# Agogic Arithmetic

This is one of my favorite integer sequences:

• 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666, 703, 741, 780, 820, 861, 903, 946, 990, 1035, 1081, 1128, 1176, 1225, 1275, 1326, 1378, 1431, ... — A000217 at OEIS

And it’s easy to work out the rule that generates the sequence. It’s the sequence of triangular numbers, of course, which you get by summing the integers:

1
1 + 2 = 3
3 + 3 = 6
6 + 4 = 10
10 + 5 = 15
15 + 6 = 21
21 + 7 = 28
28 + 8 = 36
36 + 9 = 45
[...]

I like this sequence too, but it isn’t a sequence of integers and it’s much harder to work out the rule that generates it:

• 1, 3/2, 11/6, 25/12, 137/60, 49/20, 363/140, 761/280, 7129/2520, 7381/2520, 83711/27720, 86021/27720, 1145993/360360, 1171733/360360...

But you could say that it’s the inverse of the triangular numbers, because you generate it like this:

1
1 + 1/2 = 3/2
3/2 + 1/3 = 11/6
11/6 + 1/4 = 25/12
25/12 + 1/5 = 137/60
137/60 + 1/6 = 49/20
49/20 + 1/7 = 363/140
363/140 + 1/8 = 761/280
761/280 + 1/9 = 7129/2520
[...]

It’s the harmonic series, which is defined at Wikipedia as “the infinite series formed by summing all positive unit fractions”. I can’t understand its subtleties or make any important discoveries about it, but I thought I could ask (and begin to answer) a question that perhaps no-one else in history had ever asked: When are the leading digits of the k-th harmonic number, hs(k), equal to the digits of k in base 10?

hs(1) = 1
hs(43) = 4.349...
hs(714) = 7.1487...
hs(715) = 7.1501...
hs(9763) = 9.76362...
hs(122968) = 12.296899...
hs(122969) = 12.296907...
hs(1478366) = 14.7836639...
hs(17239955) = 17.23995590...
hs(196746419) = 19.6746419...
hs(2209316467) = 22.0931646788...

Do those numbers have any true mathematical significance? I doubt it. But they were fun to find, even though I wasn’t the first person in history to ask about them:

• 1, 43, 714, 715, 9763, 122968, 122969, 1478366, 17239955, 196746419, 2209316467, 24499118645, 268950072605 — A337904 at OEIS, Numbers k such that the decimal expansion of the k-th harmonic number starts with the digits of k, in the same order.

# The Devil’s Digits

As I’ve said before, I love the way that numbers can come in many different guises. For example, take the number 21. It comes in all these guises:

21 = 10101 in base 2 = 210 in base 3 = 111 in b4 = 41 in b5 = 33 in b6 = 30 in b7 = 25 in b8 = 23 in b9 = 21 in b10 = 1A in b11 = 19 in b12 = 18 in b13 = 17 in b14 = 16 in b15 = 15 in b16 = 14 in b17 = 13 in b18 = 12 in b19 = 11 in b20 = 10 in b21

But I’ve not chosen 21 at random. If you sum the 1s in the representations of 21 in bases 2 to 21, look what you get:

21 = 10101 in base 2 = 210 in base 3 = 111 in b4 = 41 in b5 = 33 in b6 = 30 in b7 = 25 in b8 = 23 in b9 = 21 in b10 = 1A in b11 = 19 in b12 = 18 in b13 = 17 in b14 = 16 in b15 = 15 in b16 = 14 in b17 = 13 in b18 = 12 in b19 = 11 in b20 = 10 in b21

21 = 1s=101s=201s=3 in base 2 = 21s=40 in base 3 = 111s=7 in b4 = 41s=8 in b5 = 33 in b6 = 30 in b7 = 25 in b8 = 23 in b9 = 21s=9 in b10 = 1s=10A in b11 = 1s=119 in b12 = 1s=128 in b13 = 1s=137 in b14 = 1s=146 in b15 = 1s=155 in b16 = 1s=164 in b17 = 1s=173 in b18 = 1s=182 in b19 = 11s=20 in b20 = 1s=210 in b21

In other words, 21 = digcount(21,dig=1,base=2..21). But n = digcount(n,dig,b=2..n) doesn’t happen for any other digit and doesn’t happen often with 1:

3 = digcount(3,d=1,b=2..3) = 11 in b2 = 10 in b3
4 = digcount(4,d=1,b=2..4) = 100 in b2 = 11 in b3 = 10 in b4
6 = digcount(6,d=1,b=2..6) = 110 in b2 = 20 in b3 = 12 in b4 = 11 in b5 = 10 in b6
10 = digcount(10,d=1) = 1010 in b2 = 101 in b3 = 22 in b4 = 20 in b5 = 14 in b6 = 13 in b7 = 12 in b8 = 11 in b9 = 10 in b10
15 = digcount(15,d=1) = 1111 in b2 = 120 in b3 = 33 in b4 = 30 in b5 = 23 in b6 = 21 in b7 = 17 in b8 = 16 in b9 = 15 in b10 = 14 in b11 = 13 in b12 = 12 in b13 = 11 in b14 = 10 in b15
21 = digcount(21,d=1) = 10101 in b2 = 210 in b3 = 111 in b4 = 41 in b5 = 33 in b6 = 30 in b7 = 25 in b8 = 23 in b9 = 21 in b10 = 1A in b11 = 19 in b12 = 18 in b13 = 17 in b14 = 16 in b15 = 15 in b16 = 14 in b17 = 13 in b18 = 12 in b19 = 11 in b20 = 10 in b21

After that, the digcount(n,d=1,b=2..n) → n/2 (see “Digital Dissection” for further discussion). But I decided to look for the first n where digcount(n,dig,b=2..n) = 666:

digcount(1270,1) = 666
digcount(3770,2) = 666
digcount(7667,3) = 666
digcount(12184,4) = 666
digcount(18845,5) = 666
digcount(25806,6) = 666
digcount(34195,7) = 666
digcount(43352,8) = 666
digcount(54693,9) = 666

It doesn’t stop there, of course. You can carry on for ever, looking for digcount(n,A) = 666, digcount(n,B) = 666, digcount(n,C) = 666, where A = 10, B = 11 and C=12, and so on. But it doesn’t start there, either. What about digcount(n,0) = 666? That isn’t easy to find, because 0 usually occurs far less often than other digits in the representation of n. Here are the integers setting records for digcount(n,0,b=2..n):

2 → digcount(2,0) = 1 ← 2= 10 in base 2
4 → digcount(4,0) = 3; ← 4 = 100 in base 2, 11 in base 3, 10 in base 4
8 → digcount(8,0) = 5 ← 8 = 1000 in base 2, 22 in base 3, 20 in base 4, 13 in base 5, 12 in base 6, 11 in base 7, 10 in base 8
12 → digcount(12,0) = 6
16 → digcount(16,0) = 8
18 → digcount(18,0) = 9
32 → digcount(32,0) = 11
36 → digcount(36,0) = 13
64 → digcount(64,0) = 15
72 → digcount(72,0) = 18
128 → digcount(128,0) = 20
144 → digcount(144,0) = 24
252 → digcount(252,0) = 25
264 → digcount(264,0) = 27
288 → digcount(288,0) = 29
360 → digcount(360,0) = 30
504 → digcount(504,0) = 33
540 → digcount(540,0) = 36
720 → digcount(720,0) = 40
900 → digcount(900,0) = 42
1080 → digcount(1080,0) = 47
1680 → digcount(1680,0) = 48
1800 → digcount(1800,0) = 53
2160 → digcount(2160,0) = 56
2520 → digcount(2520,0) = 61
3600 → digcount(3600,0) = 64
4320 → digcount(4320,0) = 66

So what is the first n for which digcount(n,0) = 666? Watch this space.

# Reciprocal Recipes

Here’s a sequence. What’s the next number?

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1...

Here’s another sequence. What’s the next number?

0, 1, 1, 2, 3, 5, 8, 13, 21, 34...

Those aren’t trick questions, so the answers are 1 and 55, respectively. The second sequence is the famous Fibonacci sequence, where each number after [0,1] is the sum of the previous two numbers.

Now try dividing each of those sequences by powers of 2 and summing the results, like this:

1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512 + 1/1024 + 1/2048 + 1/4096 + 1/8192 + 1/16384 + 1/32768 + 1/65536 + 1/131072 + 1/262144 + 1/524288 + 1/1048576 +... = ?

0/2 + 1/4 + 1/8 + 2/16 + 3/32 + 5/64 + 8/128 + 13/256 + 21/512 + 34/1024 + 55/2048 + 89/4096 + 144/8192 + 233/16384 + 377/32768 + 610/65536 + 987/131072 + 1597/262144 + 2584/524288 + 4181/1048576 +... = ?

