Fine as Nine

Thursday, 20th Jul, 2023 by Krilling for Company in Geometry, Integer Sequences, Mathematics, Polygons, Reciprocals, Triangular Numbers and tagged enneagonal numbers, icosikaihenagonal, Integer Sequences, nonagons, pentadecagonal numbers, Polygonal numbers, reciprocals, recreational math, recreational mathematics, recreational maths, Triangular Numbers | Leave a comment

This is a regular nonagon (a polygon with nine sides):

A nonagon or enneagon (from Wikipedia)

And this is the endlessly repeating decimal of the reciprocal of 7:

1/7 = 0.142857142857142857142857…

What is the curious connection between 1/7 and nonagons? If I’d been asked that a week ago, I’d’ve had no answer. Then I found a curious connection when I was looking at the leading digits of polygonal numbers. A polygonal number is a number that can be represented in the form of a polygon. Triangular numbers look like this:

* = 1
* ** = 3 * ** *** = 6 * ** *** **** = 10
* ** *** **** ***** = 15

By looking at the shapes rather than the numbers, it’s easy to see that you generate the triangular numbers by simply summing the integers:

1 = 1 1+2=3 1+2+3=6 1+2+3+4=10 1+2+3+4+5=15

Now try the square numbers:

* = 1
** ** = 4 *** *** *** = 9 **** **** **** **** = 16 ***** ***** ***** ***** ***** = 25

You generate the square numbers by summing the odd integers:

1 = 1 1+3 = 4 1+3+5 = 9 1+3+7 = 16 1+3+7+9 = 25

Next come the pentagonal numbers, the hexagonal numbers, the heptagonal numbers, and so on. I was looking at the leading digits of these numbers and trying to find patterns. For example, when do the leading digits of the k-th triangular number, tri(k), match the digits of k? This is when:

tri(1) = 1 tri(19) = 190 tri(199) = 19900 tri(1999) = 1999000 tri(19999) = 199990000 tri(199999) = 19999900000 [...]

That pattern is easy to explain. The formula for the k-th polygonal number is k * ((pn-2)*k + (4-pn)) / 2, where pn = 3 for the triangular numbers, 4 for the square numbers, 5 for the pentagonal numbers, and so on. Therefore the k-th triangular number is k * (k + 1) / 2. When k = 19, the formula is 19 * (19 + 1) / 2 = 19 * 20 / 2 = 19 * 10 = 190. And so on. Now try the pol(k) = leaddig(pol(k)) for higher polygonal numbers. The patterns are easy to predict until you get to the nonagonal numbers:

square(10) = 100 square(100) = 10000 square(1000) = 1000000 square(10000) = 100000000 square(100000) = 10000000000 [...]
pentagonal(7) = 70 pentagonal(67) = 6700 pentagonal(667) = 667000 pentagonal(6667) = 66670000 pentagonal(66667) = 6666700000 [...] hexagonal(6) = 66 hexagonal(51) = 5151 hexagonal(501) = 501501 hexagonal(5001) = 50015001 hexagonal(50001) = 5000150001 [...] heptagonal(5) = 55 heptagonal(41) = 4141 heptagonal(401) = 401401 heptagonal(4001) = 40014001 heptagonal(40001) = 4000140001 [...] octagonal(4) = 40 octagonal(34) = 3400 octagonal(334) = 334000 octagonal(3334) = 33340000 octagonal(33334) = 3333400000 [...]
nonagonal(4) = 46 nonagonal(30) = 3075 nonagonal(287) = 287574 nonagonal(2858) = 28581429 nonagonal(28573) = 2857385719 nonagonal(285715) = 285715000000 nonagonal(2857144) = 28571444285716 nonagonal(28571430) = 2857143071428575 nonagonal(285714287) = 285714287571428574 nonagonal(2857142858) = 28571428581428571429 nonagonal(28571428573) = 2857142857385714285719 nonagonal(285714285715) = 285714285715000000000000 nonagonal(2857142857144) = 28571428571444285714285716 nonagonal(28571428571430) = 2857142857143071428571428575 nonagonal(285714285714287) = 285714285714287571428571428574 nonagonal(2857142857142858) = 28571428571428581428571428571429 nonagonal(28571428571428573) = 2857142857142857385714285714285719 nonagonal(285714285714285715) = 285714285714285715000000000000000000 nonagonal(2857142857142857144) = 28571428571428571444285714285714285716 nonagonal(28571428571428571430) = 2857142857142857143071428571428571428575 [...]

