Tag Archives: base 11

Matching Fractions

by in Arithmetic, Base Behavior, Mathematics and tagged , , , , , , , , , , , | Leave a comment

0.1666… = 1/6
0.0273972… = 2/73
0.0379746… = 3/79
0.0016181229… = 1/618
0.0027322404… = 2/732 → 1/366
0.0058548009… = 5/854
0.01393354769… = 13/933
0.07598784194… = 75/987 → 25/329
0.08998988877… = 89/989
0.141993957703… = 141/993 → 47/331
0.0005854115443… = 5/8541
0.00129282482223… = 12/9282 → 2/1547
0.00349722279366… = 34/9722 → 17/4861
0.013599274705349… = 135/9927 → 15/1103
0.0000273205382146… = 2/73205

0.0465103… = 4/65 in base 8 = 4/53 in base 10
0.13735223… = 13/73 in b8 = 11/59 in b10
0.0036256353… = 3/625 → 1/207 in b8 = 3/405 → 1/135 in b10
0.01172160236… = 11/721 → 3/233 in b8 = 9/465 → 3/155 in b10
0.01272533117… = 12/725 in b8 = 10/469 in b10
0.03175523464… = 31/755 in b8 = 25/493 in b10
0.06776766655… = 67/767 in b8 = 55/503 in b10
0.251775771755… = 251/775 in b8 = 169/509 in b10
0.0003625152504… = 3/6251 in b8 = 3/3241 in b10
0.00137303402723… = 13/7303 in b8 = 11/3779 in b10
0.00267525714052… = 26/7525 in b8 = 22/3925 in b10
0.035777577356673… = 357/7757 in b8 = 239/4079 in b10

0.3763… = 3/7 in b9 = 3/7 in b10
0.0155187… = 1/55 in b9 = 1/50 in b10
0.0371482… = 3/71 in b9 = 3/64 in b10
0.0474627… = 4/74 in b9 = 4/67 in b10
0.43878684… = 43/87 in b9 = 39/79 in b10
0.07887877766… = 78/878 in b9 = 71/719 in b10
0.01708848667… = 17/0884 → 4/221 in b9 = 16/724 → 4/181 in b10
0.170884866767… = 170/884 → 40/221 in b9 = 144/724 → 36/181 in b10

0.2828… = 2/8 → 1/4 in b11 = 2/8 → 1/4 in b10
0.4986… = 4/9 in b11 = 4/9 in b10
0.54A9A8A6… = 54/A9 in b11 = 59/119 in b10
0.0010A17039… = 1/A17 in b11 = 1/1228 in b10
0.010A170392A… = 10/A17 in b11 = 11/1228 in b10
0.01AA5854872… = 1A/A58 in b11 = 21/1273 in b10
0.027A716A416… = 27/A71 in b11 = 29/1288 in b10
0.032A78032A7… = 32/A78 → 1/34 in b11 = 35/1295 → 1/37 in b10
0.0190AA5A829… = 19/0AA5 → 4/221 in b11 = 20/1325 → 4/265 in b10
0.190AA5A829… = 190/AA5 → 40/221 in b11 = 220/1325 → 44/265 in b10

0.23B7A334… = 23/B7 in b12 = 27/139 in b10
0.075BA597224… = 75/BA5 in b12 = 89/1709 in b10
0.0ABBABAAA99… = AB/BAB in b12 = 131/1715 in b10
0.185BB5B859B4… = 185/BB5 in b12 = 245/1721 in b10

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)

Terminal Transgressivity

by in Art, English Usage, Guardianistas, In Terms Of, Keyly Committed Core Components of the Counter-Cultural Community, Transgression... and tagged , , , , , , , , , , , | Leave a comment

“If this work is about hell,” he says, “it’s not only about hell in terms of content. It’s also about hell in terms of its hellishness in terms of production.” — maximally maverick artist Jake Chapman describes how he and his brother Dinos made the transgressive sculpture Hell (2000), as quoted in Simon Garfield’s In Miniature: How Small Things Illuminate the World (2018)

Elsewhere Other-Accessible

• Ex-term-in-nate! — incendiarily interrogating issues around “in terms of”…
• All O.o.t.Ü.-F. posts interrogating issues around “in terms of”…

Peri-Performative Post-Scriptum…

Yes, this was an über-ideal quote for posting on the 23rd in terms of the month… But I was so taken with it that I couldn’t delay any longer. And anyway: it is the 23rd of the months in base 11. (I.e., 2111 = 2 * 11 + 1 = 22 + 1 = 23.)

Reverssum

by in Base Behavior, Digit-sums, Mathematics and tagged , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , | Leave a comment

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

1, 2, 4, 8, 16, 68, 100, ?

The rule I’m using is this: Reverse the number, then add the sum of the digits. So 1 doubles till it becomes 16. Then 16 becomes 61 + 6 + 1 = 68. Then 68 becomes 86 + 8 + 6 = 100. Then 100 becomes 001 + 1 = 2. And the sequence falls into a loop.

