4, 1, 4, 215, 4, 3, 4, 4, 4, 5, 4, 6, 4, 7, 4, 8, 4, 9, 5, 05, 5, 1, 5, 216, 5, 3, 5,4, 5, 5, 5, 6, 5, 7, 5, 8, 5, 9, 6, 06, 6, 1, 6, 217, 6, 3, 6, 4, 6, 5, 6, 6, 6, 7, 6, 8, 6, 9, 7, 07, 7, 1, 7, 218, 7, 3, 7, 4, 7, 5, 7, 6, 7, 7, 7, 8, 7, 9, 8, 08, 8, 1, 8, 219, 8, 3, 8, 4, 8, 5, 8, 6, 8, 7, 8, 8, 8, 9, 9, 09, 9, 1, 9, 220, 9, 3, 9, 4, 9, 5, 9, 6, 9, 7, 9, 8, 9, 9, 1, 010, 011, 1, 012, 1, 1, 013, 221, 1, 014, 3, 1, 015, 4, 1, 016, 5, 1, 017, 6, 1, 018, 7, 1, 019, 8, 1, 020, 9, 1, 1, 021

So count(2) = count(0) = 1 at digit 11 of the digit-line in the 0 of what was originally 10. And count(2) = count(0) = 21 @ digit 222 in the 0 of what was originally 110. Is a pattern starting to emerge? Yes, it is. Here are the first few points at which the count(2) = count(0) in the digit-line of base 10:

1 @ 11 in 10
21 @ 222 in 110
321 @ 3333 in 1110
4321 @ 44444 in 11110
54321 @ 555555 in 111110
654321 @ 6666666 in 1111110
7654321 @ 77777777 in 11111110
87654321 @ 888888888 in 111111110
987654321 @ 9999999999 in 1111111110
10987654321 @ 111111111110 in 11111111110
120987654321 @ 1222222222221 in 111111111110
[...]

The count(2) = count(0) = 321 at position 3333 in the digit-line, and 4321 at position 44444, and 54321 at position 555555, and so on. I don’t understand why these patterns occur, but you can predict the count-and-position of 2s and 0s easily until position 9999999999, after which things become more complicated. Related patterns for 2 and 0 occur in all other bases except binary (which doesn’t have a 2 digit). Here’s base 6:

1 @ 11 in 10 (1 @ 7 in 6)
21 @ 222 in 110 (13 @ 86 in 42)
321 @ 3333 in 1110 (121 @ 777 in 258)
4321 @ 44444 in 11110 (985 @ 6220 in 1554)
54321 @ 555555 in 111110 (7465 @ 46655 in 9330)
1054321 @ 11111110 in 1111110 (54121 @ 335922 in 55986)
12054321 @ 122222221 in 11111110 (380713 @ 2351461 in 335922)
132054321 @ 1333333332 in 111111110 (2620201 @ 16124312 in 2015538)
1432054321 @ 14444444443 in 1111111110 (17736745 @ 108839115 in 12093234)
15432054321 @ 155555555554 in 11111111110 (118513705 @ 725594110 in 72559410)
205432054321 @ 2111111111105 in 111111111110 (783641641 @ 4788921137 in 435356466)
2205432054321 @ 22222222222220 in 1111111111110 (5137206313 @ 31345665636 in 2612138802)

And what about comparing other pairs of digits? In fact, the count of all digits except 0 matches infinitely often. To write the numbers 1..9 takes one of each digit (except 0). To write the numbers 1 to 99 takes twenty of each digit (except 0). Here’s the proof:

11, 21, 31, 41, 51, 61, 71, 81, 91, 12, 01, 13, 14, 15, 22, 16, 32, 17, 42, 18, 52, 19, 62, 110, 72, 111, 82, 112, 92, 23, 02, 24, 113, 25, 26, 27, 33, 28, 43, 29, 53, 210, 63, 211, 73, 212, 83, 213, 93, 34, 03, 35, 114, 36, 214, 37, 38, 39, 44, 310, 54, 311, 64, 312, 74, 313, 84, 314, 94, 45, 04, 46, 115, 47, 215, 48, 315, 49, 410, 411, 55, 412, 65, 413, 75, 414, 85, 415, 95, 56, 05, 57, 116, 58, 216, 59, 316, 510, 416, 511, 512, 513, 66, 514, 76, 515, 86, 516, 96, 67, 06, 68, 117, 69, 217, 610, 317, 6
11
, 417, 612, 517, 613, 614, 615, 77, 616, 87, 617, 97, 78, 07, 79, 118, 710, 218, 711, 318, 712, 418, 713, 518, 714, 618, 715, 716, 717, 88, 718, 98, 89, 08, 810, 119, 811, 219, 812, 319, 813, 419, 814, 519, 815, 619, 816, 719, 817, 818, 819, 99, 910, 09, 911, 120, 912, 220, 913, 320, 914, 420, 915, 520, 916, 620, 917, 720, 918, 820, 919, 920

