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.

The Glamor of Gamma

The factorial function, n!, is easy to understand. You simply take an integer and multiply it by all integers smaller than it (by convention, 0! = 1):

0! = 1
1! = 1
2! = 2 = 2*1
3! = 6 = 3*2*1
4! = 24 = 4*3*2*1
5! = 120 = 5*4*3*2*1
6! = 720 = 6*120 = 6*5!
7! = 5040
8! = 40320
9! = 362880
10! = 3628800
11! = 39916800
12! = 479001600
13! = 6227020800
14! = 87178291200
15! = 1307674368000
16! = 20922789888000
17! = 355687428096000
18! = 6402373705728000
19! = 121645100408832000
20! = 2432902008176640000

The gamma function, Γ(n), isn’t so easy to understand. It allows you to find the factorials of not just the integers, but everything between the integers, like fractions, square roots, and transcendental numbers like π. Don’t ask me how! And don’t ask me how you get this very beautiful and unexpected result:

Γ(1/2) = √π = 1.77245385091...

But a blog called Mathematical Enchantments can tell you more:

The Square Root of Pi


Post-Performative Post-Scriptum

glamour | glamor, n. Originally Scots, introduced into the literary language by Scott. A corrupt form of grammar n.; for the sense compare gramarye n. (and French grimoire ), and for the form glomery n. 1. Magic, enchantment, spell; esp. in the phrase to cast the glamour over one. 2. a. A magical or fictitious beauty attaching to any person or object; a delusive or alluring charm. b. Charm; attractiveness; physical allure, esp. feminine beauty; frequently attributive colloquial (originally U.S.). — Oxford English Dictionary

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

Pi and By

Here’s √2 in base 2:

√2 = 1.01101010000010011110... (base=2)

And in base 3:

√2 = 1.10201122122200121221... (base=3)

And in bases 4, 5, 6, 7, 8, 9 and 10:

√2 = 1.12220021321212133303... (b=4)
√2 = 1.20134202041300003420... (b=5)
√2 = 1.22524531420552332143... (b=6)
√2 = 1.26203454521123261061... (b=7)
√2 = 1.32404746317716746220... (b=8)
√2 = 1.36485805578615303608... (b=9)
√2 = 1.41421356237309504880... (b=10)

And here’s π in the same bases:

π = 11.00100100001111110110... (b=2)
π = 10.01021101222201021100... (b=3)
π = 03.02100333122220202011... (b=4)
π = 03.03232214303343241124... (b=5)
π = 03.05033005141512410523... (b=6)
π = 03.06636514320361341102... (b=7)
π = 03.11037552421026430215... (b=8)
π = 03.12418812407442788645... (b=9)
π = 03.14159265358979323846... (b=10)

Mathematicians know that in all standard bases, the digits of √2 and π go on for ever, without falling into any regular pattern. These numbers aren’t merely irrational but transcedental. But are they also normal? That is, in each base b, do the digits 0 to [b-1] occur with the same frequency 1/b? (In general, a sequence of length l will occur in a normal number with frequency 1/(b^l).) In base 2, are there as many 1s as 0s in the digits of √2 and π? In base 3, are there as many 2s as 1s and 0s? And so on.

It’s a simple question, but so far it’s proved impossible to answer. Another question starts very simple but quickly gets very difficult. Here are the answers so far at the Online Encyclopedia of Integer Sequences (OEIS):

2, 572, 8410815, 59609420837337474 – A049364

The sequence is defined as the “Smallest number that is digitally balanced in all bases 2, 3, … n”. In base 2, the number 2 is 10, which has one 1 and one 0. In bases 2 and 3, 572 = 1000111100 and 210012, respectively. 1000111100 has five 1s and five 0s; 210012 has two 2s, two 1s and two 0s. Here are the numbers of A049364 in the necessary bases:

10 (n=2)
1000111100, 210012 (n=572)
100000000101011010111111, 120211022110200, 200011122333 (n=8410815)
11010011110001100111001111010010010001101011100110000010, 101201112000102222102011202221201100, 3103301213033102101223212002, 1000001111222333324244344 (n=59609420837337474)

But what number, a(6), satisfies the definition for bases 2, 3, 4, 5 and 6? According to the notes at the OEIS, a(6) > 5^434. That means finding a(6) is way beyond the power of present-day computers. But I assume a quantum computer could crack it. And maybe someone will come up with a short-cut or even an algorithm that supplies a(b) for any base b. Either way, I think we’ll get there, π and by.

