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.

Gyp Cip

Abundance often overwhelms, but restriction reaps riches. That’s true in mathematics and science, where you can often understand the whole better by looking at only a part of it first — restriction reaps riches. Egyptian fractions are one example in maths. In ancient Egypt, you could have any kind of fraction you liked so long as it was a reciprocal like 1/2, 1/3, 1/4 or 1/5 (well, there were two exceptions: 2/3 and 3/4 were also allowed).

So when mathematicians speak of “Egyptian fractions”, they mean those fractions that can be represented as a sum of reciprocals. Egyptian fractions are restricted and that reaps riches. Here’s one example: how many ways can you add n distinct reciprocals to make 1? When n = 1, there’s one way to do it: 1/1. When n = 2, there’s no way to do it, because 1 – 1/2 = 1/2. Therefore the summed reciprocals aren’t distinct: 1/2 + 1/2 = 1. After that, 1 – 1/3 = 2/3, 1 – 1/4 = 3/4, and so on. By the modern meaning of “Egyptian fraction”, there’s no solution for n = 2.

However, when n = 3, there is a way to do it:

• 1/2 + 1/3 + 1/6 = 1

But that’s the only way. When n = 4, things get better:

• 1/2 + 1/4 + 1/6 + 1/12 = 1
• 1/2 + 1/3 + 1/10 + 1/15 = 1
• 1/2 + 1/3 + 1/9 + 1/18 = 1
• 1/2 + 1/4 + 1/5 + 1/20 = 1
• 1/2 + 1/3 + 1/8 + 1/24 = 1
• 1/2 + 1/3 + 1/7 + 1/42 = 1

What about n = 5, n = 6 and so on? You can find the answer at the Online Encyclopedia of Integer Sequences (OEIS), where sequence A006585 is described as “Egyptian fractions: number of solutions to 1 = 1/x1 + … + 1/xn in positive integers x1 < … < xn”. The sequence is one of the shortest and strangest at the OEIS:

• 1, 0, 1, 6, 72, 2320, 245765, 151182379

When n = 1, there’s one solution: 1/1. When n = 2, there’s no solution, as I showed above. When n = 3, there’s one solution again. When n = 4, there are six solutions. And the OEIS tells you how many solutions there are for n = 5, 6, 7, 8. But n >= 9 remains unknown at the time of writing.

To understand the problem, consider the three reciprocals, 1/2, 1/3 and 1/5. How do you sum them? They have different denominators, 2, 3 and 5, so you have to create a new denominator, 30 = 2 * 3 * 5. Then you have to adjust the numerators (the numbers above the fraction bar) so that the new fractions have the same value as the old:

• 1/2 = 15/30 = (2*3*5 / 2) / 30
• 1/3 = 10/30 = (2*3*5 / 3) / 30
• 1/5 = 06/30 = (2*3*5 / 5) / 30
• 15/30 + 10/30 + 06/30 = (15+10+6) / 30 = 31/30 = 1 + 1/30

Those three reciprocals don’t sum to 1. Now try 1/2, 1/3 and 1/6:

• 1/2 = 18/36 = (2*3*6 / 2) / 36
• 1/3 = 12/36 = (2*3*6 / 3) / 36
• 1/6 = 06/36 = (2*3*6 / 6) / 36
• 18/36 + 12/36 + 06/36 = (18+12+6) / 36 = 36/36 = 1

So when n = 3, the problem consists of finding three reciprocals, 1/a, 1/b and 1/c, such that for a, b, and c:

• a*b*c = a*b + a*c + b*c

There is only one solution: a = 2, b = 3 and c = 6. When n = 4, the problem consists of finding four reciprocals, 1/a, 1/b, 1/c and 1/d, such that for a, b, c and d:

• a*b*c*d = a*b*c + a*b*d + a*c*d + b*c*d

For example:

• 2*4*6*12 = 576
• 2*4*6 + 2*4*12 + 2*6*12 + 4*6*12 = 48 + 96 + 144 + 288 = 576
• 2*4*6*12 = 2*4*6 + 2*4*12 + 2*6*12 + 4*6*12 = 576

Therefore:

• 1/2 + 1/4 + 1/6 + 1/12 = 1

When n = 5, the problem consists of finding five reciprocals, 1/a, 1/b, 1/c, 1/d and 1/e, such that for a, b, c, d and e:

• a*b*c*d*e = a*b*c*d + a*b*c*e + a*b*d*e + a*c*d*e + b*c*d*e

There are 72 solutions and here they are:

• 1/2 + 1/4 + 1/10 + 1/12 + 1/15 = 1 (#1)
• 1/2 + 1/4 + 1/9 + 1/12 + 1/18 = 1 (#2)
• 1/2 + 1/5 + 1/6 + 1/12 + 1/20 = 1 (#3)
• 1/3 + 1/4 + 1/5 + 1/6 + 1/20 = 1 (#4)
• 1/2 + 1/4 + 1/8 + 1/12 + 1/24 = 1 (#5)
• 1/2 + 1/3 + 1/12 + 1/21 + 1/28 = 1 (#6)
• 1/2 + 1/4 + 1/6 + 1/21 + 1/28 = 1 (#7)
• 1/2 + 1/4 + 1/7 + 1/14 + 1/28 = 1 (#8)
• 1/2 + 1/3 + 1/12 + 1/20 + 1/30 = 1 (#9)
• 1/2 + 1/4 + 1/6 + 1/20 + 1/30 = 1 (#10)
• 1/2 + 1/5 + 1/6 + 1/10 + 1/30 = 1 (#11)
• 1/2 + 1/3 + 1/11 + 1/22 + 1/33 = 1 (#12)
• 1/2 + 1/3 + 1/14 + 1/15 + 1/35 = 1 (#13)
• 1/2 + 1/3 + 1/12 + 1/18 + 1/36 = 1 (#14)
• 1/2 + 1/4 + 1/6 + 1/18 + 1/36 = 1 (#15)
• 1/2 + 1/3 + 1/10 + 1/24 + 1/40 = 1 (#16)
• 1/2 + 1/4 + 1/8 + 1/10 + 1/40 = 1 (#17)
• 1/2 + 1/4 + 1/7 + 1/12 + 1/42 = 1 (#18)
• 1/2 + 1/3 + 1/9 + 1/30 + 1/45 = 1 (#19)
• 1/2 + 1/4 + 1/5 + 1/36 + 1/45 = 1 (#20)
• 1/2 + 1/5 + 1/6 + 1/9 + 1/45 = 1 (#21)
• 1/2 + 1/3 + 1/12 + 1/16 + 1/48 = 1 (#22)
• 1/2 + 1/4 + 1/6 + 1/16 + 1/48 = 1 (#23)
• 1/2 + 1/3 + 1/9 + 1/27 + 1/54 = 1 (#24)
• 1/2 + 1/3 + 1/8 + 1/42 + 1/56 = 1 (#25)
• 1/2 + 1/3 + 1/8 + 1/40 + 1/60 = 1 (#26)
• 1/2 + 1/3 + 1/10 + 1/20 + 1/60 = 1 (#27)
• 1/2 + 1/3 + 1/12 + 1/15 + 1/60 = 1 (#28)
• 1/2 + 1/4 + 1/5 + 1/30 + 1/60 = 1 (#29)
• 1/2 + 1/4 + 1/6 + 1/15 + 1/60 = 1 (#30)
• 1/2 + 1/4 + 1/5 + 1/28 + 1/70 = 1 (#31)
• 1/2 + 1/3 + 1/8 + 1/36 + 1/72 = 1 (#32)
• 1/2 + 1/3 + 1/9 + 1/24 + 1/72 = 1 (#33)
• 1/2 + 1/4 + 1/8 + 1/9 + 1/72 = 1 (#34)
• 1/2 + 1/3 + 1/12 + 1/14 + 1/84 = 1 (#35)
• 1/2 + 1/4 + 1/6 + 1/14 + 1/84 = 1 (#36)
• 1/2 + 1/3 + 1/8 + 1/33 + 1/88 = 1 (#37)
• 1/2 + 1/3 + 1/10 + 1/18 + 1/90 = 1 (#38)
• 1/2 + 1/3 + 1/7 + 1/78 + 1/91 = 1 (#39)
• 1/2 + 1/3 + 1/8 + 1/32 + 1/96 = 1 (#40)
• 1/2 + 1/3 + 1/9 + 1/22 + 1/99 = 1 (#41)
• 1/2 + 1/4 + 1/5 + 1/25 + 1/100 = 1 (#42)
• 1/2 + 1/3 + 1/7 + 1/70 + 1/105 = 1 (#43)
• 1/2 + 1/3 + 1/11 + 1/15 + 1/110 = 1 (#44)
• 1/2 + 1/3 + 1/8 + 1/30 + 1/120 = 1 (#45)
• 1/2 + 1/4 + 1/5 + 1/24 + 1/120 = 1 (#46)
• 1/2 + 1/5 + 1/6 + 1/8 + 1/120 = 1 (#47)
• 1/2 + 1/3 + 1/7 + 1/63 + 1/126 = 1 (#48)
• 1/2 + 1/3 + 1/9 + 1/21 + 1/126 = 1 (#49)
• 1/2 + 1/3 + 1/7 + 1/60 + 1/140 = 1 (#50)
• 1/2 + 1/4 + 1/7 + 1/10 + 1/140 = 1 (#51)
• 1/2 + 1/3 + 1/12 + 1/13 + 1/156 = 1 (#52)
• 1/2 + 1/4 + 1/6 + 1/13 + 1/156 = 1 (#53)
• 1/2 + 1/3 + 1/7 + 1/56 + 1/168 = 1 (#54)
• 1/2 + 1/3 + 1/8 + 1/28 + 1/168 = 1 (#55)
• 1/2 + 1/3 + 1/9 + 1/20 + 1/180 = 1 (#56)
• 1/2 + 1/3 + 1/7 + 1/54 + 1/189 = 1 (#57)
• 1/2 + 1/3 + 1/8 + 1/27 + 1/216 = 1 (#58)
• 1/2 + 1/4 + 1/5 + 1/22 + 1/220 = 1 (#59)
• 1/2 + 1/3 + 1/11 + 1/14 + 1/231 = 1 (#60)
• 1/2 + 1/3 + 1/7 + 1/51 + 1/238 = 1 (#61)
• 1/2 + 1/3 + 1/10 + 1/16 + 1/240 = 1 (#62)
• 1/2 + 1/3 + 1/7 + 1/49 + 1/294 = 1 (#63)
• 1/2 + 1/3 + 1/8 + 1/26 + 1/312 = 1 (#64)
• 1/2 + 1/3 + 1/7 + 1/48 + 1/336 = 1 (#65)
• 1/2 + 1/3 + 1/9 + 1/19 + 1/342 = 1 (#66)
• 1/2 + 1/4 + 1/5 + 1/21 + 1/420 = 1 (#67)
• 1/2 + 1/3 + 1/7 + 1/46 + 1/483 = 1 (#68)
• 1/2 + 1/3 + 1/8 + 1/25 + 1/600 = 1 (#69)
• 1/2 + 1/3 + 1/7 + 1/45 + 1/630 = 1 (#70)
• 1/2 + 1/3 + 1/7 + 1/44 + 1/924 = 1 (#71)
• 1/2 + 1/3 + 1/7 + 1/43 + 1/1806 = 1 (#72)