What’s going on with the leading digits of the nonagonals? Well, they’re generating a different reciprocal. Or rather, they’re generating the multiple of a different reciprocal:

1/7 * 2 = 2/7 = 0.285714285714285714285714285714...

And why does 1/7 have this curious connection with the nonagonal numbers? Because the nonagonal formula is k * (7k-5) / 2 = k * ((9-2) * k + (4-pn)) / 2. Now look at the pentadecagonal numbers, where pn = 15:

pentadecagonal(1538461538461538461540) = 15384615384615384615406923076923076923076930
2/13 = 0.153846153846153846153846153846...
pentadecagonal formula = k * (13k - 11) / 2 = k * ((15-2)*k + (4-15)) / 2

Penultimately, let’s look at the icosikaihenagonal numbers, where pn = 21:

icosikaihenagonal(2) = 21 icosikaihenagonal(12) = 1266 icosikaihenagonal(107) = 107856 icosikaihenagonal(1054) = 10544743 icosikaihenagonal(10528) = 1052878960 icosikaihenagonal(105265) = 105265947385 icosikaihenagonal(1052633) = 10526335263165 icosikaihenagonal(10526317) = 1052631731578951 icosikaihenagonal(105263159) = 105263159210526318 icosikaihenagonal(1052631580) = 10526315801578947370 icosikaihenagonal(10526315791) = 1052631579163157894746 icosikaihenagonal(105263157896) = 105263157896368421052636 icosikaihenagonal(1052631578949) = 10526315789497368421052643 icosikaihenagonal(10526315789475) = 1052631578947542105263157900 icosikaihenagonal(105263157894738) = 105263157894738263157894736845 icosikaihenagonal(1052631578947370) = 10526315789473706842105263157905 icosikaihenagonal(10526315789473686) = 1052631578947368689473684210526331 icosikaihenagonal(105263157894736843) = 105263157894736843000000000000000000 icosikaihenagonal(1052631578947368422) = 10526315789473684220526315789473684211 icosikaihenagonal(10526315789473684212) = 1052631578947368421257894736842105263166
2/19 = 0.1052631578947368421052631579
icosikaihenagonal formula = k * (19k - 17) / 2 = k * ((21-2)*k + (4-21)) / 2

And ultimately, let’s look at this other pattern in the leading digits of the triangular numbers, which I can’t yet explain at all:

tri(904) = 409060 tri(6191) = 19167336 tri(98984) = 4898965620 tri(996694) = 496699963165 tri(9989894) = 49898996060565 tri(99966994) = 4996699994681515 tri(999898994) = 499898999601055515 tri(9999669994) = 49996699999451815015 tri(99998989994) = 4999898999960055555015 tri(999996699994) = 499996699999945018150015 tri(9999989899994) = 49999898999996005055550015 tri(99999966999994) = 4999996699999994500181500015 tri(999999898999994) = 499999898999999600500555500015 [...]

A Walk on the Wide Side

Friday, 23rd Jun, 2023 by Krilling for Company in Base Behavior, Digit-sums, Fibonacci Sequence, Integer Sequences, Irrational numbers, φ, Mathematics, Phi and tagged Arithmetic, base 10, base 11, base 12, carries, carry, digit-sums of Fibonacci numbers, duodecimal, logarithms, powers of 2, recreational math, recreational mathematics, recreational maths, undecimal, undenary, unodecimal | 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…

Saturday, 17th Jun, 2023 by Krilling for Company in Base Behavior, Digit-sums, Fibonacci Sequence, Integer Sequences, Mathematics, Phi and tagged base 10, base 11, digit-sums, Fibonacci index, Fibonacci numbers, Integer Sequences, recreational math, recreational mathematics, recreational maths, undecimal, undenary, unodecimal | 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?

Primal Stream

Sunday, 29th Jan, 2023 by Krilling for Company in Integer Sequences, Mathematics, Prime numbers and tagged consecutive integers, highly composite odd numbers, primes, recreational math, recreational mathematics, recreational maths, sums of consecutive integers | 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.