And what about writing 1..999, 1..9999, and so on? If you think about it, for every pair of non-zero digits, d1 and d2, all numbers containing one digit can be matched with a number containing the other. 100 → 200, 111 → 222, 314 → 324, 561189571 → 562289572, and so on. So in counting 1..999, 1..9999, 1..99999, you use the same number of non-zero digits. And once again a pattern emerges:

count(0) = 0; count(1) = 1; count(2) = 1; count(3) = 1; count(4) = 1; count(5) = 1; count(6) = 1; count(7) = 1; count(8) = 1; count(9) = 1 (writing 1..9)
count(0) = 9; count(1) = 20; count(2) = 20; count(3) = 20; count(4) = 20; count(5) = 20; count(6) = 20; count(7) = 20; count(8) = 20; count(9) = 20 (writing 1..99)
0: 189; 1: 300; 2: 300; 3: 300; 4: 300; 5: 300; 6: 300; 7: 300; 8: 300; 9: 300 (writing 1..999)
0: 2889; 1: 4000; 2: 4000; 3: 4000; 4: 4000; 5: 4000; 6: 4000; 7: 4000; 8: 4000; 9: 4000 (writing 1..9999)
0: 38889; 1: 50000; 2: 50000; 3: 50000; 4: 50000; 5: 50000; 6: 50000; 7: 50000; 8: 50000; 9: 50000 (writing 1..99999)
0: 488889; 1: 600000; 2: 600000; 3: 600000; 4: 600000; 5: 600000; 6: 600000; 7: 600000; 8: 600000; 9: 600000 (writing 1..999999)
0: 5888889; 1: 7000000; 2: 7000000; 3: 7000000; 4: 7000000; 5: 7000000; 6: 7000000; 7: 7000000; 8: 7000000; 9: 7000000 (writing 1..9999999)
[...]

And here’s base 6 again:

0: 0; 1: 1; 2: 1; 3: 1; 4: 1; 5: 1 (writing 1..5)
0: 5; 1: 20; 2: 20; 3: 20; 4: 20; 5: 20 (writing 1..55 in base 6)
0: 145; 1: 300; 2: 300; 3: 300; 4: 300; 5: 300 (writing 1..555)
0: 2445; 1: 4000; 2: 4000; 3: 4000; 4: 4000; 5: 4000 (writing 1..5555)
0: 34445; 1: 50000; 2: 50000; 3: 50000; 4: 50000; 5: 50000 (writing 1..55555)
0: 444445; 1: 1000000; 2: 1000000; 3: 1000000; 4: 1000000; 5: 1000000 (writing 1..555555)
0: 5444445; 1: 11000000; 2: 11000000; 3: 11000000; 4: 11000000; 5: 11000000 (writing 1..5555555)
0: 104444445; 1: 120000000; 2: 120000000; 3: 120000000; 4: 120000000; 5: 120000000 (writing 1..55555555)
0: 1144444445; 1: 1300000000; 2: 1300000000; 3: 1300000000; 4: 1300000000; 5: 1300000000 (writing 1..555555555)

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)

B a Pal

As a keyly committed core component of the counter-cultural community (I wish!), I like to post especially edgy and esoteric material to Overlord In Terms of Core Issues Around Maximal Engagement with Key Notions of the Über-Feral on the 23rd of each month. And today I may be posting the especially edgiest and esoterickest material ever dot dot dot

After all, this entry at the Online Encyclopedia of Integer Sequences is about numbers that are palindromes in two particularly pertinent bases:

A060792 Numbers that are palindromic in bases 2 and 3.

0, 1, 6643, 1422773, 5415589, 90396755477, 381920985378904469, 1922624336133018996235, 2004595370006815987563563, 8022581057533823761829436662099, 392629621582222667733213907054116073, 32456836304775204439912231201966254787, 428027336071597254024922793107218595973 (A060792 at OEIS, with more entries)


And here are the underlying palindromes:

0: 0 ↔ 0
1: 1 ↔ 1
6643: 1100111110011 ↔ 100010001
1422773: 101011011010110110101 ↔ 2200021200022
5415589: 10100101010001010100101 ↔ 101012010210101
90396755477: 1010100001100000100010000011000010101 ↔ 22122022220102222022122
381920985378904469: 10101001100110110110001110011011001110001101101100110010101 ↔ 2112200222001222121212221002220022112
1922624336133018996235: 11010000011100111000101110001110011011001110001110100011100111000001011 ↔
122120102102011212112010211212110201201021221
2004595370006815987563563: 110101000011111010101010100101111011110111011110111101001010101010111110000101011 ↔ 221010112100202002120002212200021200202001211010122
8022581057533823761829436662099: 1100101010000100101101110000011011011111111011000011100001101111111101101100000111011010010000101010011 ↔ 21000020210011222122220212010000100001021202222122211001202000012
392629621582222667733213907054116073: 10010111001111000100010100010100000011011011000101011011100000111011010100011011011000000101000101000100011110011101001 ↔ 122102120011102000101101000002010021111120010200000101101000201110021201221
32456836304775204439912231201966254787: 11000011010101111010110010100010010011011010101001101000001000100010000010110010101011011001001000101001101011110101011000011 ↔ 1222100201002211120110022121002012121101011212102001212200110211122001020012221
428027336071597254024922793107218595973: 101000010000000110001000011111100101011110011100001110100011100010001110001011100001110011110101001111110000100011000000010000101 ↔ 222001200110022102121001000200200202022111220202002002000100121201220011002100222