Mod’s Chosen

When you divide one integer by another, one of two things happens. Either the second number goes perfectly into the first or there’s a remainder:


15 / 5 = 3
18 / 5 = 3⅗

In the first case, there’s no remainder, that is, the remainder is 0. In the second case, there’s a remainder of 3. And all that gives you the basis for what’s called modular arithmetic. It returns the remainder when one number is divided by another:


15 mod 5 = 0
16 mod 5 = 1
17 mod 5 = 2
18 mod 5 = 3
19 mod 5 = 4
20 mod 5 = 0
21 mod 5 = 1
22 mod 5 = 2...

It looks simple but a lot of mathematics is built on it. I don’t know much of that maths, but I know one thing I like: the patterns you can get from modular arithmetic. Suppose you draw a square, then find a point and measure the distances from that point to all the vertices of the square. Then add the distances up, turn the result into an integer if necessary, and test whether the result is divisible by 2 or not. If it is divisible, colour the point in. If it isn’t, leave the point blank.

Then move on to another point and perform the same test. This is modular arithematic, because for each point you’re asking whether d mod 2 = 0. The result looks like this:

d mod 2 = 0


Here are more divisors:

d mod 3 = 0


d mod 4 = 0


d mod 5 = 0


d mod 6 = 0


d mod 7 = 0


d mod 8 = 0


d mod 9 = 0


d mod 10 = 0


d mod various = 0 (animated)


You can also use modular arithmetic to determine the colour of the points. For example, if d mod n = 0, the point is black; if d mod n = 1, the point is red; if d mod n = 2, the point is green; and so on.

d mod 3 = 0, 1, 2 (coloured)


d mod 4 = 0, 1, 2, 3 (coloured)


d mod 5 = 0, 1, 2, 3, 4 (coloured)



d mod 5 = 0, 1, 2, 3, 4 (animated and expanding)


Zequality Now

Here are the numbers one to eight in base 2:

1, 10, 11, 100, 101, 110, 111, 1000…

Now see what happens when you count the zeroes:


1, 10[1], 11, 10[2]0[3], 10[4]1, 110[5], 111, 10[6]0[7]0[8]...

In base 2, the numbers one to eight contain exactly eight zeroes, that is, zerocount(1..8,b=2) = 8. But it doesn’t work out so exactly in base 3:


1, 2, 10[1], 11, 12, 20[2], 21, 22, 10[3]0[4], 10[5]1, 10[6]2, 110[7], 111, 112, 120[8], 121, 122, 20[9]0[10], 20[11]1, 20[12]2, 210[13], 211, 212, 220[14], 221, 222, 10[15]0[16]0[17], 10[18]0[19]1, 10[20]0[21]2, 10[22]10[23], 10[24]11, 10[25]12, 10[26]20[27], 10[28]21, 10[29]22, 110[30]0[31], 110[32]1, 110[33]2, 1110[34], 1111, 1112, 1120[35], 1121, 1122, 120[36]0[37], 120[38]1, 120[39]2, 1210[40], 1211, 1212, 1220[41], 1221, 1222, 20[42]0[43]0[44], 20[45]0[46]1, 20[47]0[48]2, 20[49]10[50], 20[51]11, 20[52]12, 20[53]20[54], 20[55]21, 20[56]22, 210[57]0[58], 210[59]1, 210[60]2, 2110[61], 2111, 2112, 2120[62], 2121, 2122, 220[63]0[64], 220[65]1, 220[66]2, 2210[67], 2211, 2212, 2220[68], 2221, 2222, 10[69]0[70]0[71]0[72], 10[73]0[74]0[75]1, 10[76]0[77]0[78]2, 10[79]0[80]10[81], 10[82]0[83]11, 10[84]0[85]12, 10[86]0[87]20[88]...

In base 3, 10020 = 87 and zerocount(1..87,b=3) = 88. And what about base 4? zerocount(1..1068,b=4) = 1069 (n=100,230 in base 4). After that, zerocount(1..16022,b=5) = 16023 (n=1,003,043 in base 5) and zerocount(1..284704,b=6) = 284,705 (n=10,034,024 in base 6).

