What Comes Next

Many flowers have 3, 5, 8, 13, 21, 34, 55, or 89 petals. These are all Fibonacci numbers, which are calculated by starting with two 1s, and finding further terms by adding the last two terms. The sequence begins as 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946....

Another place in nature where you can find the Fibonacci sequence is in pinecones. If you count the spirals, you will more often than not end up with a Fibonacci number. And yet another instance of the Fibonacci numbers in nature is in trees.

Pianos also make use of Fibonacci numbers. One octave of piano keys is 13 keys, which are divided into 8 white keys and 5 black keys, which are divided into groups of 3 and 2. 5/3 octaves contain 21 keys, and the total number of keys on the piano is one less than 89.

The Fibonacci sequence grows quite slowly at first, but it exhibits exponential growth. The 1,000th term is a number of 209 decimal digits. Notice that Fn is approximated by the golden ratio to the n-th power. A better approximation for Fn is ϕn/√5, where ϕ is the asymptotic ratio of two successive Fibonacci numbers, which is approximately 1.618 033 988 749 894 848. In fact, this is almost exact for sufficient values of n.

The last digit of each Fibonacci number is cyclic with period 60:

1, 1, 2, 3, 5, 8, 3, 1, 4, 5, 9, 4, 3, 7, 0, 7, 7, 4, 1, 5, 6, 1, 7, 8, 5, 3, 8, 1, 9, 0, 9, 9, 8, 7, 5, 2, 7, 9, 6, 5, 1, 6, 7, 3, 0, 3, 3, 6, 9, 5, 4, 9, 3, 2, 5, 7, 2, 9, 1, 0, …

If a prime p divides the n-th Fibonacci number, p will also divide all Fibonacci numbers whose index is a multiple of n. For example, every third Fibonacci number is even, every fourth is divisible by 3, every fifth is divisible by 5, every seventh is divisible by 13, and every eighth is divisible by 7.

And note that the 66th Fibonacci number has 4 identical digits starting at the 2nd digit. 27777890035288.

To calculate the leading digits of Fibonacci(n) for large n, multiply the logarithm of the golden ratio by n, then subtract the logarithm of the square root of 5, because the effect of the second term in the formula for calculating Fibonacci numbers is negligible for very large terms.

Below are all known digits of the googolth term, which has over 2.08987x1099 digits!

6244991128646068764887850066130550466730800975346075154177950982927588026342880062773094011757812153242934820175549809664488449123561932528719437050471...8299560546875

The leading digits are calculated using logarithms and the terminating digits are computed using the addition method (fibonacci(3x10n) will end in 709796 followed by n + 1 zeroes).

2046711111473984623691759, which is the 118th Fibonacci number, is the first Fibonacci number to contain a run of 5 consecutive 1s in its decimal expansion.

Also, if you start with 11, and continue by multiplying by 10 and adding the sum of the last two digits, you get numbers that start with the single digit Fibonacci numbers, and in fact, eventually you get to a point where the digits cycle with a period of 30. The full sequence of digits is below.

112359426953920224707852806730

If you do the same, only with subtraction, you eventually get numbers with digits that cycle after 44 digits:

11008164220173384403567788072466871550448532

Notice the last digits are Fibonacci numbers.

The Lucas numbers are defined by a similar sequence to the Fibonacci numbers, except the starting terms are 2, 1 instead of 1, 1. The sequence of Lucas numbers starts 2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, …

Below are more Lucas numbers.

2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843, 1364, 2207, 3571, 5778, 9349, 15127, 24476, 39603, 64079, …

If you round a power of 1.618 033 988 749 894 848 204 586... (ϕ) to the nearest integer, you will get a Lucas number. The powers of the golden ratio get closer and closer to being an integer as they get larger (their fractional parts eventually alternate between .000… and .999…).

Another way to generalize the Fibonacci numbers is to increase the number of terms added. The Tribonacci sequence is like the regular Fibonacci sequence, only the last three terms are added instead of just the last two. The sequence begins 1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927, 1705, 3136, 5768, 10609, …

The Tribonacci constant is approximately 1.839286755214161…, and it is very closely approximated by 103/56, the ratio of two numbers whose squares appear two terms apart in the sequence.