Performativizing the Polygonic #2

Suppose a café offers you free drinks for three days. You can have tea or coffee in any order and any number of times. If you want tea every day of the three, you can have it. So here’s a question: how many ways can you choose from two kinds of drink in three days? One simple way is to number each drink, tea = 1, coffee = 2, then count off the choices like this:


1: 111
2: 112
3: 121
4: 122
5: 211
6: 212
7: 221
8: 222

Choice #1 is 111, which means tea every day. Choice #6 is 212, which means coffee on day 1, tea on day 2 and coffee on day 3. Now look at the counting again and the way the numbers change: 111, 112, 121, 122, 211… It’s really base 2 using 1 and 2 rather than 0 and 1. That’s why there are 8 ways to choose two drinks over three days: 8 = 2^3. Next, note that you use the same number of 1s to count the choices as the number of 2s. There are twelve 1s and twelve 2s, because each number has a mirror: 111 has 222, 112 has 221, 121 has 212, and so on.

Now try the number of ways to choose from three kinds of drink (tea, coffee, orange juice) over two days:


11, 12, 13, 21, 22, 23, 31, 32, 33 (c=9)

There are 9 ways to choose, because 9 = 3^2. And each digit, 1, 2, 3, is used exactly six times when you write the choices. Now try the number of ways to choose from three kinds of drink over three days:


111, 112, 113, 121, 122, 123, 131, 132, 133, 211, 212, 213, 221, 222, 223, 231, 232, 233, 311, 312, 313, 321, 322, 323, 331, 332, 333 (c=27)

There are 27 ways and (by coincidence) each digit is used 27 times to write the choices. Now try three drinks over four days:


1111, 1112, 1113, 1121, 1122, 1123, 1131, 1132, 1133, 1211, 1212, 1213, 1221, 1222, 1223, 1231, 1232, 1233, 1311, 1312, 1313, 1321, 1322, 1323, 1331, 1332, 1333, 2111, 2112, 2113, 2121, 2122, 2123, 2131, 2132, 2133, 2211, 2212, 2213, 2221, 2222, 2223, 2231, 2232, 2233, 2311, 2312, 2313, 2321, 2322, 2323, 2331, 2332, 2333, 3111, 3112, 3113, 3121, 3122, 3123, 3131, 3132, 3133, 3211, 3212, 3213, 3221, 3222, 3223, 3231, 3232, 3233, 3311, 3312, 3313, 3321, 3322, 3323, 3331, 3332, 3333 (c=81)

There are 81 ways to choose and each digit is used 108 times. But the numbers don’t have represent choices of drink in a café. How many ways can a point inside an equilateral triangle jump four times half-way towards the vertices of the triangle? It’s the same as the way to choose from three drinks over four days. And because the point jumps toward each vertex in a symmetrical way the same number of times, you get a nice even pattern, like this:

vertices = 3, jump = 1/2


Every time the point jumps half-way towards a particular vertex, its position is marked in a unique colour. The fractal, also known as a Sierpiński triangle, actually represents all possible choices for an indefinite number of jumps. Here’s the same rule applied to a square. There are four vertices, so the point is tracing all possible ways to choose four vertices for an indefinite number of jumps:

v = 4, jump = 1/2


As you can see, it’s not an obvious fractal. But what if the point jumps two-thirds of the way to its target vertex and an extra target is added at the centre of the square? This attractive fractal appears:

v = 4 + central target, jump = 2/3


If the central target is removed and an extra target is added on each side, this fractal appears:

v = 4 + 4 midpoints, jump = 2/3


That fractal is known as a Sierpiński carpet. Now up to the pentagon. This fractal of endlessly nested contingent pentagons is created by a point jumping 1/φ = 0·6180339887… of the distance towards the five vertices:

v = 5, jump = 1/φ


With a central target in the pentagon, this fractal appears:

v = 5 + central, jump = 1/φ


The central red pattern fits exactly inside the five that surround it:

v = 5 + central, jump = 1/φ (closeup)


v = 5 + c, jump = 1/φ (animated)


For a fractal of endlessly nested contingent hexagons, the jump is 2/3:

v = 6, jump = 2/3


With a central target, you get a filled variation of the hexagonal fractal:

v = 6 + c, jump = 2/3


And for a fractal of endlessly nested contingent octagons, the jump is 1/√2 = 0·7071067811… = √½:

v = 8, jump = 1/√2


Previously pre-posted:

Performativizing the Polygonic