The numbers are getting bigger fast and it’s becoming increasingly impractible to count the zeroes individually. What you need is an algorithm that will take any given n and work out how many zeroes are required to write the numbers 1 to n. The simplest way to do this is to work out how many times 0 has appeared in each position of the number. The 1s position is easy: you simply divide the number by the base and discard the remainder. For example, in base 10, take the number 25. The 0 must have appeared in the 1s position twice, for 10 and 20, so zerocount(1..25) = 25 \ 10 = 2. In 2017, the 0 must have appeared in the 1s position 201 times = 2017 \ 10. And so on.

It gets a little trickier for the higher positions, the 10s, 100s, 1000s and so on, but the same basic principle applies. And so you can easily create an algorithm that takes a number, n, and produces zerocount(1..n) in a particular base. With this algorithm, you can quickly find zerocount(1..n) >= n in higher bases:


zerocount(1..1000,b=2) = 1,000 (n=8)*
zerocount(1..10020,b=3) = 10,021 (n=87)
zerocount(1..100230,b=4) = 100,231 (n=1,068)
zerocount(1..1003042,b=5) = 1,003,043 (n=16,022)
zerocount(1..10034024,b=6) = 10,034,025 (n=284,704)
zerocount(1..100405550,b=7) = 100,405,551 (n=5,834,024)
zerocount(1..1004500236,b=8) = 1,004,500,237 (n=135,430,302)
zerocount(1..10050705366,b=9) = 10,050,705,367 (n=3,511,116,537)
zerocount(1..100559404366,b=10) = 100,559,404,367
zerocount(1..1006083A68919,b=11) = 1,006,083,A68,919 (n=3,152,738,985,031)*
zerocount(1..10066AA1430568,b=12) = 10,066,AA1,430,569 (n=107,400,330,425,888)
zerocount(1..1007098A8719B81,b=13) = 100,709,8A8,719,B81 (n=3,950,024,143,546,664)*
zerocount(1..10077C39805D81C7,b=14) = 1,007,7C3,980,5D8,1C8 (n=155,996,847,068,247,395)
zerocount(1..10080B0034AA5D16D,b=15) = 10,080,B00,34A,A5D,171 (n=6,584,073,072,068,125,453)
zerocount(1..10088DBE29597A6C77,b=16) = 100,88D,BE2,959,7A6,C77 (n=295,764,262,988,176,583,799)*
zerocount(1..10090C5309AG72CBB3F,b=17) = 1,009,0C5,309,AG7,2CB,B3G (n=14,088,968,131,538,370,019,982)
zerocount(1..10099F39070FC73C1G73,b=18) = 10,099,F39,070,FC7,3C1,G75 (n=709,394,716,006,812,244,474,473)
zerocount(1..100A0DC1258614CA334EB,b=19) = 100,A0D,C12,586,14C,A33,4EC (n=37,644,984,315,968,494,382,106,708)
zerocount(1..100AAGDEEB536IBHE87006,b=20) = 1,00A,AGD,EEB,536,IBH,E87,008 (n=2,099,915,447,874,594,268,014,136,006)

And you can also easily find the zequal numbers, that is, the numbers n for which, in some base, zerocount(1..n) exactly equals n:


zerocount(1..1000,b=2) = 1,000 (n=8)
zerocount(1..1006083A68919,b=11) = 1,006,083,A68,919 (n=3,152,738,985,031)
zerocount(1..1007098A8719B81,b=13) = 100,709,8A8,719,B81 (n=3,950,024,143,546,664)
zerocount(1..10088DBE29597A6C77,b=16) = 100,88D,BE2,959,7A6,C77 (n=295,764,262,988,176,583,799)
zerocount(1..100CCJFFAD4MI409MI0798CJB3,b=24) = 10,0CC,JFF,AD4,MI4,09M,I07,98C,JB3 (n=32,038,681,563,209,056,709,427,351,442,469,835)
zerocount(1..100DDL38CIO4P9K0AJ7HK74EMI7L,b=26) = 1,00D,DL3,8CI,O4P,9K0,AJ7,HK7,4EM,I7L (n=160,182,333,966,853,031,081,693,091,544,779,177,187)
zerocount(1..100EEMHG6OE8EQKO0BF17LCCIA7GPE,b=28) = 100,EEM,HG6,OE8,EQK,O0B,F17,LCC,IA7,GPE (n=928,688,890,453,756,699,447,122,559,347,771,300,777,482)
zerocount(1..100F0K7MQO6K9R1S616IEEL2JRI73PF,b=29) = 1,00F,0K7,MQO,6K9,R1S,616,IEE,L2J,RI7,3PF (n=74,508,769,042,363,852,559,476,397,161,338,769,391,145,562)
zerocount(1..100G0LIL0OQLF2O0KIFTK1Q4DC24HL7BR,b=31) = 100,G0L,IL0,OQL,F2O,0KI,FTK,1Q4,DC2,4HL,7BR (n=529,428,987,529,739,460,369,842,168,744,635,422,842,585,510,266)
zerocount(1..100H0MUTQU3A0I5005WL2PD7T1ASW7IV7NE,b=33) = 10,0H0,MUT,QU3,A0I,500,5WL,2PD,7T1,ASW,7IV,7NE (n=4,262,649,311,868,962,034,947,877,223,846,561,239,424,294,726,563,632)
zerocount(1..100HHR387RQHK9OP6EDBJEUDAK35N7MN96LB,b=34) = 100,HHR,387,RQH,K9O,P6E,DBJ,EUD,AK3,5N7,MN9,6LB (n=399,903,937,958,473,433,782,862,763,628,747,974,628,490,691,628,136,485)
zerocount(1..100IISLI0CYX2893G9E8T4I7JHKTV41U0BKRHT,b=36) = 10,0II,SLI,0CY,X28,93G,9E8,T4I,7JH,KTV,41U,0BK,RHT (n=3,831,465,379,323,568,772,890,827,210,355,149,992,132,716,389,119,437,755,185)
zerocount(1..100LLX383BPWE[40]ZL0G1M[40]1OX[39]67KOPUD5C[40]RGQ5S6W9[36],b=42) = 10,0LL,X38,3BP,WE[40],ZL0,G1M,[40]1O,X[39]6,7KO,PUD,5C[40],RGQ,5S6,W9[36] (n=6,307,330,799,917,244,669,565,360,008,241,590,852,337,124,982,231,464,556,869,653,913,711,854)
zerocount(1..100MMYPJ[38]14KDV[37]OG[39]4[42]X75BE[39][39]4[43]CK[39]K36H[41]M[37][43]5HIWNJ,b=44) = 1,00M,MYP,J[38]1,4KD,V[37]O,G[39]4,[42]X7,5BE,[39][39]4,[43]CK,[39]K3,6H[41],M[37][43],5HI,WNJ (n=90,257,901,046,284,988,692,468,444,260,851,559,856,553,889,199,511,017,124,021,440,877,333,751,943)
zerocount(1..100NN[36]3813[38][37]16F6MWV[41]UBNF5FQ48N0JRN[40]E76ZOHUNX2[42]3[43],b=46) = 100,NN[36],381,3[38][37],16F,6MW,V[41]U,BNF,5FQ,48N,0JR,N[40]E,76Z,OHU,NX2,[42]3[43] (n=1,411,636,908,622,223,745,851,790,772,948,051,467,006,489,552,352,013,745,000,752,115,904,961,213,172,605)
zerocount(1..100O0WBZO9PU6O29TM8Y0QE3I[37][39]A7E4YN[44][42]70[44]I[46]Z[45][37]Q2WYI6,b=47) = 1,00O,0WB,ZO9,PU6,O29,TM8,Y0Q,E3I,[37][39]A,7E4,YN[44],[42]70,[44]I[46],Z[45][37],Q2W,YI6 (n=182,304,598,281,321,725,937,412,348,242,305,189,665,300,088,639,063,301,010,710,450,793,661,266,208,306,996)
zerocount(1..100PP[39]37[49]NIYMN[43]YFE[44]TDTJ00EAEIP0BIDFAK[46][36]V6V[45]M[42]1M[46]SSZ[40],b=50) = 1,00P,P[39]3,7[49]N,IYM,N[43]Y,FE[44],TDT,J00,EAE,IP0,BID,FAK,[46][36]V,6V[45],M[42]1,M[46]S,SZ[40] (n=444,179,859,561,011,965,929,496,863,186,893,220,413,478,345,535,397,637,990,204,496,296,663,272,376,585,291,071,790)
zerocount(1..100Q0Y[46][44]K[49]CKG[45]A[47]Z[43]SPZKGVRN[37]2[41]ZPP[36]I[49][37]EZ[38]C[44]E[46]00CG[38][40][48]ROV,b=51) = 10,0Q0,Y[46][44],K[49]C,KG[45],A[47]Z,[43]SP,ZKG,VRN,[37]2[41],ZPP,[36]I[49],[37]EZ,[38]C[44],E[46]0,0CG,[38][40][48],ROV (n=62,191,970,278,446,971,531,566,522,791,454,395,351,613,891,150,548,291,266,262,575,754,206,359,828,753,062,692,619,547)
zerocount(1..100QQ[40]TL[39]ZA[49][41]J[41]7Q[46]4[41]66A1E6QHHTM9[44]8Z892FRUL6V[46]1[38][41]C[40][45]KB[39],b=52) = 100,QQ[40],TL[39],ZA[49],41]J[41],7Q[46],4[41]6,6A1,E6Q,HHT,M9[44],8Z8,92F,RUL,6V[46],1[38][41],C[40][45],KB[39] (n=8,876,854,501,927,007,077,802,489,292,131,402,136,556,544,697,945,824,257,389,527,114,587,644,068,732,794,430,403,381,731)
zerocount(1..100S0[37]V[53]Y6G[51]5J[42][38]X[40]XO[38]NSZ[42]XUD[47]1XVKS[52]R[39]JAHH[49][39][50][54]5PBU[42]H3[45][46]DEJ,b=55) = 100,S0[37],V[53]Y,6G[51],5J[42],[38]X[40],XO[38],NSZ,[42]XU,D[47]1,XVK,S[52]R,[39]JA,HH[49],[39][50][54],5PB,U[42]H,3[45][46],DEJ (n=28,865,808,580,366,629,824,612,818,017,012,809,163,332,327,132,687,722,294,521,718,120,736,868,268,650,080,765,802,786,141,387,114)