The square of 9, or 81, appears in the Tribonacci sequence. The next two terms are not very interesting, but the term after that, 504, is equal to 9*56. Note how one of the 9s in 9*9 changes to a 56. The next term, 927, is equal to 9*103.

Similar sequences to the Fibonacci sequence can be formed by adding a multiplication. The Pell numbers are formed by starting with a1 = 1, a2 = 2, and an = 2an-1 + an-2 (n >= 3). The first few Pell numbers are 1, 2, 5, 12, 29, 70, 169, 408, 985, 2378… The ratio of two consecutive terms converges to the silver ratio 2.414 213 562 373 095 048…, one more than the square root of 2.

A Fibonacci or Pell-like sequence can be formed by multiplying by any other number.

The 3rd term of the n-bonacci sequence is equal to n2 + 1, and the 4th term is equal to n(n2 + 2) or n3 + 2n. The fifth term is equal to (n2 + 1)(n2 + 2) – 1.

and we can even diagonalize over the number multiplied, to form the following sequence:

1, 1, 3, 10, 43, 225, 1393, 9976, 81201, 740785, 7489051, 83120346, 1004933203, 13147251985 …

The n-th term in the sequence can be lower-bounded by nn-5. The sequence in the OEIS is A001040.

A sequence I would definitely recamánd

And here’s another one for you: 1, 3, 6, 2, 7, 13, 20, 12, …, what comes next?

This sequence is known as Recamán’s sequence, named after Colombian Bernardo Recamán Santos. In this sequence, to get to the n-th term, we either subtract n from the previous term (Siempre atrás, always back), unless the result is already in the sequence; then, we add n.

The sequence starts with a 0, then add 1, then add 2 to obtain 3, then add 3 since 0 is already in the sequence, then subtract 4 as 2 is not yet in the sequence, then add 5 as we can’t go back another 5 steps, then add 6 as 1 is already in the sequence, and so on.

When visualized on a number line, this sequence produces some amazing circles.

And, when translated into music, the sequence will sound almost like someone is actually playing. The sequence will sound like several chromatic scales put together, but in a pleasant-sounding way.

It is conjectured that every number appears in the sequence, but it is unproven. 1015 terms were calculated as of 2018, but the number 852,655 has so far never appeared.

A similar sequence exists, where to find the next term, we divide the current term by its index, unless the current term isn’t evenly divisible by the index, in which case we multiply. The sequence begins 1, 1, 2, 6, 24, 120, 20, 140, 1120, 10080, 1008, 11088, 924, 12012, 858, …

The 2,732nd term in the sequence (the highest the OEIS gives) is equal to 173418537797154024279365729699128354411166316923761964994158137229514063144377850434582316289108329995953685037938652990071781190331220455422652537982780806240041759198468347986103880198801094771152207242231004765082776688270322260094652667670309563969326638251243885105970467163095805711950767870034809520727123859662730179386138068453638444629164107980522624464801015123170133963234527745235460012766689197227593102020255123377474749479543455895139257439630416870300408206999048064156803438550406380120837999091199707959364671023590773732351385199779858268947839882032468082645153897630222459660861470775892351528796612655122338446567351564377204035917594454063939146416527325471298020016255603182349302815107019699215879918783994275206306974633112980291162436125075104616902657174273791170835215586514855967693287869546833047753773670596908206675291512875758060244966630197649490166853739533979396319468999378851995813947188642094197473106014289580070550416587594849278863468815275148922643125.

Generating an extremely fast-growing sequence is quite easy.

Suppose we have a list of functions, say n2, n2 + 1, n2 + 2, n2 + 3, …. We apply each function to the corresponding value of n and obtain 0, 2, 6, 12, 20, 30, 42, …, the oblong numbers. So far, not very impressive.

But what if we apply the function again? We then get 0, 5, 38, 147, 404 (error: term not found), 905, 1800, …. This is on par with n4 and is equal to n4 + 2n3 + n2 + n.

We could keep on applying the function, but there is a better way: diagonalization. Instead of applying the function a fixed number of times, we diagonalize the number of iterations and iterate n times.