What are the sums? I was surprised to learn that they’re identical:

1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512 + 1/1024 + 1/2048 + 1/4096 + 1/8192 + 1/16384 + 1/32768 + 1/65536 + 1/131072 + 1/262144 + 1/524288 + 1/1048576 +... = 1

0/2 + 1/4 + 1/8 + 2/16 + 3/32 + 5/64 + 8/128 + 13/256 + 21/512 + 34/1024 + 55/2048 + 89/4096 + 144/8192 + 233/16384 + 377/32768 + 610/65536 + 987/131072 + 1597/262144 + 2584/524288 + 4181/1048576 +... = 1

I discovered this when I was playing with an old scientific calculator and calculated these sums:

5^2 + 2^2 = 29
5^2 + 4^2 = 41
5^2 + 6^2 = 61
5^2 + 8^2 = 89

The sums are all prime numbers. Then I idly calculated the reciprocal of 1/89:

1/89 = 0·011235955056179775...

The digits 011235… are the start of the Fibonacci sequence. It seems to go awry after that, but I remembered what David Wells had said in his wonderful Penguin Dictionary of Curious and Interesting Numbers (1986): “89 is the 11th Fibonacci number, and the period of its reciprocal is generated by the Fibonacci sequence: 1/89 = 0·11235…” He means that the Fibonacci sequence generates the digits of 1/89 like this, when you sum the columns and move carries left as necessary:

0
1
↓↓1
↓↓↓2
↓↓↓↓3
↓↓↓↓↓5
↓↓↓↓↓↓8
↓↓↓↓↓↓13
↓↓↓↓↓↓↓21
↓↓↓↓↓↓↓↓34
↓↓↓↓↓↓↓↓↓55
↓↓↓↓↓↓↓↓↓↓89...
↓↓↓↓↓↓↓↓↓↓
0112359550...

I tried this method of summing the Fibonacci sequence in other bases. Although it was old, the scientific calculator was crudely programmable. And it helpfully converted the sum into a final fraction once there were enough decimal digits:

0/3 + 1/32 + 1/33 + 2/34 + 3/35 + 5/36 + 8/37 + 13/38 + 21/39 + 34/310 + 55/311 + 89/312 + 144/313 + 233/314 + 377/315 + 610/316 + 987/317 + 1597/318 + 2584/319 + 4181/320 +... = 1/5 = 0·012101210121012101210 in b3

0/4 + 1/42 + 1/43 + 2/44 + 3/45 + 5/46 + 8/47 + 13/48 + 21/49 + 34/410 + 55/411 + 89/412 + 144/413 + 233/414 + 377/415 + 610/416 + 987/417 + 1597/418 + 2584/419 + 4181/420 +... = 1/11 = 0·011310113101131011310 in b4

0/5 + 1/52 + 1/53 + 2/54 + 3/55 + 5/56 + 8/57 + 13/58 + 21/59 + 34/510 + 55/511 + 89/512 + 144/513 + 233/514 + 377/515 + 610/516 + 987/517 + 1597/518 + 2584/519 + 4181/520 +... = 1/19 = 0·011242141011242141011 in b5

0/6 + 1/62 + 1/63 + 2/64 + 3/65 + 5/66 + 8/67 + 13/68 + 21/69 + 34/610 + 55/611 + 89/612 + 144/613 + 233/614 + 377/615 + 610/616 + 987/617 + 1597/618 + 2584/619 + 4181/620 +... = 1/29 = 0·011240454431510112404 in b6

0/7 + 1/72 + 1/73 + 2/74 + 3/75 + 5/76 + 8/77 + 13/78 + 21/79 + 34/710 + 55/711 + 89/712 + 144/713 + 233/714 + 377/715 + 610/716 + 987/717 + 1597/718 + 2584/719 + 4181/720 +... = 1/41 = 0·011236326213520225056 in b7

It was interesting to see that all the reciprocals so far were of primes. I carried on:

0/8 + 1/82 + 1/83 + 2/84 + 3/85 + 5/86 + 8/87 + 13/88 + 21/89 + 34/810 + 55/811 + 89/812 + 144/813 + 233/814 + 377/815 + 610/816 + 987/817 + 1597/818 + 2584/819 + 4181/820 +... = 1/55 = 0·011236202247440451710 in b8

Not a prime reciprocal, but a reciprocal of a Fibonacci number. Here are some more sums:

0/9 + 1/92 + 1/93 + 2/94 + 3/95 + 5/96 + 8/97 + 13/98 + 21/99 + 34/910 + 55/911 + 89/912 + 144/913 + 233/914 + 377/915 + 610/916 + 987/917 + 1597/918 + 2584/919 + 4181/920 +... = 1/71 (another prime) = 0·011236067540450563033 in b9