Reversing the number means that small numbers can get big and big numbers can get small, but the second tendency is stronger for the first few seeds:

• 1 → 2 → 4 → 8 → 16 → 68 → 100 → 2
• 2 → 4 → 8 → 16 → 68 → 100 → 2
• 3 → 6 → 12 → 24 → 48 → 96 → 84 → 60 → 12
• 4 → 8 → 16 → 68 → 100 → 2 → 4
• 5 → 10 → 2 → 4 → 8 → 16 → 68 → 100 → 2
• 6 → 12 → 24 → 48 → 96 → 84 → 60 → 12
• 7 → 14 → 46 → 74 → 58 → 98 → 106 → 608 → 820 → 38 → 94 → 62 → 34 → 50 → 10 → 2 → 4 → 8 → 16 → 68 → 100 → 2
• 8 → 16 → 68 → 100 → 2 → 4 → 8
• 9 → 18 → 90 → 18
• 10 → 2 → 4 → 8 → 16 → 68 → 100 → 2

An 11-seed is a little more interesting:

11 → 13 → 35 → 61 → 23 → 37 → 83 → 49 → 107 → 709 → 923 → 343 → 353 → 364 → 476 → 691 → 212 → 217 → 722 → 238 → 845 → 565 → 581 → 199 → 1010 → 103 → 305 → 511 → 122 → 226 → 632 → 247 → 755 → 574 → 491 → 208 → 812 → 229 → 935 → 556 → 671 → 190 → 101 → 103 (11 leads to an 18-loop from 103 at step 26; total steps = 44)

Now try some higher bases:

• 1 → 2 → 4 → 8 → 15 → 57 → 86 → 80 → 15 (base=11)
• 1 → 2 → 4 → 8 → 14 → 46 → 72 → 34 → 4A → B6 → 84 → 58 → 96 → 80 → 14 (base=12)
• 1 → 2 → 4 → 8 → 13 → 35 → 5B → C8 → A6 → 80 → 13 (base=13)
• 1 → 2 → 4 → 8 → 12 → 24 → 48 → 92 → 36 → 6C → DA → C8 → A4 → 5A → B6 → 80 → 12 (base=14)
• 1 → 2 → 4 → 8 → 11 → 13 → 35 → 5B → C6 → 80 → 11 (base=15)
• 1 → 2 → 4 → 8 → 10 → 2 (base=16)

Does the 1-seed always create a short sequence? No, it gets pretty long in base-19 and base-20:

• 1 → 2 → 4 → 8 → [16] → 1D → DF → [17]3 → 4[18] → 107 → 709 → 914 → 424 → 42E → E35 → 54[17] → [17]5C → C7D → D96 → 6B3 → 3C7 → 7D6 → 6EE → E[16]2 → 2[18]8 → 90B → B1A → A2E → E3[17] → [17]5A → A7B → B90 → AC→ DD → F1 → 2C → C[16] → [18]2 → 40 → 8 (base=19)
• 1 → 2 → 4 → 8 → [16] → 1C → CE → F[18] → 108 → 80A → A16 → 627 → 731 → 13[18] → [18]43 → 363 → 36F → F77 → 794 → 4A7 → 7B5 → 5CA → ADC → CF5 → 5[17]4 → 4[18]B → B[19][17] → [18]1[18] → [18]3F → F5E → E79 → 994 → 4AB → BB9 → 9D2 → 2ED → DFB → B[17]C → C[19]B → C1E → E2[19] → [19]49 → 96B → B7F → F94 → 4B3 → 3C2 → 2D0 → D[17] → [19]3 → 51 → 1B → BD → EF → [17]3 → 4[17] → [18]5 → 71 → 1F → F[17] → [19]7 → 95 → 63 → 3F → [16]1 → 2D → D[17] (base=20)

Then it settles down again:

• 1 → 2 → 4 → 8 → [16] → 1B → BD → EE → [16]0 → 1B (base=21)
• 1 → 2 → 4 → 8 → [16] → 1A → AC → DA → BE → FE → [16]0 → 1A (base=22)
• 1 → 2 → 4 → 8 → [16] → 19 → 9B → C6 → 77 → 7[21] → [22]C → EA → BF → [16]E → [16]0 → 19 (base=23)

Base-33 is also short:

1 → 2 → 4 → 8 → [16] → [32] → 1[31] → [32]0 → 1[31] (base=33)

And so is base-35:

1 → 2 → 4 → 8 → [16] → [32] → 1[29] → [29][31] → [33][19] → [21]F → [16][22] → [23][19] → [20][30] → [32]0 → 1[29] (base=35)

So what about base-34?