This forms the sequence 0, 2, 38, 21612, 26640768404, 449986245140985954219030, …, which is already growth-equivalent to repeated squaring. This makes sense, since we are iterating a function slightly above n2 n times.

But what if we don’t stop there? What if we iterate the number within the function once? We get the sequence 0, 6, 1802, 218546201670519621, 253731261674114007572110097505136152892655016917940760033127812965931735386252080804 …, which grows at roughly the rate of an iterated 4th power!

What if we iterate enough times for n steps? Then we get the sequence 0, 2, 1802, 2281250891080340926326737418121311216425715930812454720992447468743062, …. Whoa! We just went from the weight of a pile of bricks to the number of particles in a typical galaxy!

Later in the number book, we will encounter many more examples of the immense power of diagonalization and recursion.

Did you know that a sequence may have a pattern that’s not apparent at first? Take this sequence, which is somewhat reminiscent of Avatar, where the protagonist tries to fly the Banshee, but the Banshee doesn’t fly. Eventually, the protagonist says “Fly straight, d@mn it”, and the Banshee begins flying.

We begin with a0 = 1. To find an, we either add n+1 to the previous term if n and the previous term have no common divisors other than 1, otherwise divide by the greatest common divisor shared. The sequence begins 1, 1, 4, 8, 2, 8, 4, 12, 3, 1, 12, 24, 2, …

If graphed, the sequence looks very erratic, with the sequence jumping up and down in an unpredictable fashion. That is, until the seemingly arbitrary value of 638, which is the first time a 1 occurs at an even number. After that, the graph will look like three straight lines going off to infinity, like the Banshee beginning to fly.

The reason for this is because when we reach a2n = 1, we find that a2n+1 = 1 + 2n+1 + 1 = 2n+3, then a2n+2 = 2n+3 + 2n+2 + 1 = 4n+6, as 2n+2 and 2n+3 are surely coprime. Then, since 4n+6 is exactly twice 2n+3, we go all the way back down to 2 at a2n+3, and reach 1 again at a2n+4.

Blast off!

Here’s another one: 11, 12, 14, 18, 26, 58, …, what comes next?

Given the first few terms, you would think that this sequence would take eons to get into the land of very large numbers, as the first six terms are all 2-digit numbers. The sequence seems at best to be exponentially-growing, as the difference from 10 seems to keep doubling in the first few terms. However, you’re in for a shock.

The next term is 298, which is still in the range of everyday numbers. The sequence still appears to only be growing barely faster than exponentially.

Not convinced yet? The next term is already 180298 (notice how the last 3 digits are the same). We already leapt from a few hundred to over a hundred thousand. Try to guess how large the next term is…

The next value after that is 190534583862796232642707594, a number about 32 times greater than the Earth’s mass in kilograms. As if that wasn’t enough, the next term is a 684-digit number approximately equal to 1.90 247 610 511 7*10683, and its exact value is below.

190247610511749325712830235380677606907660597821972511966612655558640928842520854126652085616844022890490264929774168945288820052619117073448996137588309420057029307298653425375545268353859650901737297032596673982334499077814263706893786410452053355498435709872772254737988126258119973927968540054993166882115774290442313389725893427136299582512205099731400938091475987348176517269561281705319651587753980666822058293432286629429026792024754941185392550882780686382435585400348812586571314498622821842703113456869008993189788992362962319496437875846049979709324208704066490508156865096445067193340904572222556332376680028464674827519379512802923419145265056204841429781369353980176650

In this sequence, each term, when written in the base of the previous term, looks identical to the previous term written in base 10. This means that we roughly take each term to the power of the number of digits it has minus 1 to get to the next term, which is almost the same as squaring the number of digits. Thus, the sequence exhibits triple exponential growth, which is equivalent to repeatedly raising the number of digits to a certain power.

The next term after that is about 5.956909032385*10466,679, much larger than the googolgong, and the next term after that is approximately 5.1206478700774*10217,789,650,727, which is large enough that storing all the digits would take a decent portion of a typical hard drive. And that’s just the 12th term!

By merely the 16th term, we’ve surpassed the mighty googolplex (its first 500 and last few digits are a few paragraphs below), and the 39th term is larger than even the largest nontrivial integer power whose first digits I’ve calculated.