0/10 + 1/102 + 1/103 + 2/104 + 3/105 + 5/106 + 8/107 + 13/108 + 21/109 + 34/1010 + 55/1011 + 89/1012 + 144/1013 + 233/1014 + 377/1015 + 610/1016 + 987/1017 + 1597/1018 + 2584/1019 + 4181/1020 +... = 1/89 (and another) = 0·011235955056179775280 in b10

0/11 + 1/112 + 1/113 + 2/114 + 3/115 + 5/116 + 8/117 + 13/118 + 21/119 + 34/1110 + 55/1111 + 89/1112 + 144/1113 + 233/1114 + 377/1115 + 610/1116 + 987/1117 + 1597/1118 + 2584/1119 + 4181/1120 +... = 1/109 (and another) = 0·011235942695392022470 in b11

0/12 + 1/122 + 1/123 + 2/124 + 3/125 + 5/126 + 8/127 + 13/128 + 21/129 + 34/1210 + 55/1211 + 89/1212 + 144/1213 + 233/1214 + 377/1215 + 610/1216 + 987/1217 + 1597/1218 + 2584/1219 + 4181/1220 +... = 1/131 (and another) = 0·011235930336A53909A87 in b12

0/13 + 1/132 + 1/133 + 2/134 + 3/135 + 5/136 + 8/137 + 13/138 + 21/139 + 34/1310 + 55/1311 + 89/1312 + 144/1313 + 233/1314 + 377/1315 + 610/1316 + 987/1317 + 1597/1318 + 2584/1319 + 4181/1320 +... = 1/155 (not a prime or a Fibonacci number) = 0·01123591ACAA861794044 in b13

The reciprocals go like this:

1/1, 1/5, 1/11, 1/19, 1/29, 1/41, 1/55, 1/71, 1/89, 1/109, 1/131, 1/155...

And it should be easy to see the rule that generates them:

5 = 1 + 4
11 = 5 + 6
19 = 11 + 8
29 = 19 + 10
41 = 29 + 12
55 = 41 + 14
71 = 55 + 16
89 = 17 + 18
109 = 89 + 20
131 = 109 + 22
155 = 131 + 24
[...]

But I don’t understand why the rule applies, let alone why the Fibonacci sequence generates these reciprocals in the first place.

# Root Pursuit

Roots are hard, powers are easy. For example, the square root of 2, or √2, is the mysterious and never-ending number that is equal to 2 when multiplied by itself:

• √2 = 1·414213562373095048801688724209698078569671875376948073...

It’s hard to calculate √2. But the powers of 2, or 2^p, are the straightforward numbers that you get by multiplying 2 repeatedly by itself. It’s easy to calculate 2^p:

• 2 = 2^1
• 4 = 2^2
• 8 = 2^3
• 16 = 2^4
• 32 = 2^5
• 64 = 2^6
• 128 = 2^7
• 256 = 2^8
• 512 = 2^9
• 1024 = 2^10
• 2048 = 2^11
• 4096 = 2^12
• 8192 = 2^13
• 16384 = 2^14
• 32768 = 2^15
• 65536 = 2^16
• 131072 = 2^17
• 262144 = 2^18
• 524288 = 2^19
• 1048576 = 2^20
[...]

But there is a way to find √2 by finding 2^p, as I discovered after I asked a simple question about 2^p and 3^p. What are the longest runs of matching digits at the beginning of each power?

131072 = 2^17
129140163 = 3^17
1255420347077336152767157884641... = 2^193
1214512980685298442335534165687... = 3^193
2175541218577478036232553294038... = 2^619
2177993962169082260270654106078... = 3^619
7524389324549354450012295667238... = 2^2016
7524012611682575322123383229826... = 3^2016

There’s no obvious pattern. Then I asked the same question about 2^p and 5^p. And an interesting pattern appeared:

32 = 2^5
3125 = 5^5
316912650057057350374175801344 = 2^98
3155443620884047221646914261131... = 5^98
3162535207926728411757739792483... = 2^1068
3162020133383977882730040274356... = 5^1068
3162266908803418110961625404267... = 2^127185
3162288411569894029343799063611... = 5^127185

The digits 31622 rang a bell. Isn’t that the start of √10? Yes, it is:

• √10 = 3·1622776601683793319988935444327185337195551393252168268575...