Autonomata

“Describe yourself.” You can say it to people. And you can say it to numbers too. For example, here’s the number 3412 describing the positions of its own digits, starting at 1 and working upward:


3412 – the 1 is in the 3rd position, the 2 is in the 4th position, the 3 is in the 1st position, and the 4 is in the 2nd position.

In other words, the positions of the digits 1 to 4 of 3412 recreate its own digits:


3412 → (3,4,1,2) → 3412

The number 3412 describes itself – it’s autonomatic (from Greek auto, “self” + onoma, “name”). So are these numbers:


1
21
132
2143
52341
215634
7243651
68573142
321654798

More precisely, they’re panautonomatic numbers, because they describe the positions of all their own digits (Greek pan or panto, “all”). But what if you use the positions of only, say, the 1s or the 3s in a number? In base ten, only one number describes itself like that: 1. But we’re not confined to base 10. In base 2, the positions of the 1s in 110 (= 6) are 1 and 10 (= 2). So 110 is monautonomatic in binary (Greek mono, “single”). 10 is also monautonomatic in binary, if the digit being described is 0: it’s in 2nd position or position 10 in binary. These numbers are monoautonomatic in binary too:


110100 = 52 (digit = 1)
10100101111 = 1327 (d=0)

In 110100, the 1s are in 1st, 2nd and 4th position, or positions 1, 10, 100 in binary. In 10100101111, the 0s are in 2nd, 4th, 5th and 7th position, or positions 10, 100, 101, 111 in binary. Here are more monautonomatic numbers in other bases:


21011 in base 4 = 581 (digit = 1)
11122122 in base 3 = 3392 (d=2)
131011 in base 5 = 5131 (d=1)
2101112 in base 4 = 9302 (d=1)
11122122102 in base 3 = 91595 (d=2)
13101112 in base 5 = 128282 (d=1)
210111221 in base 4 = 148841 (d=1)

For example, in 131011 the 1s are in 1st, 3rd, 5th and 6th position, or positions 1, 3, 10 and 11 in quinary. But these numbers run out quickly and the only monautonomatic number in bases 6 and higher is 1. However, there are infinitely long monoautonomatic integer sequences in all bases. For example, in binary this sequence at the Online Encyclopedia of Integer Sequences describes itself using the positions of its 1s:


A167502: 1, 10, 100, 111, 1000, 1001, 1010, 1110, 10001, 10010, 10100, 10110, 10111, 11000, 11010, 11110, 11111, 100010, 100100, 100110, 101001, 101011, 101100, 101110, 110000, 110001, 110010, 110011, 110100, 111000, 111001, 111011, 111101, 11111, …

In base 10, it looks like this:


A167500: 1, 2, 4, 7, 8, 9, 10, 14, 17, 18, 20, 22, 23, 24, 26, 30, 31, 34, 36, 38, 41, 43, 44, 46, 48, 49, 50, 51, 52, 56, 57, 59, 61, 62, 63, 64, 66, 67, 68, 69, 70, 71, 75, 77, 80, 83, 86, 87, 89, 91, 94, 95, 97, 99, 100, 101, 103, 104, 107, 109, 110, 111, 113, 114, 119, 120, 124, … (see A287515 for a similar sequence using 0s)

Square on a Three String

222 A.D. was the year in which the Emperor Heliogabalus was assassinated by his own soldiers. Exactly 1666 years later, the Anglo-Dutch classicist Sir Lawrence Alma-Tadema exhibited his painting The Roses of Heliogabalus (1888). I suggested in “Roses Are Golden” that Alma-Tadema must have chosen the year as deliberately as he chose the dimensions of his canvas, which, at 52″ x 84 1/8“, is an excellent approximation to the golden ratio.

But did Alma-Tadema know that lines at 0º and 222º divide a circle in the golden ratio? He could easily have done, just as he could easily have known that 222 precedes the 48th prime, 223. But it is highly unlikely that he knew that 223 yields a magic square whose columns, rows and diagonals all sum to 222. To create the square, simply list the 222 multiples of the reciprocal 1/223 in base 3, or ternary. The digits of the reciprocal repeat after exactly 222 digits and its multiples begin and end like this:

001/223 = 0.00001002102101021212111012022211122022... in base 3
002/223 = 0.00002011211202120201222101122200021121...
003/223 = 0.00010021021010212121110120222111220221...
004/223 = 0.00011100200112011110221210022100120020...
005/223 = 0.00012110002220110100102222122012012120...

[...]

218/223 = 0.22210112220002112122120000100210210102... in base 3
219/223 = 0.22211122022110211112001012200122102202...
220/223 = 0.22212201201212010101112102000111002001...
221/223 = 0.22220211011020102021000121100022201101...
222/223 = 0.22221220120121201010111210200011100200...

Each column, row and diagonal of ternary digits sums to 222. Here is the full n/223 square represented with 0s in grey, 1s in white and 2s in red:

(Click for larger)


It isn’t difficult to see that the white squares are mirror-symmetrical on a horizontal axis. Here is the symmetrical pattern rotated by 90º:

(Click for larger)


But why should the 1s be symmetrical? This isn’t something special to 1/223, because it happens with prime reciprocals like 1/7 too:

1/7 = 0.010212... in base 3
2/7 = 0.021201...
3/7 = 0.102120...
4/7 = 0.120102...
5/7 = 0.201021...
6/7 = 0.212010...

And you can notice something else: 0s mirror 2s and 2s mirror 0s. A related pattern appears in base 10:

1/7 = 0.142857...
2/7 = 0.285714...
3/7 = 0.428571...
4/7 = 0.571428...
5/7 = 0.714285...
6/7 = 0.857142...

The digit 1 in the decimal digits of n/7 corresponds to the digit 8 in the decimal digits of (7-n)/7; 4 corresponds to 5; 2 corresponds to 7; 8 corresponds to 1; 5 corresponds to 4; and 7 corresponds to 2. In short, if you’re given the digits d1 of n/7, you know the digits d2 of (n-7)/7 by the rule d2 = 9-d1.

