Category Archives: Mathematics

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?

Hour Power

by in Fractals, Geometry, Mathematics and tagged , , , | Leave a comment

How do you get an hourglass from this shape?

Rep-4 L-tromino

In fact, it’s easy. You simply divide the shape into four identical copies of itself, discard one copy, and repeat the process with each of the sub-copies:

l-triomino_124

Constructing an hourglass (animated)

↓

l-triomino_124_upright_static1

Hourglass (static)

Here are some more posts about what I call the hourglass fractal:

• The Hourglass Fractal at Overlord of the Über-feral

Tright Sights

by in Circles, Fractals, Geometry, Mathematics, Square Roots, Triangles and tagged , , , , , , | Leave a comment

Here’s a right triangle, where a^2 + b^2 = c^2. But what are the exact values of a, b, and c?

You might be able to guess by eye, but could you prove your guess? Now try the same right triangle tiled with three identical copies of itself:

1-√3-2 triangle as rep3 rep-tile

Now you can prove the exact values of a, b, and c. If the vertical side, a, is 1, then the hypotenuse, c, is 2, because the length that fits once into a fits twice into c. Therefore 2^2 = 1^2 + b^2 → 4 = 1 + b^2 → 4-1 = b^2 → 3 = b^2 → √3 = b. The horizontal side, b, has a length of √3 = 1.73205080757… So the right triangle is 1-√3-2. And if it’s rep3, that is, can be divided into three identical copies of itself, then it’s also rep9, rep27, and so on:

1-√3-2 triangle as rep9 rep-tile

1-√3-2 triangle as rep27 rep-tile

1-√3-2 triangle as rep81 rep-tile

1-√3-2 triangle as rep243 rep-tile

1-√3-2 triangle as rep729 rep-tile

Once you’ve got a rep-tile, you can create fractals. But the 1-√3-2 triangle is cramped. You need more space to work with. And it’s easy to find that space when you realize that a standard equilateral triangle can be divided into six 1-√3-2 triangles:

Equilateral triangle divided into six 1-√3-2 triangles

Equilateral triangle tiled with 1-√3-2 triangles (stage 1)

(please open in new window if image is distorted)

Equilateral triangle tiled with 1-√3-2 triangles (stage 2)

Equilateral triangle tiled with 1-√3-2 triangles (stage 3)

Here are variant colorings of the stage-3 tiled triangle:

But where are the fractals? In one way, you’ve already seen them. But they get more obvious like this:

Fractal stage 1

Fractal #2

Fractal #3

Fractal #4

Fractal #5

Fractal #6

Fractal #7

Fractal #8

Fractal (animated)

Another fractal stage 1

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Another fractal #8

Another fractal (animated)

And when you have a fractal created using an equilateral triangle, it’s easy to expand the fractal into a circle, like this:

Original fractal

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Fractal expanded into circle

Triangular fractal to circular fractal (animated)

Extra Tetra

by in Fractals, Geometry, Mathematics and tagged , , , , , , , , , , , , | Leave a comment

Construction of a Sierpiński tetrahedron (from WikiMedia)

Post-Performative Post-Scriptum

The toxic title of this incendiary intervention radically references George Harrison’s album Extra Texture (1975).

First Whirled Warp

by in Circles, Geometry, Mathematics, Squares, Triangles and tagged , , , , , , | Leave a comment

Imagine two points moving clockwise around the circumference of a circle. Find the midpoint between the two points when one point is moving twice as fast as the other. The midpoint will trace this shape:

Midpoint of two points moving around circle at speeds s and s*2

(n.b. to make things easier to see, the red circle shown here and elsewhere is slightly larger than the virtual circle used to calculate the midpoints)

Now suppose that one point is moving anticlockwise. The midpoint will now trace this shape:

Midpoint for s, -s*2

Now try three points, two moving at the same speed and one moving twice as fast:

Midpoint for s, s, s*2

When the point moving twice as fast is moving anticlockwise, this shape appears:

Midpoint for s, s, -s*2

Here are more of these midpoint-shapes:

