Tag Archives: undecimal

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?

Mötley Vüe

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

Here’s the Fibonacci sequence, where each term (after the first two) is created by adding the two previous numbers:


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

In “Fib and Let Tri”, I described how my eye was caught by 55, which is a palindrome, reading the same backwards and forwards. “Were there any other Fibonacci palindromes?” I wondered. So I looked to see. Now my eye has been caught by 55 again, but for another reason. It should be easy to spot another interesting aspect to 55 when the Fibonacci numbers are set out like this:


fib(1) = 1
fib(2) = 1
fib(3) = 2
fib(4) = 3
fib(5) = 5
fib(6) = 8
fib(7) = 13
fib(8) = 21
fib(9) = 34
fib(10) = 55
fib(11) = 89
fib(12) = 144
fib(13) = 233
fib(14) = 377
fib(15) = 610
fib(16) = 987
fib(17) = 1597
fib(18) = 2584
fib(19) = 4181
fib(20) = 6765
[...]

55 is fib(10), the 10th Fibonacci number, and 5+5 = 10. That is, digsum(fib(10)) = 10. What other Fibonacci numbers work like that? I soon found some and confirmed my answer at the Online Encyclopedia of Integer Sequences:


1, 5, 10, 31, 35, 62, 72, 175, 180, 216, 251, 252, 360, 494, 504, 540, 946, 1188, 2222 — A020995 at OEIS

And that seems to be the lot, according to the OEIS. In base 10, at least, but why stop at base 10? When I looked at base 11, the numbers of digsum(fib(k)) = k didn’t stop coming, because I couldn’t take the Fibonacci numbers very high on my computer. But the OEIS gives a much longer list, starting like this:


1, 5, 13, 41, 53, 55, 60, 61, 90, 97, 169, 185, 193, 215, 265, 269, 353, 355, 385, 397, 437, 481, 493, 617, 629, 630, 653, 713, 750, 769, 780, 889, 905, 960, 1013, 1025, 1045, 1205, 1320, 1405, 1435, 1501, 1620, 1650, 1657, 1705, 1735, 1769, 1793, 1913, 1981, 2125, 2153, 2280, 2297, 2389, 2413, 2460, 2465, 2509, 2533, 2549, 2609, 2610, 2633, 2730, 2749, 2845, 2893, 2915, 3041, 3055, 3155, 3209, 3360, 3475, 3485, 3521, 3641, 3721, 3749, 3757, 3761, 3840, 3865, 3929, 3941, 4075, 4273, 4301, 4650, 4937, 5195, 5209, 5435, 5489, 5490, 5700, 5917, 6169, 6253, 6335, 6361, 6373, 6401, 6581, 6593, 6701, 6750, 6941, 7021, 7349, 7577, 7595, 7693, 7740, 7805, 7873, 8009, 8017, 8215, 8341, 8495, 8737, 8861, 8970, 8995, 9120, 9133, 9181, 9269, 9277, 9535, 9541, 9737, 9935, 9953, 10297, 10609, 10789, 10855, 11317, 11809, 12029, 12175... — A025490 at OEIS

The list ends with 1636597 = A18666[b11] and the OEIS says that 1636597 almost certainly completes the list. According to David C. Terr’s paper “On the Sums of Fibonacci Numbers” (pdf), published in the Fibonacci Quarterly in 1996, the estimated digit-sum for the k-th Fibonacci number in base b is given by the formula (b-1)/2 * k * log(b,φ), where log(b,φ) is the logarithm in base b of the golden ratio, 1·61803398874… Terr then notes that the simplified formula (b-1)/2 * log(b,φ) gives the estimated average ratio digsum(fib(k)) / k in base b. Here are the estimates for bases 2 to 20:


b02 = 0.3471209568153086...
b03 = 0.4380178794859424...
b04 = 0.5206814352229629...
b05 = 0.5979874356654401...
b06 = 0.6714235829697111...
b07 = 0.7418818776805580...
b08 = 0.8099488992357201...
b09 = 0.8760357589718848...
b10 = 0.9404443811249043...
b11 = 1.0034045909311624...
b12 = 1.0650963641043091...
b13 = 1.1256639207937723...
b14 = 1.1852250528196852...
b15 = 1.2438775226715552...
b16 = 1.3017035880574074...
b17 = 1.3587732842474014...
b18 = 1.4151468584732730...
b19 = 1.4708766105122322...
b20 = 1.5260083080264088...

In base 2, you can expect digsum(fib(k)) to be much smaller than k; in base 20, you can expect digsum(fib(k)) to be much larger. But as you can see, the estimate for base 11, 1.0034045909311624…, is very nearly 1. That’s why base 11 produces so many results for digsum(fib(k)) = k, because only a slight deviation from the estimate might create a perfect ratio of 1 for digsum(fib(k)) / k, i.e. digsum(fib(k)) = k. But in the end the results run out in base 11 too, because as k gets higher and fib(k) gets bigger, the estimate becomes more and more accurate and digsum(fib(k)) > k. With lower k, digsum(fib(k)) can easily fall below k or match k. That happens in other bases, but because their estimates are further from 1, results for digsum(fib(k)) = k run out much more quickly.

To see this base behavior represented visually, I’ve created Ulam-like spirals for k using three colors: blue for digsum(fib(k)) < k, yellow for digsum(fib(k)) > k, and red for digsum(fib(k)) = k (with the green square at the center representing fib(1) = 1). As you can see below, the spiral for base 11 immediately stands out. It’s motley, not dominated by blue or yellow like the other spirals:

Spiral for digsum(fib(k)) in base 9
(blue for digsum(fib(k)) < k, yellow for digsum(fib(k)) > k, red for digsum(fib(k)) = k, green for fib(1))

Spiral for digsum(fib(k)) in base 10

Spiral for digsum(fib(k)) in base 11 — a motley view of blue, yellow and red

Spiral for digsum(fib(k)) in base 12

Spiral for digsum(fib(k)) in base 13

Finally, here are spirals at higher and higher resolution for digsum(fib(k)) = k in base 11:

digsum(fib(k)) = k in base 11 (low resolution)
(green square is fib(1))

digsum(fib(k)) = k in base 11 (x2 resolution)

digsum(fib(k)) = k in base 11 (x4)

digsum(fib(k)) = k in base 11 (x8)

digsum(fib(k)) = k in base 11 (x16)

digsum(fib(k)) = k in base 11 (x32)

digsum(fib(k)) = k in base 11 (x64)

digsum(fib(k)) = k in base 11 (x128)

digsum(fib(k)) = k in base 11 (animated)