Why does that happen? Examine these sums:

 1/7 = 0.142857142857142857142857142857142857142857...
+6/7 = 0.857142857142857142857142857142857142857142...
 7/7 = 0.999999999999999999999999999999999999999999... = 1.0

 2/7 = 0.285714285714285714285714285714285714285714...
+5/7 = 0.714285714285714285714285714285714285714285...
 7/7 = 0.999999999999999999999999999999999999999999... = 1.0

 3/7 = 0.428571428571428571428571428571428571428571...
+4/7 = 0.571428571428571428571428571428571428571428...
 7/7 = 0.999999999999999999999999999999999999999999... = 1.0

And here are the same sums in ternary (where the first seven integers are 1, 2, 10, 11, 12, 20, 21):

  1/21 = 0.010212010212010212010212010212010212010212...
+20/21 = 0.212010212010212010212010212010212010212010...
 21/21 = 0.222222222222222222222222222222222222222222... = 1.0

  2/21 = 0.021201021201021201021201021201021201021201...
+12/21 = 0.201021201021201021201021201021201021201021...
 21/21 = 0.222222222222222222222222222222222222222222... = 1.0

 10/21 = 0.102120102120102120102120102120102120102120...
+11/21 = 0.120102120102120102120102120102120102120102...
 21/21 = 0.222222222222222222222222222222222222222222... = 1.0

Accordingly, in base b with the prime p, the digits d1 of n/p correspond to the digits (p-n)/p by the rule d2 = (b-1)-d1. This explains why the 1s mirror themselves in ternary: 1 = 2-1 = (3-1)-1. In base 5, the 2s mirror themselves by the rule 2 = 4-2 = (5-1) – 2. In all odd bases, some digit will mirror itself; in all even bases, no digit will. The mirror-digit will be equal to (b-1)/2, which is always an integer when b is odd, but never an integer when b is even.

Here are some more examples of the symmetrical patterns found in odd bases:

Patterns of 1s in 1/19 in base 3


Patterns of 6s in 1/19 in base 13


Patterns of 7s in 1/19 in base 15


Elsewhere other-posted:

Roses Are Golden — more on The Roses of Heliogabalus (1888)
Three Is The Key — more on the 1/223 square

T4K1NG S3LF13S

It’s like watching a seed grow. You take a number and count how many 0s it contains, then how many 1s, how many 2s, 3s, 4s and so on. Then you create a new number by writing the count of each digit followed by the digit itself. Then you repeat the process with the new number.

Here’s how it works if you start with the number 1:

1

The count of digits is one 1, so the new number is this:

→ 11

The count of digits for 11 is two 1s, so the next number is:

→ 21

The count of digits for 21 is one 1, one 2, so the next number is:

→ 1112

The count of digits for 1112 is three 1s, one 2, so the next number is:

→ 3112

The count of digits for 3112 is two 1s, one 2, one 3, so the next number is:

→ 211213

What happens after that? Here are the numbers as a sequence:

1 → 11 → 21 → 1112 → 3112 → 211213 → 312213 → 212223 → 114213 → 31121314 → 41122314 → 31221324 → 21322314

That’s all you need, because something interesting happens with 21322314. The digit count is two 1s, three 2s, two 3s, one 4, so the next number is:

→ 21322314

In other words, 21322314 is what might be called a self-descriptive number: it describes the count of its own digits. That’s why I think this procedure is like watching a seed grow. You start with the tiny seed of 1 and end in the giant oak of 21322314, whose factorization is 2 * 3^2 * 13 * 91121. But there are many more self-descriptive numbers in base ten and some of them are much bigger than 21322314. A047841 at the Online Encyclopedia of Integer Sequences lists all 109 of them (and calls them “autobiographical numbers”). Here are a few, starting with the simplest possible:

22 → two 2s → 22
10213223 → one 0, two 1s, three 2s, two 3s → 10213223
10311233 → one 0, three 1s, one 2, three 3s → 10311233
21322314 → two 1s, three 2s, two 3s, one 4 → 21322314
21322315 → two 1s, three 2s, two 3s, one 5 → 21322315
21322316 → two 1s, three 2s, two 3s, one 6 → 21322316*
1031223314 → one 0, three 1s, two 2s, three 3s, one 4 → 10
31223314
3122331415 → three 1s, two 2s, three 3s, one 4, one 5
→ 3122331415
3122331416 → three 1s, two 2s, three 3s, one 4, one 6
→ 3122331416*

*And for 21322317, 21322318, 21322319; 3122331417, 3122331418, 3122331419.