1 → 2 → 4 → 8 → [16] → [32] → 1[30] → [30][32] → 10[24] → [24]0[26] → [26]26 → 63[26] → [26]47 → 75[29] → [29]6E → E8A → A9C → CA7 → 7B7 → 7B[32] → [32]C[23] → [23]E[31] → [31][16][23] → [23][18][33] → [33][20][29] → [29][23]D → D[25][26] → [26][27]9 → 9[29][20] → [20][30][33] → [33][33]1 → 21[32] → [32]23 → 341 → 14B → B4[17] → [17]59 → 96E → E74 → 485 → 58[21] → [21]95 → 5A[22] → [22]B8 → 8C[29] → [29]D[23] → [23]F[26] → [26][17][19] → [19][19][20] → [20][21]9 → 9[23]2 → 2[24]9 → 9[25]3 → 3[26]C → C[27]A → A[28][27] → [27][30]7 → 7[32][23] → [24]01 → 11F → F1[18] → [18]2F → F3[19] → [19]4[18] → [18]5[26] → [26]6[33] → [33]8[23] → [23]A[29] → [29]C[17] → [17]E[19] → [19]F[33] → [33][17][18] → [18][19][33] → [33][21][20] → [20][24]5 → 5[26]1 → 1[27]3 → 3[27][32] → [32][28][31] → [31][31][21] → [22]0C → C1[22] → [22]2D → D3[25] → [25]4[20] → [20]66 → 67[18] → [18]83 → 39D → D9[28] → [28]A[29] → [29]C[27] → [27]E[29] → [29][16][29] → [29][19]1 → 1[21]A → A[21][33] → [33][23]6 → 6[25][27] → [27][26][30] → [30][29]8 → 8[31][29] → [29][33]8 → 91[31] → [31]2[16] → [16]4C → C5E → E69 → 979 → 980 → 8[26] → [27]8 → 9[28] → [29]C → E2 → 2[30] → [31]0 → 1[28] → [28][30] → [32][18] → [20]E → F[20] → [21][16] → [17][24] → [25][24] → [26]6 → 7[24] → [25]4 → 5[20] → [20][30] → [32]2 → 3[32] → [33]4 → 62 → 2E → E[18] → [19]C → D[16] → [17]8 → 98 → 8[26] (1 leads to a 30-loop from 8[26] / 298 in base-34 at step 111; total steps = 141)

An alternative rule is to add the digit-sum first and then reverse the result. Now 8 becomes 8 + 8 = 16 and 16 becomes 61. Then 61 becomes 61 + 6 + 1 = 68 and 68 becomes 86. Then 86 becomes 86 + 8 + 6 = 100 and 100 becomes 001 = 1:

• 1 → 2 → 4 → 8 → 61 → 86 → 1
• 2 → 4 → 8 → 61 → 86 → 1 → 2
• 3 → 6 → 21 → 42 → 84 → 69 → 48 → 6
• 4 → 8 → 61 → 86 → 1 → 2 → 4
• 5 → 1 → 2 → 4 → 8 → 62 → 7 → 48 → 6 → 27 → 63 → 27
• 6 → 21 → 42 → 84 → 69 → 48 → 6
• 7 → 41 → 64 → 47 → 85 → 89 → 601 → 806 → 28 → 83 → 49 → 26 → 43 → 5 → 6 → 27 → 63 → 27
• 8 → 61 → 86 → 1 → 2 → 4 → 8
• 9 → 81 → 9
• 10 → 11 → 31 → 53 → 16 → 32 → 73 → 38 → 94 → 701 → 907 → 329 → 343 → 353 → 463 → 674 → 196 → 212 → 712 → 227 → 832 → 548 → 565 → 185 → 991 → 101 → 301 → 503 → 115 → 221 → 622 → 236 → 742 → 557 → 475 → 194→ 802 → 218 → 922 → 539 → 655 → 176 → 91 → 102 → 501 → 705 → 717 → 237 → 942 → 759 → 87 → 208 → 812 → 328 → 143 → 151 → 851 → 568 → 785 → 508 → 125 → 331 → 833 → 748 → 767 → 787 → 908 → 529 → 545 → 955 → 479 → 994 → 6102 → 1116 → 5211 → 225 → 432 → 144 → 351 → 63 → 27 → 63

Block and Goal

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

123456789. How many ways are there to insert + and – between the numbers and create a formula for 100? With pen and ink it takes a long time to answer. With programming, the answer will flash up in an instant:

01. 1 + 2 + 3 - 4 + 5 + 6 + 78 + 9 = 100
02. 1 + 2 + 34 - 5 + 67 - 8 + 9 = 100
03. 1 + 23 - 4 + 5 + 6 + 78 - 9 = 100
04. 1 + 23 - 4 + 56 + 7 + 8 + 9 = 100
05. 12 - 3 - 4 + 5 - 6 + 7 + 89 = 100
06. 12 + 3 + 4 + 5 - 6 - 7 + 89 = 100
07. 12 + 3 - 4 + 5 + 67 + 8 + 9 = 100
08. 123 - 4 - 5 - 6 - 7 + 8 - 9 = 100
09. 123 + 4 - 5 + 67 - 89 = 100
10. 123 + 45 - 67 + 8 - 9 = 100
11. 123 - 45 - 67 + 89 = 100