All the sums start with 1/2 except for one:

• 1/2 + 1/5 + 1/6 + 1/12 + 1/20 = 1 (#3)
• 1/3 + 1/4 + 1/5 + 1/6 + 1/20 = 1 (#4)

Here are the solutions in another format:

(2,4,10,12,15), (2,4,9,12,18), (2,5,6,12,20), (3,4,5,6,20), (2,4,8,12,24), (2,3,12,21,28), (2,4,6,21,28), (2,4,7,14,28), (2,3,12,20,30), (2,4,6,20,30), (2,5,6,10,30), (2,3,11,22,33), (2,3,14,15,35), (2,3,12,18,36), (2,4,6,18,36), (2,3,10,24,40), (2,4,8,10,40), (2,4,7,12,42), (2,3,9,30,45), (2,4,5,36,45), (2,5,6,9,45), (2,3,12,16,48), (2,4,6,16,48), (2,3,9,27,54), (2,3,8,42,56), (2,3,8,40,60), (2,3,10,20,60), (2,3,12,15,60), (2,4,5,30,60), (2,4,6,15,60), (2,4,5,28,70), (2,3,8,36,72), (2,3,9,24,72), (2,4,8,9,72), (2,3,12,14,84), (2,4,6,14,84), (2,3,8,33,88), (2,3,10,18,90), (2,3,7,78,91), (2,3,8,32,96), (2,3,9,22,99), (2,4,5,25,100), (2,3,7,70,105), (2,3,11,15,110), (2,3,8,30,120), (2,4,5,24,120), (2,5,6,8,120), (2,3,7,63,126), (2,3,9,21,126), (2,3,7,60,140), (2,4,7,10,140), (2,3,12,13,156), (2,4,6,13,156), (2,3,7,56,168), (2,3,8,28,168), (2,3,9,20,180), (2,3,7,54,189), (2,3,8,27,216), (2,4,5,22,220), (2,3,11,14,231), (2,3,7,51,238), (2,3,10,16,240), (2,3,7,49,294), (2,3,8,26,312), (2,3,7,48,336), (2,3,9,19,342), (2,4,5,21,420), (2,3,7,46,483), (2,3,8,25,600), (2,3,7,45,630), (2,3,7,44,924), (2,3,7,43,1806)


Note

Strictly speaking, there are two solutions for n = 2 in genuine Egyptian fractions, because 1/3 + 2/3 = 1 and 1/4 + 3/4 = 1. As noted above, 2/3 and 3/4 were permitted as fractions in ancient Egypt.

