Category Archives: Base Behavior

Bi-Bell Basics

by in Arithmetic, Base Behavior, Binary numbers, Blancmange Curves, Fibonacci Sequence, Fractals, Geometry, Mathematics, Powers and tagged , , , , , , , , , , , , , , , , , | Leave a comment

Here’s what you might call a Sisyphean sequence. It struggles upward, then slips back, over and over again:

1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 1, 2, 2, 3, 2, 3, 3, 4, 2...


The struggle goes on for ever. Every time it reaches a new maximum, it will fall back to 1 at the next step. And in fact 1, 2, 3 and all other integers occur infinitely often in the sequence, because it represents the digit-sums of binary numbers:

1 ← 1
1 = 1+0 ← 10 in binary = 2 in base ten
2 = 1+1 ← 11 = 3
1 = 1+0+0 ← 100 = 4
2 = 1+0+1 ← 101 = 5
2 = 1+1+0 ← 110 = 6
3 = 1+1+1 ← 111 = 7
1 = 1+0+0+0 ← 1000 = 8
2 = 1+0+0+1 ← 1001 = 9
2 = 1+0+1+0 ← 1010 = 10
3 = 1+0+1+1 ← 1011 = 11
2 = 1+1+0+0 ← 1100 = 12
3 = 1+1+0+1 ← 1101 = 13
3 = 1+1+1+0 ← 1110 = 14
4 = 1+1+1+1 ← 1111 = 15
1 = 1+0+0+0+0 ← 10000 = 16
2 = 1+0+0+0+1 ← 10001 = 17
2 = 1+0+0+1+0 ← 10010 = 18
3 = 1+0+0+1+1 ← 10011 = 19
2 = 1+0+1+0+0 ← 10100 = 20


Now here’s a related sequence in which all integers do not occur infinitely often:

1, 2, 3, 3, 4, 5, 6, 4, 5, 6, 7, 7, 8, 9, 10, 5, 6, 7, 8, 8, 9, 10, 11, 9, 10, 11, 12, 12, 13, 14, 15, 6, 7, 8, 9, 9, 10, 11, 12, 10, 11, 12, 13, 13, 14, 15, 16, 11, 12, 13, 14, 14, 15, 16, 17, 15, 16, 17, 18, 18, 19, 20, 21, 7, 8, 9, 10, 10, 11, 12, 13, 11, 12, 13, 14, 14, 15, 16, 17, 12, 13, 14, 15, 15, 16, 17, 18, 16, 17, 18, 19, 19, 20, 21, 22, 13, 14, 15, 16, 16, 17, 18, 19, 17, 18, 19, 20, 20, 21, 22, 23, 18, 19, 20, 21, 21, 22, 23, 24, 22, 23, 24, 25, 25, 26, 27, 28, 8, 9, 10, 11, 11, 12, 13, 14, 12, 13, 14, 15, 15...


The sequence represents the sum of the values of occupied columns in the binary numbers, reading from right to left:

10 in binary = 2 in base ten
21 (column values from right to left)
2*1 + 1*0 = 2

11 = 3
21
2*1 + 1*1 = 3

100 = 4
321 (column values from right to left)
3*1 + 2*0 + 1*0 = 3

101 = 5
321
3*1 + 2*0 + 1*1 = 4

110 = 6
321
3*1 + 2*1 + 1*0 = 5

111 = 7
321
3*1 + 2*1 + 1*1 = 6

1000 = 8
4321
4*1 + 3*0 + 2*0 + 1*0 = 4

1001 = 9
4321
4*1 + 3*0 + 2*0 + 1*1 = 5

1010 = 10
4321
4*1 + 3*0 + 2*1 + 1*0 = 6

1011 = 11
4321
4*1 + 3*0 + 2*1 + 1*1 = 7

1100 = 12
4321
4*1 + 3*1 + 2*0 + 1*0 = 7

1101 = 13
4321
4*1 + 3*1 + 2*0 + 1*1 = 8

1110 = 14
4321
4*1 + 3*1 + 2*1 + 1*0 = 9

1111 = 15
4321
4*1 + 3*1 + 2*1 + 1*1 = 10

10000 = 16
54321
5*1 + 4*0 + 3*0 + 2*0 + 1*0 = 5


In that sequence, although no number occurs infinitely often, some numbers occur more often than others. If you represent the count of sums up to a certain digit-length as a graph, you get a famous shape:

Bell curve formed by the count of column-sums in base 2

Bi-bell curves for 1 to 16 binary digits (animated)

In “Pi in the Bi”, I looked at that way of forming the bell curve and called it the bi-bell curve. Now I want to go further. Suppose that you assign varying values to the columns and try other bases. For example, what happens if you assign the values 2^p + 1 to the columns, reading from right to left, then use base 3 to generate the sums? These are the values of 2^p + 1:

2, 3, 5, 9, 17, 33, 65, 129, 257, 513, 1025...


And here’s an example of how you generate a column-sum in base 3:

2102 in base 3 = 65 in base ten
9532 (column values from right to left)
2*9 + 1*5 + 0*3 + 2*2 = 27


The graphs for these column-sums using base 3 look like this as the digit-length rises. They’re no longer bell-curves (and please note that widths and heights have been normalized so that all graphs fit the same space):

Graph for the count of column-sums in base 3 using 2^p + 1 (digit-length <= 7)

(width and height are normalized)

Graph for base 3 and 2^p + 1 (dl<=8)

Graph for base 3 and 2^p + 1 (dl<=9)

Graph for base 3 and 2^p + 1 (dl<=10)

Graph for base 3 and 2^p + 1 (dl<=11)

Graph for base 3 and 2^p + 1 (dl<=12)

Graph for base 3 and 2^p + 1 (animated)

Now try base 3 and column-values of 2^p + 2 = 3, 4, 6, 10, 18, 34, 66, 130, 258, 514, 1026…

Graph for base 3 and 2^p + 2 (dl<=7)

Graph for base 3 and 2^p + 2 (dl<=8)

Graph for base 3 and 2^p + 2 (dl<=9)

Graph for base 3 and 2^p + 2 (dl<=10)

Graph for base 3 and 2^p + 2 (animated)

Now try base 5 and 2^p + 1 for the columns. The original bell curve has become like a fractal called the blancmange curve:

Graph for base 5 and 2^p + 1 (dl<=7)

Graph for base 5 and 2^p + 1 (dl<=8)

Graph for base 5 and 2^p + 1 (dl<=9)

Graph for base 5 and 2^p + 1 (dl<=10)

Graph for base 5 and 2^p + 1 (animated)

And finally, return to base 2 and try the Fibonacci numbers for the columns:

Graph for base 2 and Fibonacci numbers = 1,1,2,3,5… (dl<=7)

Graph for base 2 and Fibonacci numbers (dl<=9)

Graph for base 2 and Fibonacci numbers (dl<=11)

Graph for base 2 and Fibonacci numbers (dl<=13)

Graph for base 2 and Fibonacci numbers (dl<=15)

Graph for base 2 and Fibonacci numbers (animated)

Previously Pre-Posted…

• Pi in the Bi — bell curves generated by binary digits

A Walk on the Wide Side

by in Base Behavior, Digit-sums, Fibonacci Sequence, Integer Sequences, Irrational numbers, φ, Mathematics, Phi and tagged , , , , , , , , , , , , , , , | Leave a comment

How wide is a number? The obvious answer is to count digits and say that 1 and 9 are one digit wide, 11 and 99 are two digits wide, 111 and 999 are three digits wide, and so on. But that isn’t a very good answer. 111 and 999 are both three digits wide, but 999 is nine larger times than 111. And although 111 and 999 are both one digit wider than 11 and 99, 111 is much closer to 99 than 999 is to 111.