And the beauty of programming is that you can easily generalize the problem to other bases. In base b, how many ways are there to insert + and – in the block [12345…b-1] to create a formula for b^2? When b = 10, the answer is 11. When b = 11, it’s 42. Here are two of those formulae in base-11:

123 - 45 + 6 + 7 - 8 + 9 + A = 100[b=11]
146 - 49 + 6 + 7 - 8 + 9 + 10 = 121

123 + 45 + 6 + 7 - 89 + A = 100[b=11]
146 + 49 + 6 + 7 - 97 + 10 = 121

When b = 12, it’s 51. Here are two of the formulae:

123 + 4 + 5 + 67 - 8 - 9A + B = 100[b=12]
171 + 4 + 5 + 79 - 8 - 118 + 11 = 144

123 + 4 + 56 + 7 - 89 - A + B = 100[b=12]
171 + 4 + 66 + 7 - 105 - 10 + 11 = 144

So that’s 11 formulae in base-10, 42 in base-11 and 51 in base-12. So what about base-13? The answer may be surprising: in base-13, there are no +/- formulae for 13^2 = 169 using the numbers 1 to 12. Nor are there any formulae in base-9 for 9^2 = 81 using the numbers 1 to 8. If you reverse the block, 987654321, the same thing happens. Base-10 has 15 formulae, base-11 has 54 and base-12 has 42. Here are some examples:

9 - 8 + 7 + 65 - 4 + 32 - 1 = 100
98 - 76 + 54 + 3 + 21 = 100

A9 + 87 - 65 + 4 - 3 - 21 = 100[b=11]
119 + 95 - 71 + 4 - 3 - 23 = 121

BA - 98 + 76 - 5 - 4 + 32 - 1 = 100[b=12]
142 - 116 + 90 - 5 - 4 + 38 - 1 = 144

But base-9 and base-13 again have no formulae. What’s going on? Is it a coincidence that 9 and 13 are each one more than a multiple of 4? No. Base-17 also has no formulae for b^2 = 13^2 = 169. Here is the list of formulae for bases-7 thru 17:

1, 2, 0, 11, 42, 51, 0, 292, 1344, 1571, 0 (block = 12345...)
3, 2, 0, 15, 54, 42, 0, 317, 1430, 1499, 0 (block = ...54321)

To understand what’s going on, consider any sequence of consecutive integers starting at 1. The number of odd integers in the sequence must always be greater than or equal to the number of even integers:

1, 2 (1 odd : 1 even)
1, 2, 3 (2 odds : 1 even)
1, 2, 3, 4 (2 : 2)
1, 2, 3, 4, 5 (3 : 2)
1, 2, 3, 4, 5, 6 (3 : 3)
1, 2, 3, 4, 5, 6, 7 (4 : 3)
1, 2, 3, 4, 5, 6, 7, 8 (4 : 4)

The odd numbers in a sequence determine the parity of the sum, that is, whether it is odd or even. For example:

1 + 2 = 3 (1 odd number)
1 + 2 + 3 = 6 (2 odd numbers)
1 + 2 + 3 + 4 = 10 (2 odd numbers)
1 + 2 + 3 + 4 + 5 = 15 (3 odd numbers)
1 + 2 + 3 + 4 + 5 + 6 = 21 (3 odd numbers)
1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 (4 odd numbers)

If there is an even number of odd numbers, the sum will be even; if there is an odd number, the sum will be odd. Consider sequences that end in a multiple of 4:

1, 2, 3, 4 → 2 odds : 2 evens
1, 2, 3, 4, 5, 6, 7, 8 → 4 : 4
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 → 6 : 6

Such sequences always contain an even number of odd numbers. Now, consider these formulae in base-10:

1. 12 + 3 + 4 + 56 + 7 + 8 + 9 = 99
2. 123 - 45 - 67 + 89 = 100
3. 123 + 4 + 56 + 7 - 89 = 101

They can be re-written like this:

1. 1×10^1 + 2×10^0 + 3×10^0 + 4×10^0 + 5×10^1 + 6×10^0 + 7×10^0 + 8×10^0 + 9×10^0 = 99

2. 1×10^2 + 2×10^1 + 3×10^0 – 4×10^1 – 5×10^0 – 6×10^1 – 7×10^0 + 8×10^1 + 9×10^0 = 100

3. 1×10^2 + 2×10^1 + 3×10^0 + 4×10^0 + 5×10^1 + 6×10^1 + 7×10^0 – 8×10^1 – 9×10^0 = 101

In general, the base-10 formulae will take this form:

1×10^a +/- 2×10^b +/- 3×10^c +/– 4×10^d +/– 5×10^e +/– 6×10^f +/– 7×10^g +/– 8×10^h +/– 9×10^i = 100

It’s important to note that the exponent of 10, or the power to which it is raised, determines whether an odd number remains odd or becomes even. For example, 3×10^0 = 3×1 = 3, whereas 3×10^1 = 3×10 = 30 and 3×10^2 = 3×100 = 300. Therefore the number of odd numbers in a base-10 formula can vary and so can the parity of the sum. Now consider base-9. When you’re trying to find a block-formula for 9^2 = 81, the formula will have to take this form:

1×9^a +/- 2×9^b +/- 3×9^c +/- 4×9^d +/- 5×9^e +/- 6×9^f +/- 7×9^g +/- 8×9^h = 81

But no such formula exists for 81 (with standard exponents). It’s now possible to see why this is so. Unlike base-10, the odd numbers in the formula will remain odd what the power of 9. For example, 3×9^0 = 3×1 = 3, 3×9^1 = 3×9 = 27 and 3×9^2 = 3×81 = 243. Therefore base-9 formulae will always contain four odd numbers and will always produce an even number. Odd numbers in base-2 always end in 1, even numbers always end in 0. Therefore, to determine the parity of a sum of integers, convert the integers to base-2, discard all but the final digit of each integer, then sum the 1s. In a base-9 formula, these are the four possible results:

1 + 1 + 1 + 1 = 4
1 + 1 + 1 - 1 = 2
1 + 1 - 1 - 1 = 0
1 - 1 - 1 - 1 = -2

The sum represents the parity of the answer, which is always even. Similar reasoning applies to base-13, base-17 and all other base-[b=4n+1].

Persist List

by in Base Behavior, Mathematics, Narcissistic numbers and tagged , , , , , , , , , , , , , , , , , , , , , , , | 1 Comment

Multiplicative persistence is a complex term but a simple concept. Take a number, multiply its digits, repeat. Sooner or later the result is a single digit:

25 → 2 x 5 = 10 → 1 x 0 = 0 (mp=2)
39 → 3 x 9 = 27 → 2 x 7 = 14 → 1 x 4 = 4 (mp=3)

So 25 has a multiplicative persistence of 2 and 39 a multiplicative persistence of 3. Each is the smallest number with that m.p. in base-10. Further records are set by these numbers:

77 → 49 → 36 → 18 → 8 (mp=4)
679 → 378 → 168 → 48 → 32 → 6 (mp=5)
6788 → 2688 → 768 → 336 → 54 → 20 → 0 (mp=6)
68889 → 27648 → 2688 → 768 → 336 → 54 → 20 → 0 (mp=7)
2677889 → 338688 → 27648 → 2688 → 768 → 336 → 54 → 20 → 0 (mp=8)
26888999 → 4478976 → 338688 → 27648 → 2688 → 768 → 336 → 54 → 20 → 0 (mp=9)
3778888999 → 438939648 → 4478976 → 338688 → 27648 → 2688 → 768 → 336 → 54 → 20 → 0 (mp=10)

Now here’s base-9:

25[b=9] → 11 → 1 (mp=2)
38[b=9] → 26 → 13 → 3 (mp=3)
57[b=9] → 38 → 26 → 13 → 3 (mp=4)
477[b=9] → 237 → 46 → 26 → 13 → 3 (mp=5)
45788[b=9] → 13255 → 176 → 46 → 26 → 13 → 3 (mp=6)
2577777[b=9] → 275484 → 13255 → 176 → 46 → 26 → 13 → 3 (mp=7)

And base-11:

26[b=11] → 11 → 1 (mp=2)
3A[b=11] → 28 → 15 → 5 (mp=3)
69[b=11] → 4A → 37 → 1A → A (=10b=10) (mp=4)
269[b=11] → 99 → 74 → 26 → 11 → 1 (mp=5)
3579[b=11] → 78A → 46A → 1A9 → 82 → 15 → 5 (mp=6)
26778[b=11] → 3597 → 78A → 46A → 1A9 → 82 → 15 → 5 (mp=7)
47788A[b=11] → 86277 → 3597 → 78A → 46A → 1A9 → 82 → 15 → 5 (mp=8)
67899AAA[b=11] → 143A9869 → 299596 → 2A954 → 2783 → 286 → 88 → 59 → 41 → 4 (mp=9)
77777889999[b=11] → 2AA174996A → 143A9869 → 299596 → 2A954 → 2783 → 286 → 88 → 59 → 41 → 4 (mp=10)

I was also interested in the narcissism of multiplicative persistence. That is, are any numbers equal to the sum of the numbers created while calculating their multiplicative persistence? Yes:

86 = (8 x 6 = 48) + (4 x 8 = 32) + (3 x 2 = 6)