Can You Dij It? #2

It’s very simple, but I’m fascinated by it. I’m talking about something I call the digit-line, or the stream of digits you get when you split numbers in a particular base into individual digits. For example, here are the numbers one to ten in bases 2 and 3:

Base = 2: 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010…
Base = 3: 1, 2, 10, 11, 12, 20, 21, 22, 100, 101…


If you turn them into digit-lines, they look like this:

Base = 2: 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0… (A030190 in the Online Encyclopedia of Integer Sequences)
Base = 3: 1, 2, 1, 0, 1, 1, 1, 2, 2, 0, 2, 1, 2, 2, 1, 0, 0, 1, 0, 1… (A003137 in the OEIS)


At the tenth digit of the two digit-lines, both digits equal zero for the first time:

Base = 2: 1, 1, 0, 1, 1, 1, 0, 0, 1, 0
Base = 3: 1, 2, 1, 0, 1, 1, 1, 2, 2, 0


When the binary and ternary digits are represented together, the digit-lines look like this:

(1,1), (1,2), (0,1), (1,0), (1,1), (1,1), (0,1), (0,2), (1,2), (0,0)


But in base 4, the tenth digit of the digit-line is 1. So when do all the digits of the digit-line first equal zero for bases 2 to 4? Here the early integers in those bases:

Base 2: 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, 10000, 10001, 10010, 10011, 10100, 10101…

Base 3: 1, 2, 10, 11, 12, 20, 21, 22, 100, 101, 102, 110, 111, 112, 120, 121, 122, 200, 201, 202, 210, 211, 212, 220, 221, 222, 1000, 1001, 1002…

Base 4: 1, 2, 3, 10, 11, 12, 13, 20, 21, 22, 23, 30, 31, 32, 33, 100, 101, 102, 103, 110, 111, 112, 113, 120, 121, 122, 123, 130, 131, 132, 133, 200…


And here are the digits of the digit-line in bases 2 to 4 represented together:

(1,1,1), (1,2,2), (0,1,3), (1,0,1), (1,1,0), (1,1,1), (0,1,1), (0,2,1), (1,2,2), (0,0,1), (1,2,3), (1,1,2), (1,2,0), (0,2,2), (1,1,1), (1,0,2), (1,0,2), (1,1,2), (0,0,3), (0,1,3), (0,1,0), (1,0,3), (0,2,1), (0,1,3), (1,1,2), (1,0,3), (0,1,3), (1,1,1), (0,1,0), (1,1,0), (0,1,1), (1,2,0), (1,1,1), (1,2,1), (1,0,0), (0,1,2), (0,2,1), (1,1,0), (1,1,3), (0,2,1), (1,2,1), (1,2,0), (1,0,1), (1,0,1), (0,2,1), (1,0,1), (1,1,1), (1,2,2), (1,0,1), (1,2,1), (0,2,3), (0,1,1), (0,0,2), (0,2,0), (1,1,1), (0,1,2), (0,2,1), (0,1,1), (1,2,2), (1,2,2), (0,2,1), (0,0,2), (1,2,3), (0,2,1), (1,1,3), (0,2,0), (0,2,1), (1,2,3), (1,1,1), (1,0,1), (0,0,3), (1,0,2), (0,1,1), (0,0,3), (1,0,3), (0,1,2), (1,1,0), (0,0,0)

At the 78th digit, all three digits equal zero. But the 78th digit of the digit-line in base 5 is 1. So when are the digits first equal to zero in bases 2 to 5? It’s not difficult to find out, but the difficulty of the search increases fast as the bases get bigger. Here are the results up to base 13 (note that bases 11 and 12 both have zeroes at digit 103721663):

dig=0 in bases 2 to 3 at the 10th digit of the digit-line
dig=0 in bases 2 to 4 at the 78th digit of the digit-line
dig=0 in bases 2 to 5 at the 182nd digit of the digit-line
dig=0 in bases 2 to 6 at the 302nd digit of the digit-line
dig=0 in bases 2 to 7 at the 12149th digit of the digit-line
dig=0 in bases 2 to 8 at the 45243rd digit of the digit-line
dig=0 in bases 2 to 9 at the 255261st digit of the digit-line
dig=0 in bases 2 to 10 at the 8850623rd digit of the digit-line
dig=0 in bases 2 to 12 at the 103721663rd digit of the digit-line
dig=0 in bases 2 to 13 at the 807778264th digit of the digit-line


I assume that, for any base b > 2, you can find some point in the digit-line at which d = 0 for all bases 2 to b. Indeed, I assume that this happens infinitely often. But I don’t know any short-cut for finding the first digit at which this occurs.