So there’s got to be a better answer to the question. I came across it indirectly, when I started looking at carries in powers. I wanted to know how fast a number grew in digit-width as it was multiplied repeatedly by, say, 2. For example, 2^3 = 8 and 2^4 = 16, so there’s been a carry at the far left and 2^4 = 16 has increased in digit-width by 1 over 2^3 = 8. After that, 2^6 = 64 and 2^7 = 128, so there’s another carry and another increase in digit-width. I wrote a program to sum the carries and divide them by the power. If I were better at math, I would’ve known what the value of carries / power was going to be. Here’s the program beginning to find it (it begins with a carry of 1, to mark 2^0 = 1 as creating a digit ex nihilo, as it were):


8 = 2^3
16 = 2^4 → 2 / 4 = 0.5
64 = 2^6
128 = 2^7 → 3 / 7 = 0.4285714285714285714285714286
512 = 2^9
1024 = 2^10 → 4 / 10 = 0.4
8192 = 2^13
16384 = 2^14 → 5 / 14 = 0.3571428571428571428571428571
65536 = 2^16
131072 = 2^17 → 6 / 17 = 0.3529411764705882352941176471
524288 = 2^19
1048576 = 2^20 → 7 / 20 = 0.35
8388608 = 2^23
16777216 = 2^24 → 8 / 24 = 0.3...
67108864 = 2^26
134217728 = 2^27 → 9 / 27 = 0.3...
536870912 = 2^29
1073741824 = 2^30 → 10 / 30 = 0.3...
8589934592 = 2^33
17179869184 = 2^34 → 11 / 34 = 0.3235294117647058823529411765
68719476736 = 2^36
137438953472 = 2^37 → 12 / 37 = 0.3243243243243243243243243243
549755813888 = 2^39
1099511627776 = 2^40 → 13 / 40 = 0.325
8796093022208 = 2^43
17592186044416 = 2^44 → 14 / 44 = 0.318...
70368744177664 = 2^46
140737488355328 = 2^47 → 15 / 47 = 0.3191489361702127659574468085
562949953421312 = 2^49
1125899906842624 = 2^50 → 16 / 50 = 0.32
9007199254740992 = 2^53
18014398509481984 = 2^54 → 17 / 54 = 0.3148...
72057594037927936 = 2^56
144115188075855872 = 2^57 → 18 / 57 = 0.3157894736842105263157894737
576460752303423488 = 2^59
1152921504606846976 = 2^60 → 19 / 60 = 0.316...
9223372036854775808 = 2^63
18446744073709551616 = 2^64 → 20 / 64 = 0.3125
73786976294838206464 = 2^66
147573952589676412928 = 2^67 → 21 / 67 = 0.3134328358208955223880597015
590295810358705651712 = 2^69
1180591620717411303424 = 2^70 → 22 / 70 = 0.3142857...
9444732965739290427392 = 2^73
18889465931478580854784 = 2^74 → 23 / 74 = 0.3108...
75557863725914323419136 = 2^76
151115727451828646838272 = 2^77 → 24 / 77 = 0.3116883...
604462909807314587353088 = 2^79
1208925819614629174706176 = 2^80 → 25 / 80 = 0.3125
9671406556917033397649408 = 2^83
19342813113834066795298816 = 2^84 → 26 / 84 = 0.3095238095238095238095238095
77371252455336267181195264 = 2^86
154742504910672534362390528 = 2^87 → 27 / 87 = 0.3103448275862068965517241379
618970019642690137449562112 = 2^89
1237940039285380274899124224 = 2^90 → 28 / 90 = 0.31...
9903520314283042199192993792 = 2^93
19807040628566084398385987584 = 2^94 → 29 / 94 = 0.3085106382978723404255319149
79228162514264337593543950336 = 2^96
158456325028528675187087900672 = 2^97 → 30 / 97 = 0.3092783505154639175257731959
633825300114114700748351602688 = 2^99
1267650600228229401496703205376 = 2^100 → 31 / 100 = 0.31