I haven’t found any more in base-10 (apart from the trivial 0 to 9) and can’t prove that this is the only one. Base-9 offers this:

78[b=9] = 62 + 13 + 3

I can’t find any at all in base-11, but here are base-12 and base-27:

57[b=12] = 2B + 1A + A
A8[b=12] = 68 + 40 + 0

4[23][b=27] = 3B + 16 + 6
7[24][b=27] = 66 + 19 + 9
A[18][b=27] = 6[18] + 40 + 0
[26][24][b=27] = [23]3 + 2F + 13 + 3
[26][23][26][b=27] = [21]8[23] + 583 + 4C + 1[21] + [21]

But the richest base I’ve found so far is base-108, with fourteen narcissistic multiplicative-persistence sums:

4[92][b=108] = 3[44] + 1[24] + [24]
5[63][b=108] = 2[99] + 1[90] + [90]
7[96][b=108] = 6[24] + 1[36] + [36]
A[72][b=108] = 6[72] + 40 + 0
[19][81][b=108] = E[27] + 3[54] + 1[54] + [54]
[26][96][b=108] = [23]C + 2[60] + 1C + C
[35][81][b=108] = [26][27] + 6[54] + 30 + 0
[37][55][b=108] = [18][91] + F[18] + 2[54] + 10 + 0
[73][60][b=108] = [40][60] + [22][24] + 4[96] + 3[60] + 1[72] + [72]
[107][66][b=108] = [65][42] + [25][30] + 6[102] + 5[72] + 3[36] + 10 + 0
[71][84][b=108] = [55][24] + C[24] + 2[72] + 1[36] + [36]
[107][99][b=108] = [98]9 + 8[18] + 1[36] + [36]
5[92][96][b=108] = 3[84][96] + 280 + 0
8[107][100][b=108] = 7[36][64] + 1[41][36] + D[72] + 8[72] + 5[36] + 1[72] + [72]

Update (10/ii/14): The best now is base-180 with eighteen multiplicative-persistence sums.

5[105][b=180] = 2[165] + 1[150] + [150]
7[118][b=180] = 4[106] + 2[64] + [128]
7[160][b=180] = 6[40] + 1[60] + [60]
8[108][b=180] = 4[144] + 3[36] + [108]
A[120][b=180] = 6[120] + 40 + 0 (s=5)
[19][135][b=180] = E[45] + 3[90] + 1[90] + [90]
[21][108][b=180] = C[108] + 7[36] + 1[72] + [72]
[26][160][b=180] = [23][20] + 2[100] + 1[20] + [20]
[31][98][b=180] = [16][158] + E8 + [112]
[35][135][b=180] = [26][45] + 6[90] + 30 + 0 (s=10)
[44][96][b=180] = [23][84] + A[132] + 7[60] + 2[60] + [120]
[71][140][b=180] = [55][40] + C[40] + 2[120] + 1[60] + [60]
[73][100][b=180] = [40][100] + [22][40] + 4[160] + 3[100] + 1[120] + [120]
[107][110][b=180] = [65][70] + [25][50] + 6[170] + 5[120] + 3[60] + 10 + 0
[107][165][b=180] = [98]F + 8[30] + 1[60] + [60] (s=15)
[172][132][b=180] = [126][24] + [16][144] + C[144] + 9[108] + 5[72] + 20 + 0
5[173][145][b=180] = 3[156][145] + 2[17]0 + 0
E[170][120][b=180] = 8[146][120] + 4[58][120] + [154][120] + [102][120] + [68]0 + 0

Six Six Nix

by in Base Behavior, Digit-sums, Mathematics, Narcissistic numbers and tagged , , , , , , , , , , , , , , , , , , , , | Leave a comment

4 x 3 = 13. A mistake? Not in base-9, where 13 = 1×9^1 + 3 = 12 in base-10. This means that 13 is a sum-product number in base-9: first add its digits, then multiply them, then multiply the digit-sum by the digit-product: (1+3) x (1×3) = 13[9]. There are four more sum-product numbers in this base:

2086[9] = 17 x 116 = (2 + 8 + 6) x (2 x 8 x 6) = 1536[10] = 16 x 96
281876[9] = 35 x 7333 = (2 + 8 + 1 + 8 + 7 + 6) x (2 x 8 x 1 x 8 x 7 x 6) = 172032[10] = 32 x 5376
724856[9] = 35 x 20383 = (7 + 2 + 4 + 8 + 5 + 6) x (7 x 2 x 4 x 8 x 5 x 6) = 430080[10] = 32 x 13440
7487248[9] = 44 x 162582 = (7 + 4 + 8 + 7 + 2 + 4 + 8) x (7 x 4 x 8 x 7 x 2 x 4 x 8) = 4014080[10] = 40 x 100352

And that’s the lot, apart from the trivial 0 = (0) x (0) and 1 = (1) x (1), which are true in all bases.

What about base-10?