I wrote a fast machine-code program to find even longer runs of matching initial digits. Sure enough, the pattern continued:

• 316227... = 2^2728361
• 316227... = 5^2728361
• 3162277... = 2^15917834
• 3162277... = 5^15917834
• 31622776... = 2^73482154
• 31622776... = 5^73482154
• 3162277660... = 2^961700165
• 3162277660... = 5^961700165

But why are powers of 2 and 5 generating the digits of √10? If you’re good at math, that’s a trivial question about a trivial discovery. Here’s the answer: We use base ten and 10 = 2 * 5, 10^2 = 100 = 2^2 * 5^2 = 4 * 25, 10^3 = 1000 = 2^3 * 5^3 = 8 * 125, and so on. When the initial digits of 2^p and 5^p match, those matching digits must come from the digits of √10. Otherwise the product of 2^p * 5^p would be too large or too small. Here are the records for matching initial digits multiplied by themselves:

32 = 2^5
3125 = 5^5
• 3^2 = 9

316912650057057350374175801344 = 2^98
3155443620884047221646914261131... = 5^98
• 31^2 = 961

3162535207926728411757739792483... = 2^1068
3162020133383977882730040274356... = 5^1068
• 3162^2 = 9998244

3162266908803418110961625404267... = 2^127185
3162288411569894029343799063611... = 5^127185
• 31622^2 = 999950884

• 316227... = 2^2728361
• 316227... = 5^2728361
• 316227^2 = 99999515529

• 3162277... = 2^15917834
• 3162277... = 5^15917834
• 3162277^2 = 9999995824729

• 31622776... = 2^73482154
• 31622776... = 5^73482154
• 31622776^2 = 999999961946176

• 3162277660... = 2^961700165
• 3162277660... = 5^961700165
• 3162277660^2 = 9999999998935075600

The square of each matching run falls short of 10^p. And so when the digits of 2^p and 5^p stop matching, one power must fall below √10, as it were, and one must rise above:

3 162266908803418110961625404267... = 2^127185
3·162277660168379331998893544432... = √10
3 162288411569894029343799063611... = 5^127185

In this way, 2^p * 5^p = 10^p. And that’s why matching initial digits of 2^p and 5^p generate the digits of √10. The same thing, mutatis mutandis, happens in base 6 with 2^p and 3^p, because 6 = 2 * 3:

• 2.24103122055214532500432040411... = √6 (in base 6)

24 = 2^4
213 = 3^4
225522024 = 2^34 in base 6 = 2^22 in base 10
22225525003213 = 3^34 (3^22)
2241525132535231233233555114533... = 2^1303 (2^327)
2240133444421105112410441102423... = 3^1303 (3^327)
2241055222343212030022044325420... = 2^153251 (2^15007)
2241003215453455515322105001310... = 3^153251 (3^15007)
2241032233315203525544525150530... = 2^233204 (2^20164)
2241030204225410320250422435321... = 3^233204 (3^20164)
2241031334114245140003252435303... = 2^2110415 (2^102539)
2241031103430053425141014505442... = 3^2110415 (3^102539)

And in base 30, where 30 = 2 * 3 * 5, you can find the digits of √30 in three different ways, because 30 = 2 * 15 = 3 * 10 = 5 * 6:

• 5·E9F2LE6BBPBF0F52B7385PE6E5CLN... = √30 (in base 30)

55AA4 = 2^M in base 30 = 2^22 in base 10
5NO6CQN69C3Q0E1Q7F = F^M = 15^22
5E63NMOAO4JPQD6996F3HPLIMLIRL6F... = 2^K6 (2^606)
5ECQDMIOCIAIR0DGJ4O4H8EN10AQ2GR... = F^K6 (15^606)
5E9DTE7BO41HIQDDO0NB1MFNEE4QJRF... = 2^B14 (2^9934)
5E9G5SL7KBNKFLKSG89J9J9NT17KHHO... = F^B14 (15^9934)
[...]
5R4C9 = 3^E in base 30 = 3^14 in base 10
52CE6A3L3A = A^E = 10^14
5E6SOQE5II5A8IRCH9HFBGO7835KL8A = 3^3N (3^113)
5EC1BLQHNJLTGD00SLBEDQ73AH465E3... = A^3N (10^113)
5E9FI455MQI4KOJM0HSBP3GG6OL9T8P... = 3^EJH (3^13187)
5E9EH8N8D9TR1AH48MT7OR3MHAGFNFQ... = A^EJH (10^13187)
[...]
5OCNCNRAP = 5^I in base 30 = 5^18 in base 10
54NO22GI76 = 6^I (6^18)
5EG4RAMD1IGGHQ8QS2QR0S0EH09DK16... = 5^1M7 (5^1567)
5E2PG4Q2G63DOBIJ54E4O035Q9TEJGH... = 6^1M7 (6^1567)
5E96DB9T6TBIM1FCCK8A8J7IDRCTM71... = 5^F9G (5^13786)
5E9NM222PN9Q9TEFTJ94261NRBB8FCH... = 6^F9G (6^13786)
[...]