After calculating 2^p higher and higher (I discarded trailing digits of 2^p), I realized that the answer — carries / power — was converging on a value of slightly less than 0.30103. In the end (doh!), I realized that what I was calculating was the logarithm of 2 in base 10:


log(2) = 0.3010299956639811952137388947...
10^0.301029995663981... = 2

You can use then same carries-and-powers method to approximate the values of other logarithms:


log(1) = 0
log(2) = 0.3010299956639811952137388947...
log(3) = 0.4771212547196624372950279033...
log(4) = 0.6020599913279623904274777894...
log(5) = 0.6989700043360188047862611053...
log(6) = 0.7781512503836436325087667980...
log(7) = 0.8450980400142568307122162586...
log(8) = 0.9030899869919435856412166842...
log(9) = 0.9542425094393248745900558065...

I also realized logarithms are a good answer to the question I raised above: How wide is a number? The logs of the powers of 2 are multiples of log(2):


    log(2^1) = log(2) = 0.301029995663981195213738894
    log(2^2) = log(4) = 0.602059991327962390427477789 = 2 * log(2)
    log(2^3) = log(8) = 0.903089986991943585641216684 = 3 * log(2)
   log(2^4) = log(16) = 1.204119982655924780854955579 = 4 * log(2)
   log(2^5) = log(32) = 1.505149978319905976068694474 = 5 * log(2)
   log(2^6) = log(64) = 1.806179973983887171282433368 = 6 * log(2)
  log(2^7) = log(128) = 2.107209969647868366496172263 = 7 * log(2)
  log(2^8) = log(256) = 2.408239965311849561709911158 = 8 * log(2)
  log(2^9) = log(512) = 2.709269960975830756923650053 = 9 * log(2)
log(2^10) = log(1024) = 3.010299956639811952137388947 = 10 * log(2)

4 is 2 times larger than 2 and, in a sense, the width of 4 is 0.301029995663981… greater than the width of 2. As you can see, when the integer part of the log-sum increases by 1, so does the digit-width of the power:


 log(2^3) = log(8) = 0.903089986991943585641216684 = 3 * log(2)
log(2^4) = log(16) = 1.204119982655924780854955579 = 4 * log(2)

[...]

 log(2^6) = log(64) = 1.806179973983887171282433368 = 6 * log(2)
log(2^7) = log(128) = 2.107209969647868366496172263 = 7 * log(2)

[...]

  log(2^9) = log(512) = 2.709269960975830756923650053 = 9 * log(2)
log(2^10) = log(1024) = 3.01029995663981195213738894 = 10 * log(2)

In other words, powers of 2 are increasing in width by 0.301029995663981… units. When the increase flips the integer part of the log-sum up by 1, the digit-width or digit-count also increases by 1. To find the digit-count of a number, n, in a particular base, you simply take the integer part of log(n,b) and add 1. In base 10, the log of 123456789 is 8.091514… The integer part is 8 and 8+1 = 9. But it also makes perfect sense that log(1) = 0. No matter how many times you multiply a number by 1, the number never changes. That is, its width stays the same. So you can say that 1 has a width of 0, while 2 has a width of 0.301029995663981…

Logarithms also answer a question pre-previously raised on Overlord of the Über-Feral: Why are the Fibonacci numbers so productive in base 11 for digsum(fib(k)) = k? In base 10, such numbers are quickly exhausted:


digsum(fib(1)) = 1 = digsum(1)
digsum(fib(5)) = 5 = digsum(5)
digsum(fib(10)) = 10 = digsum(55)
digsum(fib(31)) = 31 = digsum(1346269)
digsum(fib(35)) = 35 = digsum(9227465)
digsum(fib(62)) = 62 = digsum(4052739537881)
digsum(fib(72)) = 72 = digsum(498454011879264)
digsum(fib(175)) = 175 = digsum(1672445759041379840132227567949787325)
digsum(fib(180)) = 180 = digsum(18547707689471986212190138521399707760)
digsum(fib(216)) = 216 = digsum(619220451666590135228675387863297874269396512)
digsum(fib(251)) = 251 = digsum(12776523572924732586037033894655031898659556447352249)
digsum(fib(252)) = 252 = digsum(20672849399056463095319772838289364792345825123228624)
digsum(fib(360)) = 360
digsum(fib(494)) = 494
digsum(fib(540)) = 540
digsum(fib(946)) = 946
digsum(fib(1188)) = 1188
digsum(fib(2222)) = 2222