Midpoint for s, s*3

Midpoint for s, -s*3

Midpoint for s*2, s*3

Midpoint for s, -s, s*2

Midpoint for s, s*2, -s*2

Midpoint for s, s*2, s*2

Midpoint for s, -s*3, -s*5

Midpoint for s, s*2, s*3

Midpoint for s, s*2, -s*3

Midpoint for s, -s*3, s*5

Midpoint for s, s*3, s*5

Midpoint for s, s, s, s*3

Midpoint for s, s, s, -s*3

Midpoint for s, s, -s, s*3

Midpoint for s, s, -s, -s*3

But what about points moving around the perimeter of a polygon? Here are the midpoints of two points moving clockwise around the perimeter of a square, with one point moving twice as fast as the other:

Midpoint for square with s, s*2

And when one point moves anticlockwise:

Midpoint for square with s, -s*2

If you adjust the midpoints so that the square fills a circle, they look like this:

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Midpoint for square with s, s*2, with square adjusted to fill circle

When the red circle is removed, the midpoint-shape is easier to see:

Midpoint for square with s, s*2, circ-adjusted

Here are more midpoint-shapes from squares:

Midpoint for s, s*3

Midpoint for s, -s*3

Midpoint for s, s*4

And some more circularly adjusted midpoint-shapes from squares:

Midpoint for s, s*3, circ-adjusted

Midpoint for s*2, s*3, circ-adjusted

Midpoint for s, s*5, circ-adjusted

Midpoint for s, s*6, circ-adjusted

Midpoint for s, s*7, circ-adjusted

Finally (for now), let’s look at triangles. If three points are moving clockwise around the perimeter of a triangle, one moving four times as fast as the other two, the midpoint traces this shape:

Midpoint for triangle with s, s, s*4

Now try one of the points moving anticlockwise:

Midpoint for s, s, -s*4

Midpoint for s, -s, s*4

If you adjust the midpoints so that the triangular space fills a circle, they look like this:

Midpoint for s, s, s*4, with triangular space adjusted to fill circle

Midpoint for s, -s, s*4, circ-adjusted

Midpoint for s, s, -s*4, circ-adjusted

There are lots more (infinitely more!) midpoint-shapes to see, so watch this (circularly adjusted) space.

Previously pre-posted (please peruse)

• Second Whirled Warp — more on points moving around polygons
• We Can Circ It Out — more on converting polygons into circles

Pi in the Bi

by in Binary numbers, Mathematics, Pi and tagged , , , , , , , , , , , , | Leave a comment

Binary is beautiful — both simple and subtle. What could be simpler than using only two digits to count with?


0, 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, 10000, 10001, 10010, 10011, 10100, 10101, 10110, 10111, 11000, 11001, 11010, 11011, 11100, 11101, 11110, 11111, 100000, 100001, 100010, 100011, 100100, 100101, 100110, 100111, 101000, 101001, 101010, 101011, 101100, 101101, 101110, 101111, 110000, 110001, 110010, 110011, 110100, 110101, 110110, 110111, 111000, 111001, 111010, 111011, 111100, 111101, 111110, 111111, 1000000...

But the simple patterns in the two digits of binary involve two of the most important numbers in mathematics: π and e (aka Euler’s number):


π = 3.141592653589793238462643383...
e = 2.718281828459045235360287471...

It’s easy to write π and e in binary:


π = 11.00100 10000 11111 10110 10101 00010...
e = 10.10110 11111 10000 10101 00010 11000...

But how do π and e appear in the patterns of binary 1 and 0? Well, suppose you use the digits of binary to generate the sums of distinct integers. For example, here are the sums of distinct integers you can generate with four digits of binary, if you count the digits from right to left (so the rightmost digit is 1, the the next-to-rightmost digit is 2, the next-to-leftmost digit is 3, and the leftmost digit is 4):


0000 → 0*4 + 0*3 + 0*2 + 0*1 = 0
0001 → 0*4 + 0*3 + 0*2 + 1*1 = 1*1 = 1
0010 → 0*4 + 0*3 + 1*2 + 0*1 = 1*2 = 2
0011 → 0*4 + 0*3 + 1*2 + 1*1 = 1*2 + 1*1 = 3
0100 → 1*3 = 3
0101 → 1*3 + 1*1 = 4
0110 → 3 + 2 = 5
0111 → 3 + 2 + 1 = 6
1000 → 4
1001 → 4 + 1 = 5
1010 → 4 + 2 = 6
1011 → 4 + 2 + 1 = 7
1100 → 4 + 3 = 7
1101 → 4 + 3 + 1 = 8
1110 → 4 + 3 + 2 = 9
1111 → 4 + 3 + 2 + 1 = 10