135 = 9 x 15 = (1 + 3 + 5) x (1 x 3 x 5)
144 = 9 x 16 = (1 + 4 + 4) x (1 x 4 x 4)
1088 = 17 x 64 = (1 + 8 + 8) x (1 x 8 x 8)

1088 is missing from the list at Wikipedia and the Encyclopedia of Integer Sequences, but I like the look of it, so I’m including it here. Base-11 has five sum-product numbers:

419[11] = 13 x 33 = (4 + 1 + 9) x (4 x 1 x 9) = 504[10] = 14 x 36
253[11] = [10] x 28 = (2 + 5 + 3) x (2 x 5 x 3) = 300[10] = 10 x 30
2189[11] = 19 x 121 = (2 + 1 + 8 + 9) x (2 x 1 x 8 x 9) = 2880[10] = 20 x 144
7634[11] = 19 x 419 = (7 + 6 + 3 + 4) x (7 x 6 x 3 x 4) = 10080[10] = 20 x 504
82974[11] = 28 x 3036 = (8 + 2 + 9 + 7 + 4) x (8 x 2 x 9 x 7 x 4) = 120960[10] = 30 x 4032

But the record for bases below 50 is set by 7:

22[7] = 4 x 4 = (2 + 2) x (2 x 2) = 16[10] = 4 x 4
505[7] = 13 x 34 = (5 + 5) x (5 x 5) = 250[10] = 10 x 25
242[7] = 11 x 22 = (2 + 4 + 2) x (2 x 4 x 2) = 128[10] = 8 x 16
1254[7] = 15 x 55 = (1 + 2 + 5 + 4) x (1 x 2 x 5 x 4) = 480[10] = 12 x 40
2343[7] = 15 x 132 = (2 + 3 + 4 + 3) x (2 x 3 x 4 x 3) = 864[10] = 12 x 72
116655[7] = 33 x 2424 = (1 + 1 + 6 + 6 + 5 + 5) x (1 x 1 x 6 x 6 x 5 x 5) = 21600[10] = 24 x 900
346236[7] = 33 x 10362 = (3 + 4 + 6 + 2 + 3 + 6) x (3 x 4 x 6 x 2 x 3 x 6) = 62208[10] = 24 x 2592
424644[7] = 33 x 11646 = (4 + 2 + 4 + 6 + 4 + 4) x (4 x 2 x 4 x 6 x 4 x 4) = 73728[10] = 24 x 3072

And base-6? Six Nix. There are no sum-product numbers unique to that base (to the best of my far-from-infallible knowledge). Here is the full list for base-3 to base-50 (not counting 0 and 1 as sum-product numbers):

5 in base-11 4 in base-21 3 in base-31 2 in base-41
4 in base-12 5 in base-22 1 in base-32 3 in base-42
0 in base-3 3 in base-13 4 in base-23 3 in base-33 4 in base-43
2 in base-4 3 in base-14 2 in base-24 4 in base-34 5 in base-44
1 in base-5 2 in base-15 3 in base-25 2 in base-35 6 in base-45
0 in base-6 2 in base-16 6 in base-26 2 in base-36 7 in base-46
8 in base-7 6 in base-17 0 in base-27 3 in base-37 3 in base-47
1 in base-8 5 in base-18 1 in base-28 3 in base-38 7 in base-48
5 in base-9 7 in base-19 0 in base-29 1 in base-39 5 in base-49
3 in base-10 3 in base-20 2 in base-30 2 in base-40 3 in base-50

Narcissarithmetic

by in Base Behavior, Digit-sums, Mathematics, Powers and tagged , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , | Leave a comment

Why is 438,579,088 a beautiful number? Simple: it may seem entirely arbitrary, but it’s actually self-empowered:

438,579,088 = 4^4 + 3^3 + 8^8 + 5^5 + 7^7 + 9^9 + 0^0 + 8^8 + 8^8 = 256 + 27 + 16777216 + 3125 + 823543 + 387420489 + 0 + 16777216 + 16777216 (usually 0^0 = 1, but the rule is slightly varied here)

438,579,088 is so beautiful, in fact, that it’s in love with itself as a narcissistic number, or number that can be generated by manipulation of its own digits. 89 = 8^1 + 9^2 = 8 + 81 and 135 = 1^1 + 3^2 + 5^3 = 1 + 9 + 125 are different kinds of narcissistic number. 3435 is self-empowered again:

3435 = 3^3 + 4^4 + 3^3 + 5^5 = 27 + 256 + 27 + 3125

But that’s your lot: there are no more numbers in base-10 that are equal to the sum of their self-empowered digits (apart from the trivial 0 and 1). To prove this, start by considering that there is a limit to the size of a self-empowered number. 9^9 is 387,420,489, which is nine digits long. The function autopower(999,999,999) = 387,420,489 x 9 = 3,486,784,401, which is ten digits long. But autopower(999,999,999,999) = 387,420,489 x 12 = 4,649,045,868, also ten digits long.