In base 11, such numbers go on and on:


digsum(fib(1),b=11) = 1 = digsum(1) (k=1)
digsum(fib(5),b=11) = 5 = digsum(5) (k=5)
digsum(fib(12),b=11) = 12 = digsum(1A2) (k=13)
digsum(fib(38),b=11) = 38 = digsum(855138A1) (k=41)
digsum(fib(49)) = 49 = digsum(2067A724762) (k=53) (c=5)
digsum(fib(50)) = 50 = digsum(542194A6905) (k=55)
digsum(fib(55)) = 55 = digsum(54756364A280) (k=60)
digsum(fib(56)) = 56 = digsum(886283256841) (k=61)
digsum(fib(82)) = 82 = digsum(57751318A9814A6410) (k=90)
digsum(fib(89)) = 89 = digsum(140492673676A06482A2) (k=97)
digsum(fib(144)) = 144 = digsum(401631365A48A784A09392136653457871) (k=169) (c=10)
digsum(fib(159)) = 159 = digsum(67217257641069185100889658A1AA72A0805) (k=185)
digsum(fib(166)) = 166 = digsum(26466A3A88237918577363A2390343388205432) (k=193)
digsum(fib(186)) = 186 = digsum(6A963147A9599623A20A05390315140A21992A96005) (k=215)
digsum(fib(221)) = 221 (k=265) (c=15)
digsum(fib(225)) = 225 (k=269)
digsum(fib(2A1)) = 2A1 (k=353)
digsum(fib(2A3)) = 2A3 (k=355)

[...]

digsum(fib(39409)) = 39409 (k=56395)
digsum(fib(3958A)) = 3958A (k=56605) (c=295)
digsum(fib(3965A)) = 3965A (k=56693)
digsum(fib(3A106)) = 3A106 (k=57360)
digsum(fib(3AA46)) = 3AA46 (k=58493)
digsum(fib(40140)) = 40140 (k=58729)
digsum(fib(4222A)) = 4222A (k=61500) (c=300)
digsum(fib(42609)) = 42609 (k=61961)
digsum(fib(42775)) = 42775 (k=62155)
digsum(fib(4287A)) = 4287A (k=62281)
digsum(fib(430A2)) = 430A2 (k=62669)
digsum(fib(43499)) = 43499 (k=63149) (c=305)
digsum(fib(435A9)) = 435A9 (k=63281)

[...]

digsum(fib(157476)) = 157476 (k=244140) (c=525)
digsum(fib(158470)) = 158470 (k=245465)
digsum(fib(159037)) = 159037 (k=246275)
digsum(fib(159285)) = 159285 (k=246570)
digsum(fib(159978)) = 159978 (k=247409)
digsum(fib(162993)) = 162993 (k=252750) (c=530)
digsum(fib(163A32)) = 163A32 (k=254135)
digsum(fib(164918)) = 164918 (k=255329)
digsum(fib(166985)) = 166985 (k=258065)
digsum(fib(167234)) = 167234 (k=258493)
digsum(fib(167371)) = 167371 (k=258655) (c=535)
digsum(fib(1676A5)) = 1676A5 (k=259055)
digsum(fib(16992A)) = 16992A (k=261997)

[...]