There are 16 sums (16 = 2^4) generating 11 integers, 0 to 10. But some integers involve more than one sum:


3 = 2 + 1 ← 0011
3 = 3 ← 0100

4 = 3 + 1 ← 0101
4 = 4 ← 1000

5 = 3 + 2 ← 0110
5 = 4 + 1 ← 1001

6 = 3 + 2 + 1 ← 0111
6 = 4 + 2 ← 1010

7 = 4 + 2 + 1 ← 1011
7 = 4 + 3 ← 1100

Note the symmetry of the sums: the binary number 0011, yielding 3, is the mirror of 1100, yielding 7; the binary number 0100, yielding 3 again, is the mirror of 1011, yielding 7 again. In each pair of mirror-sums, the two numbers, 3 and 7, are related by the formula 10-3 = 7 and 10-7 = 3. This also applies to 4 and 6, where 10-4 = 6 and 10-6 = 4, and to 5, which is its own mirror (because 10-5 = 5). Now, try mapping the number of distinct sums for 0 to 10 as a graph:

Graph for distinct sums of the integers 0 to 4

The graph show how 0, 1 and 2 have one sum each, 3, 4, 5, 6 and 7 have two sums each, and 8, 9 and 10 have one sum each. Now look at the graph for sums derived from three digits of binary:

Graph for distinct sums of the integers 0 to 3

The single taller line of the seven lines represents the two sums of 3, because three digits of binary yield only one sum for 0, 1, 2, 4, 5 and 6:


000 → 0
001 → 1
010 → 2
011 → 2 + 1 = 3
100 → 3
101 → 3 + 1 = 4
110 → 3 + 2 = 5
111 → 3 + 2 + 1 = 6

Next, look at graphs for sums derived from one to sixteen binary digits and note how the symmetry of the lines begins to create a beautiful curve (the y axis is normalized, so that the highest number of sums reaches the same height in each graph):

Graph for sums from 1 binary digit

Graph for sums from 2 binary digits

Graph for sums from 3 binary digits

Graph for sums from 4 binary digits

Graph for sums from 5 binary digits

Graph for sums from 6 binary digits

Graph for sums from 7 binary digits

Graph for sums from 8 binary digits

Graph for sums from 9 binary digits

Graph for sums from 10 binary digits

Graph for sums from 11 binary digits

Graph for sums from 12 binary digits

Graph for sums from 13 binary digits

Graph for sums from 14 binary digits

Graph for sums from 15 binary digits

Graph for sums from 16 binary digits

Graphs for 1 to 16 binary digits (animated)

You may recognize the shape emerging above as the bell curve, whose formula is this:

Formula for the normal distribution or bell curve (image from ThoughtCo)

And that’s how you can find pi in the bi, or π in the binary digits of 0, 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101…

(And how you find e too, as promised above.)

Post-Performative Post-Scriptum

I asked this question above: What could be simpler than using only two digits? Well, using only one digit is simpler still:


1, 11, 111, 1111, 11111, 111111, 1111111, 11111111, 111111111, 1111111111...

But I don’t see an easy way to find π and e in numbers like that.

Primal Stream

by in Integer Sequences, Mathematics, Prime numbers and tagged , , , , , , | Leave a comment

It’s obvious when you think about: an even number can never be the sum of two consecutive integers. Conversely, an odd number (except 1) is always the sum of two consecutive integers: 3 = 1 + 2; 5 = 2 + 3; 7 = 3 + 4; 9 = 4 + 5; and so on. The sum of three consecutive integers can be either odd or even: 6 = 1 + 2 + 3; 9 = 2 + 3 + 4. The sum of four consecutive integers must always be even: 1 + 2 + 3 + 4 = 10; 2 + 3 + 4 + 5 = 14. And so on.