And here’s what happens when you seed a sequence with a number containing all possible digits in base ten:

1234567890 → 10111213141516171819 → 101111213141516171819 → 101211213141516171819 → 101112213141516171819

That final number is self-descriptive:

101112213141516171819 → one 0, eleven 1s, two 2s, one 3, one 4, one 5, one 6, one 7, one 8, one 9 → 101112213141516171819

So some numbers are self-descriptive and some start a sequence that ends in a self-descriptive number. But that doesn’t exhaust the possibilities. Some numbers are part of a loop:

103142132415 → 104122232415 → 103142132415
104122232415 → 103142132415 → 104122232415
1051421314152619 → 1061221324251619 → 1051421314152619…
5142131415261819 → 6122132425161819 → 5142131415261819
106142131416271819 → 107122132426171819 → 106142131416271819


10512223142518 → 10414213142518 → 10512213341518 → 10512223142518
51222314251718 → 41421314251718 → 51221334151718 →
51222314251718

But all that is base ten. What about other bases? In fact, nearly all self-descriptive numbers in base ten are also self-descriptive in other bases. An infinite number of other bases, in fact. 22 is a self-descriptive number for all b > 2. The sequence seeded with 1 is identical in all b > 4:

1 → 11 → 21 → 1112 → 3112 → 211213 → 312213 → 212223 → 114213 → 31121314 → 41122314 → 31221324 → 2132231421322314

In bases 2, 3 and 4, the sequence seeded with 1 looks like this:

1 → 11 → 101 → 10101 → 100111 → 1001001 → 1000111 → 11010011101001… (b=2) (1101001[2] = 105 in base 10)
1 → 11 → 21 → 1112 → 10112 → 1010112 → 2011112 → 10111221011122… (b=3) (1011122[3] = 854 in base 10)
1 → 11 → 21 → 1112 → 3112 → 211213 → 312213 → 212223 → 1110213 → 101011213 → 201111213 → 101112213101112213… (b=4) (101112213[4] = 71079 in base 10)

In base 2 there are only two self-descriptive numbers (and no loops):

111 → three 1s → 111… (b=2) (111 = 7 in base 10)
1101001 → three 0s, four 1s → 1101001… (b=2) (1101001 = 105 in base 10)

So if you apply the “count digits” procedure in base 2, all numbers, except 111, begin a sequence that ends in 1101001. Base 3 has a few more self-descriptive numbers and also some loops:

2222… (b >= 3)
10111 → one 0, four 1s → 10111… (b=3)
11112 → four 1s, one 2 → 11112
100101 → three 0s, three 1s → 100101… (b=3)
1011122 → one 0, four 1s, two 2s → 1011122… (b=3)
2021102 → two 0s, two 1s, three 2s → 2021102… (b=3)
10010122 → three 0s, three 1s, two 2s → 10010122


2012112 → 10101102 → 10011112 → 2012112
10011112 → 2012112 → 10101102 → 10011112
10101102 → 10011112 → 2012112 → 10101102

A question I haven’t been able to answer: Is there a base in which loops can be longer than these?

103142132415 → 104122232415 → 103142132415
10512223142518 → 10414213142518 → 10512213341518 → 10512223142518

A question I have been able to answer: What is the sequence when it’s seeded with the title of this blog-post? T4K1NGS3LF13S is a number in all bases >= 30 and its base-30 form equals 15,494,492,743,722,316,018 in base 10 (with the factorization 2 * 72704927 * 106557377767). If T4K1NGS3LF13S seeds a sequence in any b >= 30, the result looks like this:

T4K1NGS3LF13S → 2123141F1G1K1L1N2S1T → 813213141F1G1K1L1N1S1T → A1122314181F1G1K1L1N1S1T → B1221314181A1F1G1K1L1N1S1T → C1221314181A1B1F1G1K1L1N1S1T → D1221314181A1B1C1F1G1K1L1N1S1T → E1221314181A1B1C1D1F1G1K1L1N1S1T → F1221314181A1B1C1D1E1F1G1K1L1N1S1T → G1221314181A1B1C1D1E2F1G1K1L1N1S1T → F1321314181A1B1C1D1E1F2G1K1L1N1S1T → F1222314181A1B1C1D1E2F1G1K1L1N1S1T → E1421314181A1B1C1D1E2F1G1K1L1N1S1T → F1221324181A1B1C1D2E1F1G1K1L1N1S1T → E1421314181A1B1C1D1E2F1G1K1L1N1S1T