When do these numbers run out in base 11? I don’t know, but I do know why there are so many of them. The answer involves the logarithm of a special number. The most famous aspect of Fibonacci numbers is that the ratio, fib(k) / fib(k-1), of successive numbers converges on an irrational constant known as Φ. Here are the first Fibonacci numbers, where fib(k) = fib(k-2) + fib(k-1) (in other words, 1+1 = 2, 1+2 = 3, 2+3 = 5, and so on):


1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, ...

And here are the first ratios:


1 / 1 = 1
2 / 1 = 2
3 / 2 = 1.5
5 / 3 = 1.6...
8 / 5 = 1.6
13 / 8 = 1.625
21 / 13 = 1.6153846...
34 / 21 = 1.619047...
55 / 34 = 1.617647058823529411764705882
89 / 55 = 1.618...
144 / 89 = 1.617977528089887640449438202
233 / 144 = 1.61805...
377 / 233 = 1.618025751072961373390557940
610 / 377 = 1.618037135278514588859416446
987 / 610 = 1.618032786885245901639344262
1597 / 987 = 1.618034447821681864235055724
2584 / 1597 = 1.618033813400125234815278648
4181 / 2584 = 1.618034055727554179566563468
6765 / 4181 = 1.618033963166706529538387946
[...]

The ratios get closer and closer to Φ = 1.618033988749894848204586834… = (√5 + 1) / 2. In other words, fib(k) ≈ fib(k-1) * Φ = fib(k-1) * 1.618… in base 10. This means that the digit-length of fib(k) ≈ integer(k * log(&Phi)) + 1. In base b, the average value of a digit in a Fibonacci number is (b^2-b) / 2b. Therefore in base 10, the average value of a digit is (10^2-10) / 20 = 90 / 20 = 4.5. The average value of digsum(fib(k)) ≈ 4.5 * log(&Phi) * k = 4.5 * 0.20898764… * k = 0.940444… * k. It isn’t surprising that as fib(k) gets larger, digsum(fib(k)) tends to get smaller than k.

In base 10, anyway. But what about base 11? In base 11, log(Φ) = 0.20068091818623… and the average value of a base-11 digit in fib(k) is 5 = 110 / 22 = (11^2 – 11) / 22. Therefore the average value of digsum(fib(k)) in base 11 is 5 * log(&Phi) * k = 5 * 0.20068091818623… * k = 1.00340459… * k. The average value of digsum(fib(k)) is much closer to k and it’s not surprising that for so many fib(k) in base 11, digsum(fib(k)) = k. In base 11, log(Φ) ≈ 1/5 and because the average digval is 5, digsum(fib(k)) ≈ 5 * 1/5 * k = 1 * k = k. As we’ve seen, that isn’t true in base 10. Nor is it true in base 12, where log(Φ) = 0.1936538843826… and average digval is 5.5 = (12^2 – 12) / 24 = 132 / 24. Therefore the average value in base 12 of digsum(fib(k)) = 1.0650963641… * k. The function digsum(fib(k)) = k rapidly dries up in base 12, just as it does in base 10:


digsum(fib(1),b=12) = 1 = digsum(1) (k=1)
digsum(fib(5),b=12) = 5 = digsum(5) (k=5)
digsum(fib(11) = 11 = digsum(175) (k=13)
digsum(fib(12) = 12 = digsum(275) (k=14)
digsum(fib(75) = 75 = digsum(976446538A0863811) (k=89) (c=5)
digsum(fib(80) = 80 = digsum(1B3643B50939808B400) (k=96)
digsum(fib(A3) = A3 = digsum(35147A566682BB9529034402) (k=123)
digsum(fib(165) = 165 (k=221)
digsum(fib(283) = 283 (k=387)
digsum(fib(2AB) = 2AB (k=419) (c=10)
digsum(fib(39A) = 39A (k=550)
digsum(fib(460) = 460 (k=648)
digsum(fib(525) = 525 (k=749)
digsum(fib(602) = 602 (k=866)
digsum(fib(624) = 624 (k=892) (c=15)
digsum(fib(781) = 781 (k=1105)
digsum(fib(1219) = 1219 (k=2037)