This sequence defies first impressions in another way. At first, it seems as if the last digits of each term are totally unpredictable, and that knowing the last digits of an arbitrary term would require knowing every single digit of the term before it.

That is, until we hit our first term that ends in a 0, which is the 10th term. After that, we add …6*a103 + 6*a102 + 5*a10 + 0 = …000 + …000 + …250 + 0. All the terms in the sum before that also end with more than 3 zeroes, so the last 3 digits of a11 (the 466,680-digit number) are …250. In fact, all further terms end in …250, and the last digits converge to the same leftward-infinite sequence as repeatedly squaring 5, but with a 0 appended.

First and last 10 digits of an, n >= 10

a10: 1902476105………3980176650
a11: 5956909032………4566218250
a12: 5120647870………4003466250
a13: 5406334535………8701706250
a14: 6371843501………4992906250
a15: 2499938188………6448906250

Decimal expansion of a16

30916928440269127711482909640906281496302151058626699126438222580193018481424007061117715556587924978513496311254905635422162306249797502732737635141286981235467695335583682579884003973199116527720012999691001642848500611009108469209187839032477086255770203236138207811447343721307887493718895354276878330338575706557569532458764135223961527124060591005873789671636775799529847535455055566713483272461865810935277138007923890356536694191985055543980396903437229389191818493885691666883924498391725443……………………………………………………………………………………………………………………………………………………………..(25,620,984,557,551,828,354,898,460,188,478,602,239,613,057,156,993,381,321,190,812,275,940,564,170,869,182,358,086,283,639,428,169,767,789,504,576,363,875,736,950,438,133,877,911,216,128,652,715,672,892,653,191,965,112,639,952,576,267,900,560 digits)…………………………………………………………………………………………………………………………………..3728906250

In the long run, each term in the sequence is approximated by 22^2^(n-7), or, more accurately, 22^(5*2^(n-9)). That’s 2 squared 5*2n-9 times.

A similar sequence in binary becomes exactly equal to 22^2^n, while the sequence in base 3 begins 4, 5, 7, 15, 255, …, and the sequence in base 4 begins 5, 6, 8, 16, 256, 4294967296 …, while the sequence in base 100 begins 101, 102, 104, 108, 116, 132, 164, 228, 484, 2020, 40420,

A repunit is a number like 11, 111, 1,111, or 11,111 that consists of only the digit 1. Repunits also have certain properties. For instance, if you subtract 1 from a repunit and divide by 10, the answer is yet another repunit. Also, if you square a repunit, you can make all the digits appear (See Repunit squares and cubes, & higher repunit powers).

R1 = 1

R2 = 11

R3 = 111

R4 = 1,111

R5 = 11,111

R6 = 111,111

R7 = 1,111,111

R8 = 11,111,111

R9 = 111,111,111

R10 = 1,111,111,111

R11 = 11,111,111,111

R12 = 111,111,111,111

R13 = 1,111,111,111,111

R14 = 11,111,111,111,111

R15 = 111,111,111,111,111

R16 = 1,111,111,111,111,111

R17 = 11,111,111,111,111,111

R18 = 111,111,111,111,111,111

R19 = 1,111,111,111,111,111,111

R20 = 11,111,111,111,111,111,111

If n is a power of 3, or another repunit where the number of 1s is a power of 3, then n divides Rn. For example, 27 is equal to 33, and thus R27 is divisible by 27, in fact, 111111111111111111111111111 = 27 x 4115226337448559670781893.

R3 = 37 x 3

R9 = 12345679 x 9

R27 = 4115226337448559670781893 x 27

R81 = 1371742112482853223593964334705075445816186556927297668038408779149519890260631 x 81

The divisibility of a repunit is dependent on the divisibility of the number of digits. If a repunit has an even number of digits, it is divisible by 11. Otherwise, it is not. If the number of digits is divisible by 3, then the repunit is divisible by 3 and 37.

The divisors of the googolth repunit include, but are not limited to the following numbers.