So that’s √10, √6 and √30. But I said at the beginning that you can find √2 by finding 2^p. How do you do that? By offsetting the powers, as it were. With 2^p and 5^p, you can find the digits of √10. With 2^(p+1) and 5^p, you can find the digits of √2 and √20, because 2^(p+1) * 5^p = 2 * 2^p * 5^p = 2 * 10^p:

•  √2 = 1·414213562373095048801688724209698078569671875376948073...
• √20 = 4·472135954999579392818347337462552470881236719223051448...

16 = 2^4
125 = 5^3
140737488355328 = 2^47
142108547152020037174224853515625 = 5^46
1413... = 2^243
1414... = 5^242
14141... = 2^6651
14142... = 5^6650
141421... = 2^35389
141420... = 5^35388
4472136... = 2^162574
4472135... = 5^162573
141421359... = 2^3216082
141421352... = 5^3216081
447213595... = 2^172530387
447213595... = 5^172530386
[...]

# God Give Me Benf’

In “Wake the Snake”, I looked at the digits of powers of 2 and mentioned a fascinating mathematical phenomenon known as Benford’s law, which governs — in a not-yet-fully-explained way — the leading digits of a wide variety of natural and human statistics, from the lengths of rivers to the votes cast in elections. Benford’s law also governs a lot of mathematical data. It states, for example, that the first digit, d, of a power of 2 in base b (except b = 2, 4, 8, 16…) will occur with the frequency logb(1 + 1/d). In base 10, therefore, Benford’s law states that the digits 1..9 will occur with the following frequencies at the beginning of 2^p:

1: 30.102999%
2: 17.609125%
3: 12.493873%
4: 09.691001%
5: 07.918124%
6: 06.694678%
7: 05.799194%
8: 05.115252%
9: 04.575749%

Here’s a graph of the actual relative frequencies of 1..9 as the leading digit of 2^p (open images in a new window if they appear distorted):

And here’s a graph for the predicted frequencies of 1..9 as the leading digit of 2^p, as calculated by the log(1+1/d) of Benford’s law:

The two graphs agree very well. But Benford’s law applies to more than one leading digit. Here are actual and predicted graphs for the first two leading digits of 2^p, 10..99:

And actual and predicted graphs for the first three leading digits of 2^p, 100..999:

But you can represent the leading digit of 2^p in another way: using an adaptation of the famous Ulam spiral. Suppose powers of 2 are represented as a spiral of squares that begins like this, with 2^0 in the center, 2^1 to the right of center, 2^2 above 2^1, and so on:

←←←⮲
432↑
501↑
6789

If the digits of 2^p start with 1, fill the square in question; if the digits of 2^p don’t start with 1, leave the square empty. When you do this, you get this interesting pattern (the purple square at the very center represents 2^0):

Ulam-like power-spiral for 2^p where 1 is the leading digit

Here’s a higher-resolution power-spiral for 1 as the leading digit:

Power-spiral for 2^p, leading-digit = 1 (higher resolution)

And here, at higher resolution still, are power-spirals for all the possible leading digits of 2^p, 1..9 (some spirals look very similar, so you have to compare those ones carefully):

Power-spiral for 2^p, leading-digit = 1 (very high resolution)

Power-spiral for 2^p, leading-digit = 2

Power-spiral for 2^p, ld = 3

Power-spiral for 2^p, ld = 4

Power-spiral for 2^p, ld = 5

Power-spiral for 2^p, ld = 6

Power-spiral for 2^p, ld = 7

Power-spiral for 2^p, ld = 8

Power-spiral for 2^p, ld = 9

Power-spiral for 2^p, ld = 1..9 (animated)

Now try the power-spiral of 2^p, ld = 1, in some other bases:

Power-spiral for 2^p, leading-digit = 1, base = 9

Power-spiral for 2^p, ld = 1, b = 15

You can also try power-spirals for other n^p. Here’s 3^p:

Power-spiral for 3^p, ld = 1, b = 10

Power-spiral for 3^p, ld = 2, b = 10

Power-spiral for 3^p, ld = 1, b = 4

Power-spiral for 3^p, ld = 1, b = 7

Power-spiral for 3^p, ld = 1, b = 18

Elsewhere Other-Accessible…

Wake the Snake — an earlier look at the digits of 2^p

# Wake the Snake

In my story “Kopfwurmkundalini”, I imagined the square root of 2 as an infinitely long worm or snake whose endlessly varying digit-segments contained all stories ever (and never) written:

• √2 = 1·414213562373095048801688724209698078569671875376948073…

But there’s another way to get all stories ever written from the number 2. You don’t look at the root(s) of 2, but at the powers of 2:

• 2 = 2^1 = 2
• 4 = 2^2 = 2*2
• 8 = 2^3 = 2*2*2
• 16 = 2^4 = 2*2*2*2
• 32 = 2^5 = 2*2*2*2*2
• 64 = 2^6 = 2*2*2*2*2*2
• 128 = 2^7 = 2*2*2*2*2*2*2
• 256 = 2^8 = 2*2*2*2*2*2*2*2
• 512 = 2^9 = 2*2*2*2*2*2*2*2*2
• 1024 = 2^10
• 2048 = 2^11
• 4096 = 2^12
• 8192 = 2^13
• 16384 = 2^14
• 32768 = 2^15
• 65536 = 2^16
• 131072 = 2^17
• 262144 = 2^18
• 524288 = 2^19
• 1048576 = 2^20
• 2097152 = 2^21
• 4194304 = 2^22
• 8388608 = 2^23
• 16777216 = 2^24
• 33554432 = 2^25
• 67108864 = 2^26
• 134217728 = 2^27
• 268435456 = 2^28
• 536870912 = 2^29
• 1073741824 = 2^30
[...]

The powers of 2 are like an ever-lengthening snake swimming across a pool. The snake has an endlessly mutating head and a rhythmically waving tail with a regular but ever-more complex wake. That is, the leading digits of 2^p don’t repeat but the trailing digits do. Look at the single final digit of 2^p, for example:

• 02 = 2^1
• 04 = 2^2
• 08 = 2^3
• 16 = 2^4
• 32 = 2^5
• 64 = 2^6
• 128 = 2^7
• 256 = 2^8
• 512 = 2^9
• 1024 = 2^10
• 2048 = 2^11
• 4096 = 2^12
• 8192 = 2^13
• 16384 = 2^14
• 32768 = 2^15
• 65536 = 2^16
• 131072 = 2^17
• 262144 = 2^18
• 524288 = 2^19
• 1048576 = 2^20
• 2097152 = 2^21
• 4194304 = 2^22
[...]

The final digit of 2^p falls into a loop: 2 → 4 → 8 → 6 → 2 → 4→ 8…

Now try the final two digits of 2^p:

02 = 2^1
04 = 2^2
08 = 2^3
16 = 2^4
32 = 2^5
64 = 2^6
• 128 = 2^7
• 256 = 2^8
• 512 = 2^9
• 1024 = 2^10
• 2048 = 2^11
• 4096 = 2^12
• 8192 = 2^13
• 16384 = 2^14
• 32768 = 2^15
• 65536 = 2^16
• 131072 = 2^17
• 262144 = 2^18
• 524288 = 2^19
• 1048576 = 2^20
• 2097152 = 2^21
• 4194304 = 2^22
• 8388608 = 2^23
• 16777216 = 2^24
• 33554432 = 2^25
• 67108864 = 2^26
• 134217728 = 2^27
• 268435456 = 2^28
• 536870912 = 2^29
• 1073741824 = 2^30
[...]

Now there’s a longer loop: 02 → 04 → 08 → 16 → 32 → 64 → 28 → 56 → 12 → 24 → 48 → 96 → 92 → 84 → 68 → 36 → 72 → 44 → 88 → 76 → 52 → 04 → 08 → 16 → 32 → 64 → 28… Any number of trailing digits, 1 or 2 or one trillion, falls into a loop. It just takes longer as the number of trailing digits increases.

That’s the tail of the snake. At the other end, the head of the snake, the digits don’t fall into a loop (because of the carries from the lower digits). So, while you can get only 2, 4, 8 and 6 as the final digits of 2^p, you can get any digit but 0 as the first digit of 2^p. Indeed, I conjecture (but can’t prove) that not only will all integers eventually appear as the leading digits of 2^p, but they will do so infinitely often. Think of a number and it will appear as the leading digits of 2^p. Let’s try the numbers 1, 12, 123, 1234, 12345…:

16 = 2^4
128 = 2^7
12379400392853802748... = 2^90
12340799625835686853... = 2^1545
12345257952011458590... = 2^34555
12345695478410965346... = 2^63293
12345673811591269861... = 2^4869721
12345678260232358911... = 2^5194868
12345678999199154389... = 2^62759188

But what about the numbers 9, 98, 987, 986, 98765… as leading digits of 2^p? They don’t appear as quickly:

9007199254740992 = 2^53
98079714615416886934... = 2^186
98726397006685494828... = 2^1548
98768356967522174395... = 2^21257
98765563827287722773... = 2^63296
98765426081858871289... = 2^5194871
98765430693066680199... = 2^11627034
98765432584491513519... = 2^260855656
98765432109571471006... = 2^1641098748

Why do fragments of 123456789 appear much sooner than fragments of 987654321? Well, even though all integers occur infinitely often as leading digits of 2^p, some integers occur more often than others, as it were. The leading digits of 2^p are actually governed by a fascinating mathematical phenomenon known as Benford’s law, which states, for example, that the single first digit, d, will occur with the frequency log10(1 + 1/d). Here are the actual frequencies of 1..9 for all powers of 2 up to 2^101000, compared with the estimate by Benford’s law:

1: 30% of leading digits ↔ 30.1% estimated
2: 17.55% ↔ 17.6%
3: 12.45% ↔ 12.49%
4: 09.65% ↔ 9.69%
5: 07.89% ↔ 7.92%
6: 06.67% ↔ 6.69%
7: 05.77% ↔ 5.79%
8: 05.09% ↔ 5.11%
9: 04.56% ↔ 4.57%

Because (inter alia) 1 appears as the first digit of 2^p far more often than 9 does, the fragments of 123456789 appear faster than the fragments of 987654321. Mutatis mutandis, the same applies in all other bases (apart from bases that are powers of 2, where there’s a single leading digit, 1, 2, 4, 8…, followed by 0s). But although a number like 123456789 occurs much frequently than 987654321 in 2^p expressed in base 10 (and higher), both integers occur infinitely often.

As do all other integers. And because stories can be expressed as numbers, all stories ever (and never) written appear in the powers of 2. Infinitely often. You’ll just have to trim the tail of the story-snake.

# Magiciprocal

A021023 Decimal expansion of 1/19.

0, 5, 2, 6, 3, 1, 5, 7, 8, 9, 4, 7, 3, 6, 8, 4, 2, 1, 0, 5, 2, 6, 3, 1, 5, 7, 8, 9, 4, 7, 3, 6, 8, 4, 2, 1, 0, 5, 2, 6, 3, 1, 5, 7, 8, 9, 4, 7, 3, 6, 8, 4, 2, 1, 0, 5, 2, 6, 3, 1, 5, 7, 8, 9, 4, 7, 3, 6, 8, 4, 2, 1, 0, 5, 2, 6, 3, 1, 5, 7, 8, 9, 4, 7, 3, 6, 8, 4, 2, 1, 0, 5, 2, 6, 3, 1, 5, 7, 8 [...] The magic square that uses the decimals of 1/19 is fully magic. — A021023 at the Online Encyclopedia of Integer Sequences

# Tri Num Sum

The Sum of ten consecutive Triangular Numbers:

Starting with T0 = 0, in base 10,

The sum of the first 10 triangular numbers from T0 to T9 = 165
The sum of the next 10 triangular numbers from T10 to T19 = 1165
The sum of the next 10 triangular numbers from T20 to T29 = 3165
The sum of the next 10 triangular numbers from T30 to T39 = 6165
The sum of the next 10 triangular numbers from T40 to T49 = 10165
The sum of the next 10 triangular numbers from T50 to T59 = 15165

and so on.

The same pattern is evident in other bases [when summing T0 to Tbase-1 and so on].

• As submitted by Julian Beauchamp, 9v19, to Shyam Sunder Gupta’s “Fascinating Triangular Numbers”.

# 1nf1nity

Here are the natural numbers or counting numbers:

• 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, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77... — A000027 at the Online Encyclopedia of Integer Sequences (OEIS)

Here are the prime numbers, or numbers divisible only by themselves and 1:

• 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271... — A000040 at the OEIS

Here are the palindromic prime numbers, or prime numbers that read the same both forwards and backwards:

• 2, 3, 5, 7, 11, 101, 131, 151, 181, 191, 313, 353, 373, 383, 727, 757, 787, 797, 919, 929, 10301, 10501, 10601, 11311, 11411, 12421, 12721, 12821, 13331, 13831, 13931, 14341, 14741, 15451, 15551, 16061, 16361, 16561, 16661, 17471, 17971, 18181... — A002385 at the OEIS

Finally, here are the repunit primes, or palindromic primes consisting only of 1s:

• 11, 1111111111111111111, 11111111111111111111111, 11111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111, 11111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111... — A004022 at the OEIS (see A004023 for numbers of 1s in each repunit prime)

It’s obvious that there are more counting numbers than primes, isn’t it? Well, no. There are in fact as many primes as counting numbers. And there may be as many palindromic primes as primes. And as many repunit primes as palindromic primes.