Previously Pre-Posted…

• Mötley Vüe — more on digsum(fib(k)) = k

Two be Continued…

by in Base Behavior, Digit-sums, Fibonacci Sequence, Integer Sequences, Mathematics, Phi and tagged , , , , , , , , , , , | Leave a comment

Here’s a useless fact that nobody interested in mathematics would ever forget: digsum(fib(2222)) = 2222. That is, if you add the digits of the 2222nd Fibonacci number, you get 2222:


fib(2222) = 104,966,721,620,282,584,734,867,037,988,863,914,269,721,309,244,628,258,918,225,835,217,264,239,539,186,480,867,849,267,122,885,365,019,934,494,625,410,255,045,832,359,715,759,649,385,824,745,506,982,513,773,397,742,803,445,080,995,617,047,976,796,168,678,756,479,470,761,439,513,575,962,955,568,645,505,845,492,393,360,201,582,183,610,207,447,528,637,825,187,188,815,786,270,477,935,419,631,184,553,635,981,047,057,037,341,800,837,414,913,595,584,426,355,208,257,232,868,908,837,817,478,483,039,310,790,967,631,454,123,105,472,742,221,897,397,857,677,674,619,381,961,429,837,434,434,636,098,678,708,225,493,682,469,561

2222 = 1 + 0 + 4 + 9 + 6 + 6 + 7 + 2 + 1 + 6 + 2 + 0 + 2 + 8 + 2 + 5 + 8 + 4 + 7 + 3 + 4 + 8 + 6 + 7 + 0 + 3 + 7 + 9 + 8 + 8 + 8 + 6 + 3 + 9 + 1 + 4 + 2 + 6 + 9 + 7 + 2 + 1 + 3 + 0 + 9 + 2 + 4 + 4 + 6 + 2 + 8 + 2 + 5 + 8 + 9 + 1 + 8 + 2 + 2 + 5 + 8 + 3 + 5 + 2 + 1 + 7 + 2 + 6 + 4 + 2 + 3 + 9 + 5 + 3 + 9 + 1 + 8 + 6 + 4 + 8 + 0 + 8 + 6 + 7 + 8 + 4 + 9 + 2 + 6 + 7 + 1 + 2 + 2 + 8 + 8 + 5 + 3 + 6 + 5 + 0 + 1 + 9 + 9 + 3 + 4 + 4 + 9 + 4 + 6 + 2 + 5 + 4 + 1 + 0 + 2 + 5 + 5 + 0 + 4 + 5 + 8 + 3 + 2 + 3 + 5 + 9 + 7 + 1 + 5 + 7 + 5 + 9 + 6 + 4 + 9 + 3 + 8 + 5 + 8 + 2 + 4 + 7 + 4 + 5 + 5 + 0 + 6 + 9 + 8 + 2 + 5 + 1 + 3 + 7 + 7 + 3 + 3 + 9 + 7 + 7 + 4 + 2 + 8 + 0 + 3 + 4 + 4 + 5 + 0 + 8 + 0 + 9 + 9 + 5 + 6 + 1 + 7 + 0 + 4 + 7 + 9 + 7 + 6 + 7 + 9 + 6 + 1 + 6 + 8 + 6 + 7 + 8 + 7 + 5 + 6 + 4 + 7 + 9 + 4 + 7 + 0 + 7 + 6 + 1 + 4 + 3 + 9 + 5 + 1 + 3 + 5 + 7 + 5 + 9 + 6 + 2 + 9 + 5 + 5 + 5 + 6 + 8 + 6 + 4 + 5 + 5 + 0 + 5 + 8 + 4 + 5 + 4 + 9 + 2 + 3 + 9 + 3 + 3 + 6 + 0 + 2 + 0 + 1 + 5 + 8 + 2 + 1 + 8 + 3 + 6 + 1 + 0 + 2 + 0 + 7 + 4 + 4 + 7 + 5 + 2 + 8 + 6 + 3 + 7 + 8 + 2 + 5 + 1 + 8 + 7 + 1 + 8 + 8 + 8 + 1 + 5 + 7 + 8 + 6 + 2 + 7 + 0 + 4 + 7 + 7 + 9 + 3 + 5 + 4 + 1 + 9 + 6 + 3 + 1 + 1 + 8 + 4 + 5 + 5 + 3 + 6 + 3 + 5 + 9 + 8 + 1 + 0 + 4 + 7 + 0 + 5 + 7 + 0 + 3 + 7 + 3 + 4 + 1 + 8 + 0 + 0 + 8 + 3 + 7 + 4 + 1 + 4 + 9 + 1 + 3 + 5 + 9 + 5 + 5 + 8 + 4 + 4 + 2 + 6 + 3 + 5 + 5 + 2 + 0 + 8 + 2 + 5 + 7 + 2 + 3 + 2 + 8 + 6 + 8 + 9 + 0 + 8 + 8 + 3 + 7 + 8 + 1 + 7 + 4 + 7 + 8 + 4 + 8 + 3 + 0 + 3 + 9 + 3 + 1 + 0 + 7 + 9 + 0 + 9 + 6 + 7 + 6 + 3 + 1 + 4 + 5 + 4 + 1 + 2 + 3 + 1 + 0 + 5 + 4 + 7 + 2 + 7 + 4 + 2 + 2 + 2 + 1 + 8 + 9 + 7 + 3 + 9 + 7 + 8 + 5 + 7 + 6 + 7 + 7 + 6 + 7 + 4 + 6 + 1 + 9 + 3 + 8 + 1 + 9 + 6 + 1 + 4 + 2 + 9 + 8 + 3 + 7 + 4 + 3 + 4 + 4 + 3 + 4 + 6 + 3 + 6 + 0 + 9 + 8 + 6 + 7 + 8 + 7 + 0 + 8 + 2 + 2 + 5 + 4 + 9 + 3 + 6 + 8 + 2 + 4 + 6 + 9 + 5 + 6 + 1