101

137

251

271

353

401

449

641

751

1201

1409

1601

3541

4001

5051

9091

16001

21401

25601

27961

60101

162251

544001

670001

1610501

1676321

5070721

5882353

6600001

7019801

421144001

532525001

756100001

1797655751

5964848081

24592788001

128372635774581251

42051775804956304559810859008305819975199677041099230574273451704628125001 (74 digits)

100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000010000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001 (501 digits)

100000000000000000000000000000...000000000000001000000000000000...000000000000001000000000000000...000000000000001000000000000000...000000000000001000000000000000...000000000000001000000000000000...000000000000001000000000000000...000000000000001000000000000000...000000000000001000000000000000...0000000000000000001 (9x1099 digits)

In other bases

The rule that states that any number raised to a power minus 1 is divisible by the number that is one less than the number that was raised to the power can be used to extend repunits to other bases.

Below are the repunits in base 4.

22 = 4 - 1 = 3
3 / 3 = 1

42 = 16 - 1 = 15
15 / 3 = 5

43 = 64 - 1 = 63
63 / 3 = 21

44 = 256 - 1 = 255
255 / 3 = 85

45 = 1,024 - 1 = 1,023
1,023 / 3 = 341

450 = 1,267,650,600,228,229,401,496,703,205,376 - 1 = 1,267,650,600,228,229,401,496,703,205,375
1,267,650,600,228,229,401,496,703,205,375 / 3 = 422,550,200,076,076,467,165,567,735,125

Any power of 8 minus 1 is divisible by 7. The units place digit of a base-8 repunit in decimal exhibits a repeating cycle: 1, 9, 3, 5, 1, 9, 3, 5… A base-8 repunit cannot end in a 7 in decimal because …7 * 7 = …9, which would imply that there is a power of 8 (and thus 2) that is divisible by 5, which is impossible.

23 = 8 - 1 = 7
7 / 7 = 1

82 = 64 - 1 = 63
63 / 7 = 9

83 = 512 - 1 = 511
511 / 7 = 73

84 = 4,096 - 1 = 4,095
4,095 / 7 = 585

85 = 32,768 - 1 = 32,767
32,767 / 7 = 4,681

833 = 633,825,300,114,114,700,748,351,602,688 - 1 = 633,825,300,114,114,700,748,351,602,687
633,825,300,114,114,700,748,351,602,687 / 7 = 90,546,471,444,873,528,678,335,943,241

Any power of 32 minus 1 is divisible by 31. These are the base-32 repunits.

25 - 1= 32 - 1 = 31
31 / 31 = 1

322 - 1 = 1,024 - 1 = 1,023
1,023 / 31 = 33

323 - 1 = 32,768 - 1 = 32,767
32,767 / 31 = 1,057

324 - 1 = 1,048,576 - 1 = 1,048,575
1,048,575 / 31 = 33,825

325 - 1 = 33,554,432 - 1 = 33,554,431
33,554,431 / 31 = 1,082,401

326 - 1 = 1,073,741,824 - 1 = 1,073,741,823
1,073,741,823 / 31 = 34,636,833

3220 - 1 = 1,267,650,600,228,229,401,496,703,205,375
1,267,650,600,228,229,401,496,703,205,375 / 31 = 40,891,954,846,071,916,177,313,006,625

Any power of 512 minus 1 is divisible by 73. This happens because 73 x 7 = 511, which is one less than 512.

29 = 512 - 1 = 511 / 73 = 7

5122 = 262,144 - 1 = 262,143 / 73 = 3,591

5123 = 134,217,728 - 1 = 134,217,727 / 73 = 1,838,599

5124 = 68,719,476,736 - 1 = 68,719,476,735 / 73 = 941,362,695

5125 = 35,184,372,088,832 = 35,184,372,088,831 / 73 = 481,977,699,847

5127 = 9,223,372,036,854,775,808 - 1 = 9,223,372,036,854,775,807 / 73 = 126,347,562,148,695,559

51211 = 633,825,300,114,114,700,748,351,602,688 - 1 = 633,825,300,114,114,700,748,351,602,687 / 73 = 8,682,538,357,727,598,640,388,378,119

Any power of 20 minus 1 is divisible by 19. With each power of 20, the last digits of this quotient converge to the pattern 052631578947368421...052631578947368421. These are the repeating digits in 1/19. This means that the ending of the number 52631578947368421 is also the ending of all base-20 repunits.