But notice that 9 is the sum of consecutive integers in two different ways: 9 = 4 + 5 = 2 + 3 + 4. Having spotted that, I decided to look for numbers that were the sums of consecutive integers in the most different ways. These are the first few:

3 = 1 + 2 (number of sums = 1)
9 = 2 + 3 + 4 = 4 + 5 (s = 2)
15 = 1 + 2 + 3 + 4 + 5 = 4 + 5 + 6 = 8 + 7 = (s = 3)
45 (s = 5)
105 (s = 7)
225 (s = 8)
315 (s = 11)
945 (s = 15)
1575 (s = 17)
2835 (s = 19)
3465 (s = 23)
10395 (s = 31)


It was interesting that the number of different consecutive-integer sums for n was most often a prime number. Next I looked for the sequence at the Online Encyclopedia of Integer Sequences and discovered something that I hadn’t suspected:

A053624 Highly composite odd numbers: where d(n) increases to a record.

1, 3, 9, 15, 45, 105, 225, 315, 945, 1575, 2835, 3465, 10395, 17325, 31185, 45045, 121275, 135135, 225225, 405405, 675675, 1576575, 2027025, 2297295, 3828825, 6891885, 11486475, 26801775, 34459425, 43648605, 72747675, 130945815 — A053624 at OEIS

The notes add that the sequence is “Also least number k such that the number of partitions of k into consecutive integers is a record. For example, 45 = 22+23 = 14+15+16 = 7+8+9+10+11 = 5+6+7+8+9+10 = 1+2+3+4+5+6+7+8+9, six such partitions, but all smaller terms have fewer such partitions (15 has four).” When you don’t count the number n itself as a partition of n, you get 3 partitions for 15, i.e. consecutive integers sum to 15 in 3 different ways, so s = 3. I looked at more values for s and found that the stream of primes continued to flow:

3 → s = 1
9 = 3^2 → s = 2 (prime)
15 = 3 * 5 → s = 3 (prime)
45 = 3^2 * 5 → s = 5 (prime)
105 = 3 * 5 * 7 → s = 7 (prime)
225 = 3^2 * 5^2 → s = 8 = 2^3
315 = 3^2 * 5 * 7 → s = 11 (prime)
945 = 3^3 * 5 * 7 → s = 15 = 3 * 5
1575 = 3^2 * 5^2 * 7 → s = 17 (prime)
2835 = 3^4 * 5 * 7 → s = 19 (prime)
3465 = 3^2 * 5 * 7 * 11 → s = 23 (prime)
10395 = 3^3 * 5 * 7 * 11 → s = 31 (prime)
17325 = 3^2 * 5^2 * 7 * 11 → s = 35 = 5 * 7
31185 = 3^4 * 5 * 7 * 11 → s = 39 = 3 * 13
45045 = 3^2 * 5 * 7 * 11 * 13 → s = 47 (prime)
121275 = 3^2 * 5^2 * 7^2 * 11 → s = 53 (prime)
135135 = 3^3 * 5 * 7 * 11 * 13 → s = 63 = 3^2 * 7
225225 = 3^2 * 5^2 * 7 * 11 * 13 → s = 71 (prime)
405405 = 3^4 * 5 * 7 * 11 * 13 → s = 79 (prime)
675675 = 3^3 * 5^2 * 7 * 11 * 13 → s = 95 = 5 * 19
1576575 = 3^2 * 5^2 * 7^2 * 11 * 13 → s = 107 (prime)
2027025 = 3^4 * 5^2 * 7 * 11 * 13 → s = 119 = 7 * 17
2297295 = 3^3 * 5 * 7 * 11 * 13 * 17 → s = 127 (prime)
3828825 = 3^2 * 5^2 * 7 * 11 * 13 * 17 → s = 143 = 11 * 13
6891885 = 3^4 * 5 * 7 * 11 * 13 * 17 → s = 159 = 3 * 53
11486475 = 3^3 * 5^2 * 7 * 11 * 13 * 17 → s = 191 (prime)
26801775 = 3^2 * 5^2 * 7^2 * 11 * 13 * 17 → s = 215 = 5 * 43
34459425 = 3^4 * 5^2 * 7 * 11 * 13 * 17 → s = 239 (prime)
43648605 = 3^3 * 5 * 7 * 11 * 13 * 17 * 19 → s = 255 = 3 * 5 * 17
72747675 = 3^2 * 5^2 * 7 * 11 * 13 * 17 * 19 → s = 287 = 7 * 41
130945815 = 3^4 * 5 * 7 * 11 * 13 * 17 * 19 → s = 319 = 11 * 29


I can’t spot any way of predicting when n will yield a primal s, but I like the way that a simple question took an unexpected turn. When a number sets a record for the number of different ways it can be the sum of consecutive integers, that number will also be a highly composite odd number.