The Metamorphosis of Narcissus by Salvador Dalí

Salvador Dalí, La Metamorfosis de Narciso (1937)

So you don’t need to check numbers above a certain size. There still seem a lot of numbers to check: 438,579,088 is a long way above 3435. However, the search is easy to shorten if you consider that checking 3-3-4-5 is equivalent to checking 3-4-3-5, just as checking 034,578,889 is equivalent to checking 438,579,088. If you self-empower a number and the result has the same digits as the original number, you’ve found what you’re looking for. The order of digits in the original number doesn’t matter, because the result has automatically sorted them for you. The function autopower(3345) produces 3435, therefore 3435 must be self-empowered.

So the rule is simple: Check only the numbers in which any digit is greater than or equal to all digits to its left. In other words, you check 12 and skip 21, check 34 and skip 43, check 567 and skip 576, 657, 675, 756 and 765. That reduces the search-time considerably: discarding numbers is computationally simpler than self-empowering them. It’s also computationally simple to vary the base in which you’re searching. Base-10 produces only two self-empowered numbers, but its neighbours base-9 and base-11 are much more fertile:

30 = 3^3 + 0^0 = 30 + 0 (b=9)
27 = 27 + 0 (b=10)

31 = 3^3 + 1^1 = 30 + 1 (b=9)
28 = 27 + 1 (b=10)

156262 = 1^1 + 5^5 + 6^6 + 2^2 + 6^6 + 2^2 = 1 + 4252 + 71000 + 4 + 71000 + 4 (b=9)
96446 = 1 + 3125 + 46656 + 4 + 46656 + 4 (b=10)

1647063 = 1^1 + 6^6 + 4^4 + 7^7 + 0^0 + 6^6 + 3^3 = 1 + 71000 + 314 + 1484617 + 0 + 71000 + 30 (b=9)
917139 = 1 + 46656 + 256 + 823543 + 0 + 46656 + 27 (b=10)

1656547 = 1^1 + 6^6 + 5^5 + 6^6 + 5^5 + 4^4 + 7^7 = 1 + 71000 + 4252 + 71000 + 4252 + 314 + 1484617 (b=9)
923362 = 1 + 46656 + 3125 + 46656 + 3125 + 256 + 823543 (b=10)

34664084 = 3^3 + 4^4 + 6^6 + 6^6 + 4^4 + 0^0 + 8^8 + 4^4 = 30 + 314 + 71000 + 71000 + 314 + 0 + 34511011 + 314 (b=9)
16871323 = 27 + 256 + 46656 + 46656 + 256 + 0 + 16777216 + 256 (b=10)

66500 = 6^6 + 6^6 + 5^5 + 0^0 + 0^0 = 32065 + 32065 + 2391 + 0 + 0 (b=11)
96437 = 46656 + 46656 + 3125 + 0 + 0 (b=10)

66501 = 6^6 + 6^6 + 5^5 + 0^0 + 1^1 = 32065 + 32065 + 2391 + 0 + 1 (b=11)
96438 = 46656 + 46656 + 3125 + 0 + 1 (b=10)

517503 = 5^5 + 1^1 + 7^7 + 5^5 + 0^0 + 3^3 = 2391 + 1 + 512816 + 2391 + 0 + 25 (b=11)
829821 = 3125 + 1 + 823543 + 3125 + 0 + 27 (b=10)

18453278 = 1^1 + 8^8 + 4^4 + 5^5 + 3^3 + 2^2 + 7^7 + 8^8 = 1 + 9519A75 + 213 + 2391 + 25 + 4 + 512816 + 9519A75 (b=11)
34381388 = 1 + 16777216 + 256 + 3125 + 27 + 4 + 823543 + 16777216 (b=10)

18453487 = 1^1 + 8^8 + 4^4 + 5^5 + 3^3 + 4^4 + 8^8 + 7^7 = 1 + 9519A75 + 213 + 2391 + 25 + 213 + 9519A75 + 512816 (b=11)
34381640 = 1 + 16777216 + 256 + 3125 + 27 + 256 + 16777216 + 823543 (b=10)

It’s easy to extend the concept of self-empowered narcisso-numbers. The prime 71 = 131 in base-7 and the prime 83 = 146 in base-7. If 131[b=7] is empowered to the digits of 146[b=7], you get 146[b=7]; and if 146[b=7] is empowered to the digits of 131[b=7], you get 131[b=7], like this:

71 = 131[b=7] → 1^1 + 3^4 + 1^6 = 1 + 81 + 1 = 83 = 146[b=7]

83 = 146[b=7] → 1^1 + 4^3 + 6^1 = 1 + 64 + 6 = 71 = 131[b=7]

But it’s not easy to find more examples. Are there other-empowering pairs like that in base-10? I don’t know.