Numbers like this, where k = digsum(fib(k)), are rare. And 2222 is almost certainly the last of them. These are the relevant listings at the Online Encyclopedia of Integer Sequences:


0, 1, 5, 10, 31, 35, 62, 72, 175, 180, 216, 251, 252, 360, 494, 504, 540, 946, 1188, 2222 — A020995, Numbers k such that the sum of the digits of Fibonacci(k) is k.

0, 1, 5, 55, 1346269, 9227465, 4052739537881, 498454011879264, 1672445759041379840132227567949787325, 18547707689471986212190138521399707760, 619220451666590135228675387863297874269396512... — A067515, Fibonacci numbers with index = digit sum.

At least, they’re rare in base 10. What about other bases? Well, they’re rare in all other bases except one: base 11. When I looked there, I quickly found more than 450 numbers where digsum(fib(k),b=11) = k. So here’s an interesting little problem: Why is base 11 so productive? Or maybe I should say: Φ is base 11 so productive?

Agogic Arithmetic

by in Base Behavior, Integer Sequences, Mathematics, Triangular Numbers and tagged , , , , , | Leave a comment

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

by in 666, Base Behavior, Mathematics, Number of the Beast and tagged , , , , , , , | Leave a comment

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

by in Base Behavior, Fibonacci Sequence, Mathematics, Reciprocals and tagged , , , , , , , , , , | Leave a comment

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

by in √2, Base Behavior, Integer Sequences, Irrational numbers, Mathematics, Powers, Square root of 2, Square Roots and tagged , , , , , , , , , , , , , , , , , , | Leave a comment

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’

by in Base Behavior, Integer Sequences, Mathematics, Powers and tagged , , , , , , , , , , , , , , | Leave a comment

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

by in √2, Base Behavior, Infinity, Integer Sequences, Language, Mathematics, Powers, Square root of 2 and tagged , , , , , , , , , , , , , | Leave a comment

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.