Ace Base

Friday, 23rd Dec, 2022 by Krilling for Company in Base Behavior, Mathematics, Triangular Numbers and tagged base 8, math, mathematics, maths, number bases, octal, recreational math, recreational mathematics, recreational maths, Triangular Numbers | Leave a comment

Strange “S” in the Light

Wednesday, 23rd Nov, 2022 by Krilling for Company in Fractals, Geometry, Mathematics and tagged fractals, recreational math, recreational mathematics, recreational maths | Leave a comment

Unexpected discoveries are one of the joys of mathematics, even for amateurs. And computers help you make more of them, because computers make it easy to adjust variables or search faster and further through math-space than any unaided human ever could (on the downside, computers can make you lazy and blunt your intuition). Here’s an unexpected discovery I made using a computer in November 2020:

A distorted and dissected capital “S”

It’s a strange “S” that looks complex but is constructed very easily from three simple lines. And it’s a fractal, a shape that contains copies of itself at smaller and smaller scales:

Five sub-fractals of the Strange “S”

Elsewhere Other-Accessible…

• Fractangular Frolics — the Strange “S” in more light

We Can Circ It Out

Saturday, 5th Nov, 2022 by Krilling for Company in Circles, Fractals, Geometry, Mathematics, Polygons, Squares, Triangles and tagged animated gifs, Circles, circular fractals, fractal geometry, fractals, geometry, math, mathematics, maths, Polygons, recreational math, recreational mathematics, recreational maths, Sierpinski triangles, Sierpiński triangle | Leave a comment

It’s a pretty little problem to convert this triangular fractal…

Sierpiński triangle (Wikipedia)

…into its circular equivalent:

Sierpiński triangle as circle

Sierpiński triangle to circle (animated)

But once you’ve circ’d it out, as it were, you can easily adapt the technique to fractals based on other polygons:

T-square fractal (Wikipedia)
⇓

T-square fractal as circle

T-square fractal to circle (animated)

Elsewhere other-accessible…

• Dilating the Delta — more on converting polygonic fractals to circles…

The Devil’s Digits

Saturday, 29th Oct, 2022 by Krilling for Company in 666, Base Behavior, Mathematics, Number of the Beast and tagged counting digits, counting zeros, digits, Number of the Beast, recreational math, recreational mathematics, recreational maths, zero | Leave a comment

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

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

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

21 = 10101 in base 2 = 210 in base 3 = 111 in b4 = 41 in b5 = 33 in b6 = 30 in b7 = 25 in b8 = 23 in b9 = 21 in b10 = 1A in b11 = 19 in b12 = 18 in b13 = 17 in b14 = 16 in b15 = 15 in b16 = 14 in b17 = 13 in b18 = 12 in b19 = 11 in b20 = 10 in b21 21 = 1_s=101_s=201_s=3 in base 2 = 21_s=40 in base 3 = 111_s=7 in b4 = 41_s=8 in b5 = 33 in b6 = 30 in b7 = 25 in b8 = 23 in b9 = 21_s=9 in b10 = 1_s=10A in b11 = 1_s=119 in b12 = 1_s=128 in b13 = 1_s=137 in b14 = 1_s=146 in b15 = 1_s=155 in b16 = 1_s=164 in b17 = 1_s=173 in b18 = 1_s=182 in b19 = 11_s=20 in b20 = 1_s=210 in b21

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

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

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

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

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

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

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

Reciprocal Recipes

Sunday, 23rd Oct, 2022 by Krilling for Company in Base Behavior, Fibonacci Sequence, Mathematics, Reciprocals and tagged base 10, base 2, binary, David Wells, decimals, Fibonacci Sequence, Penguin Dictionary of Curious and Interesting Numbers, reciprocals, recreational math, recreational mathematics, recreational maths | Leave a comment

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

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

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

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

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

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

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

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

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

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

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

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

1/89 = 0·011235955056179775...

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

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

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

0/3 + 1/3² + 1/3³ + 2/3⁴ + 3/3⁵ + 5/3⁶ + 8/3⁷ + 13/3⁸ + 21/3⁹ + 34/3¹⁰ + 55/3¹¹ + 89/3¹² + 144/3¹³ + 233/3¹⁴ + 377/3¹⁵ + 610/3¹⁶ + 987/3¹⁷ + 1597/3¹⁸ + 2584/3¹⁹ + 4181/3²⁰ +... = 1/5 = 0·012101210121012101210 in b3 0/4 + 1/4² + 1/4³ + 2/4⁴ + 3/4⁵ + 5/4⁶ + 8/4⁷ + 13/4⁸ + 21/4⁹ + 34/4¹⁰ + 55/4¹¹ + 89/4¹² + 144/4¹³ + 233/4¹⁴ + 377/4¹⁵ + 610/4¹⁶ + 987/4¹⁷ + 1597/4¹⁸ + 2584/4¹⁹ + 4181/4²⁰ +... = 1/11 = 0·011310113101131011310 in b4 0/5 + 1/5² + 1/5³ + 2/5⁴ + 3/5⁵ + 5/5⁶ + 8/5⁷ + 13/5⁸ + 21/5⁹ + 34/5¹⁰ + 55/5¹¹ + 89/5¹² + 144/5¹³ + 233/5¹⁴ + 377/5¹⁵ + 610/5¹⁶ + 987/5¹⁷ + 1597/5¹⁸ + 2584/5¹⁹ + 4181/5²⁰ +... = 1/19 = 0·011242141011242141011 in b5 0/6 + 1/6² + 1/6³ + 2/6⁴ + 3/6⁵ + 5/6⁶ + 8/6⁷ + 13/6⁸ + 21/6⁹ + 34/6¹⁰ + 55/6¹¹ + 89/6¹² + 144/6¹³ + 233/6¹⁴ + 377/6¹⁵ + 610/6¹⁶ + 987/6¹⁷ + 1597/6¹⁸ + 2584/6¹⁹ + 4181/6²⁰ +... = 1/29 = 0·011240454431510112404 in b6 0/7 + 1/7² + 1/7³ + 2/7⁴ + 3/7⁵ + 5/7⁶ + 8/7⁷ + 13/7⁸ + 21/7⁹ + 34/7¹⁰ + 55/7¹¹ + 89/7¹² + 144/7¹³ + 233/7¹⁴ + 377/7¹⁵ + 610/7¹⁶ + 987/7¹⁷ + 1597/7¹⁸ + 2584/7¹⁹ + 4181/7²⁰ +... = 1/41 = 0·011236326213520225056 in b7