You Sixy Beast

by in 666, Integer Sequences, Mathematics, Number of the Beast, Palindromic numbers, Repdigits, Triangular Numbers and tagged , , , , , , , , , , , , | 2 Comments

666 is the Number of the Beast. But it’s much more than that. After all, it’s a number, so it has mathematical properties (everything has mathematical properties, but it’s a sine-qua-non of numbers). For example, 666 is a palindromic number, reading the same forwards and backwards. And it’s a repdigit, consisting of a single repeated digit. Now try answering this question: how many pebbles are there in this triangle?


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Counting the pebbles one by one would take a long time, but there’s a short-cut. Each line of the triangle after the first is one pebble longer than the previous line. There are 36 lines and therefore 36 pebbles in the final line. So the full number of pebbles = 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. And there’s an easy formula for that sum: (36^2 + 36) / 2 = (1296 + 36) / 2 = 1332 / 2 = 666.

So 666 is the 36th triangular number:


1 = 1
1+2 = 3
1+2+3 = 6
1+2+3+4 = 10
1+2+3+4+5 = 15
1+2+3+4+5+6 = 21
1+2+3+4+5+6+7 = 28
1+2+3+4+5+6+7+8 = 36
1+2+3+4+5+6+7+8+9 = 45
1+2+3+4+5+6+7+8+9+10 = 55
[...]
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 = 666

But what’s tri(666), the 666th triangular number? By the formula above, it equals (666^2 + 666) / 2 = (443556 + 666) / 2 = 444222 / 2 = 222111. But recall something else from above: tri(6) = 1+2+3+4+5+6 = 21. Is it a coincidence that tri(6) = 21 and tri(666) = 222111? No, it isn’t:


tri(6) = 21 = (6^2 + 6) / 2 = (36 + 6) / 2 = 42 / 2
tri(66) = 2211 = (66^2 + 66) / 2 = (4356 + 66) / 2 = 4422 / 2
tri(666) = 222111 = (666^2 + 666) / 2 = (443556 + 666) / 2 = 444222 / 2
tri(6666) = 22221111
tri(66666) = 2222211111
tri(666666) = 222222111111
tri(6666666) = 22222221111111
tri(66666666) = 2222222211111111
tri(666666666) = 222222222111111111
tri(6666666666) = 22222222221111111111
tri(66666666666) = 2222222222211111111111
tri(666666666666) = 222222222222111111111111
tri(6666666666666) = 22222222222221111111111111
tri(66666666666666) = 2222222222222211111111111111
tri(666666666666666) = 222222222222222111111111111111

So we’ve looked at tri(36) = 666 and tri(666) = 222111. Let’s go a step further: tri(222111) = 24666759216. So 666 appears again. And the sixiness carries on here:


tri(36) = 666
tri(3366) = 5666661
tri(333666) = 55666666611
tri(33336666) = 555666666666111
tri(3333366666) = 5555666666666661111
tri(333333666666) = 55555666666666666611111
tri(33333336666666) = 555555666666666666666111111
tri(3333333366666666) = 5555555666666666666666661111111
tri(333333333666666666) = 55555555666666666666666666611111111
tri(33333333336666666666) = 555555555666666666666666666666111111111
tri(3333333333366666666666) = 5555555555666666666666666666666661111111111
tri(333333333333666666666666) = 55555555555666666666666666666666666611111111111
tri(33333333333336666666666666) = 555555555555666666666666666666666666666111111111111
tri(3333333333333366666666666666) = 5555555555555666666666666666666666666666661111111111111
tri(333333333333333666666666666666) = 55555555555555666666666666666666666666666666611111111111111