Previously pre-posted:

Can You Dij It? #1

Can You Dij It? #1

The most powerful drug in the world is water. The second most powerful is language. But everyone’s on them, so nobody realizes how powerful they are. Well, you could stop drinking water. Then you’d soon realize its hold on the body and the brain.

But you can’t stop using language. Try it. No, the best way to realize the power of language is to learn a new one. Each is a feast with different flavours. New alphabets are good too. The Devanagari alphabet is one of the strongest, but if you want it in refined form, try the phonetic alphabet. It will transform the way you see the world. That’s because it will make you conscious of what you’re already subconsciously aware of.

But “language” is a bigger category that it used to be. Nowadays we have computer languages too. Learning one is another way of transforming the way you see the world. And like natural languages – French, Georgian, Tagalog – they come in different flavours. Pascal is not like Basic is not like C is not like Prolog. But all of them seem to put you in touch with some deeper aspect of reality. Computer languages are like mathemagick: a way to give commands to something immaterial and alter the world by the application of will.

That feeling is at its strongest when you program with machine code, the raw instructions used by the electronics of a computer. At its most fundamental, machine code is simply a series of binary numbers controlling how a computer processes other binary numbers. You can memorize and use those code-numbers, but it’s easier to use something like assembly language, which makes machine-code friendlier for human beings. But it still looks very odd to the uninitiated:

setupnum:
xor ax,ax
xor bp,bp
mov cx,20
clearloop:
mov [di+bp],ax
add bp,2
loop clearloop
ret

That’s almost at the binary bedrock. And machine code is fast. If a fast higher-level language like C feels like flying a Messerschmitt 262, which was a jet-plane, machine-code feels like flying a Messerschmitt 163, which was a rocket-plane. A very fast and very dangerous rocket-plane.

I’m not good at programming languages, least of all machine code, but they are fun to use, quite apart from the way they make you feel as though you’re in touch with a deeper aspect of reality. They do that because the world is mathematics at its most fundamental level, I think, and computer languages are a form of mathematics.

Their mathematical nature is disguised in a lot of what they’re used for, but I like to use them for recreational mathematics. Machine-code is useful when you need a lot of power and speed. For example, look at these digits:

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, 2, 1, 2, 2, 2, 3, 2, 4, 2, 5, 2, 6, 2, 7, 2, 8, 2, 9, 3, 0, 3, 1, 3, 2, 3, 3, 3, 4, 3, 5, 3, 6*, 3*, 7, 3, 8, 3, 9, 4, 0, 4, 1, 4, 2, 4…

They’re what the Online Encyclopedia of Integer Sequences (OEIS) calls “the almost natural numbers” (sequence A007376) and you generate them by writing the standard integers – 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13… – and then separating each digit with a comma: 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 0, 1, 1, 1, 2, 1, 3… The commas give them some interesting twists. In a list of the standard integers, the 1st entry is 1, the 10th entry is 10, the 213rd entry is 213, the 987,009,381th entry is 987,009,381, and so on.

But that doesn’t work with the almost natural numbers. The 10th entry is 1, not 10, and the 11th entry is 0, not 11. But the 10th entry does begin the sequence (1, 0). I wondered whether that happened again. It does. The 63rd entry in the almost natural numbers begins the sequence (6, 3) – see the asterisks in the sequence above.

This happens again at the 3105th entry, which begins the sequence (3, 1, 0, 5). After that the gaps get bigger, which is where machine code comes in. An ordinary computer-language takes a long time to reach the 89,012,345,679th entry in the almost natural numbers. Machine code is much quicker, which is why I know that the 89,012,345,679th entry begins the sequence (8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 9):

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 63, 3105, 43108, 77781, 367573, 13859021, 77911127, 911360799, 35924813703, 74075186297, 89012345679…

And an ordinary computer-language might give you the impression that base 9 doesn’t have numbers like these (apart from the trivial 1, 2, 3, 4, 5, 6, 7, 8, 10…). But it does. 63 in base 10 is a low-hanging fruit: you could find it working by hand. In base 9, the fruit are much higher-hanging. But machine code plucks them with almost ridiculous ease:

1, 2, 3, 4, 5, 6, 7, 8, 10, 570086565, 655267526, 2615038272, 4581347024, 5307541865, 7273850617, 7801234568…