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

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

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

0/9 + 1/9² + 1/9³ + 2/9⁴ + 3/9⁵ + 5/9⁶ + 8/9⁷ + 13/9⁸ + 21/9⁹ + 34/9¹⁰ + 55/9¹¹ + 89/9¹² + 144/9¹³ + 233/9¹⁴ + 377/9¹⁵ + 610/9¹⁶ + 987/9¹⁷ + 1597/9¹⁸ + 2584/9¹⁹ + 4181/9²⁰ +... = 1/71 (another prime) = 0·011236067540450563033 in b9 0/10 + 1/10² + 1/10³ + 2/10⁴ + 3/10⁵ + 5/10⁶ + 8/10⁷ + 13/10⁸ + 21/10⁹ + 34/10¹⁰ + 55/10¹¹ + 89/10¹² + 144/10¹³ + 233/10¹⁴ + 377/10¹⁵ + 610/10¹⁶ + 987/10¹⁷ + 1597/10¹⁸ + 2584/10¹⁹ + 4181/10²⁰ +... = 1/89 (and another) = 0·011235955056179775280 in b10 0/11 + 1/11² + 1/11³ + 2/11⁴ + 3/11⁵ + 5/11⁶ + 8/11⁷ + 13/11⁸ + 21/11⁹ + 34/11¹⁰ + 55/11¹¹ + 89/11¹² + 144/11¹³ + 233/11¹⁴ + 377/11¹⁵ + 610/11¹⁶ + 987/11¹⁷ + 1597/11¹⁸ + 2584/11¹⁹ + 4181/11²⁰ +... = 1/109 (and another) = 0·011235942695392022470 in b11 0/12 + 1/12² + 1/12³ + 2/12⁴ + 3/12⁵ + 5/12⁶ + 8/12⁷ + 13/12⁸ + 21/12⁹ + 34/12¹⁰ + 55/12¹¹ + 89/12¹² + 144/12¹³ + 233/12¹⁴ + 377/12¹⁵ + 610/12¹⁶ + 987/12¹⁷ + 1597/12¹⁸ + 2584/12¹⁹ + 4181/12²⁰ +... = 1/131 (and another) = 0·011235930336A53909A87 in b12
0/13 + 1/13² + 1/13³ + 2/13⁴ + 3/13⁵ + 5/13⁶ + 8/13⁷ + 13/13⁸ + 21/13⁹ + 34/13¹⁰ + 55/13¹¹ + 89/13¹² + 144/13¹³ + 233/13¹⁴ + 377/13¹⁵ + 610/13¹⁶ + 987/13¹⁷ + 1597/13¹⁸ + 2584/13¹⁹ + 4181/13²⁰ +... = 1/155 (not a prime or a Fibonacci number) = 0·01123591ACAA861794044 in b13

The reciprocals go like this:

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

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

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

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

Dicing with Danger

Monday, 19th Sep, 2022 by Krilling for Company in Mathematics, Probability, Quotations and tagged American mathematician, chance, math, mathematics, maths, probability, probability theory, quotes about math, quotes about mathematics, quotes about maths, recreational math, recreational mathematics, recreational maths | Leave a comment

“In no other branch of mathematics is it so easy for experts to blunder as in probability theory.” — Martin Gardner (1914-2010)

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