202 = 400 - 1 = 399 / 19 = 21

203 = 8,000 - 1 = 7,999 / 19 = 421

204 = 160,000 - 1 = 159,999 / 19 = 8,421

205 = 3,200,000 - 1 = 3,199,999 / 19 = 168,421

206 = 64,000,000 - 1 = 63,999,999 / 19 = 3,368,421

2020 = 104,857,600,000,000,000,000,000,000 - 1 = 104,857,599,999,999,999,999,999,999 / 19 = 5,518,821,052,631,578,947,368,421

Below is the sequence of numbers whose representation is 11...11 (n 1s) in base n, or (nn – 1)/(n - 1). Below the first 50 terms are given.

781

9331

137257

2396745

48427561

1111111111

28531167061

810554586205

25239592216021

854769755812155

31278135027204241

1229782938247303441

51702516367896047761

2314494592664502210319

109912203092239643840221

5518821052631578947368421

292129350919299126069056221

16258470350677121780792701123

949112181811268728834319677753

57988512036968874976047020558425

3700743415417188468078772226969401

246244783208286292431866971536008151

17054864932424529613394216562274995877

1227611967157532402317508453828318733805

91703076898614683377208150526107718802981

7099694210160310344827586206896551724137931

568972471024107865287021434301977158534824481

47145214107448481232376930087622032892126856225

4034688752742532119988000935458266735604286980641

356261785011170182164062136403344107715695209380895

32426691157474961645640777453338480828439488130457261

Repunit squares and cubes, & higher repunit powers

In the square of a repunit, all the digits will appear in the result, until they overlap:

112 = 121
1112 = 12321

11112 = 1234321

111112 = 123454321

1111112 = 12345654321

11111112 = 1234567654321

111111112 = 123456787654321

1111111112 = 12345678987654321

11111111112 = 1234567900987654321

111111111112 = 123456790120987654321

1111111111112 = 12345679012320987654321

The 8 gets bumped up to a 9 at 1,111,111,1112, which is 1234567900987654321. This also makes 12345678987654321 the largest palindromic repunit square. Also, multiplying a repunit square by 11 results in a palindrome with only odd digits, until they similarly overlap and break the pattern:

1 x 11 = 11

121 x 11 = 1331

12321 x 11 = 135531

1234321 x 11 = 13577531

123454321 x 11 = 1357997531

The next number in the sequence (12345654321 x 11), however, is not a palindrome:

12345654321 x 11 = 135802197531

And even the first three numbers that are products of repunits are palindromic. This sequence is shown below.

1221

1356531

15072415941

1674711207620451

1860790044610366931061

20675444733360738721748118771

2297271634742810443154153338805764581

2552524038347870310755413660544832496799359491

28361378203581611893021499527080870668821235178133404501

3151264244839250057201990879465316574366017145273842585091318510611

3501404716487705478641674393843995972531189570684905537085323473724989853498821

The middle digit of a repunit square will be the remainder (mod 9) of the number of 1s in the square root. Example: R132 = 1234567901234320987654321. The middle digit is 4, which equals the residue class of 13.

And note the squares of numbers that are eight times repunits.

7744

788544

78996544

7901076544

790121876544

79012329876544

7901234409876544

790123455209876544

79012345663209876544

7901234567743209876544

790123456788543209876544

79012345678996543209876544

7901234567901076543209876544

790123456790121876543209876544

79012345679012329876543209876544

7901234567901234409876543209876544

By taking the fourth powers of numbers that consist of only the digit 3, one gets a sequence similar to the repunit squares. On the left and right is where all the digits appear. In the middle, random multiples of 37 appear.

1185921

12296370321

123407414814321

1234518519259254321

12345629629703703654321

123456740740748148147654321

1234567851851852592592587654321

12345678962962963037037036987654321

123456790074074074081481481480987654321

Notice what happens when we raise a number of the form 3 x Rn to the fifth power. It converges to the pattern 411522633744855967078189300411522....

243

39135393

4094691316893

411316913576131893

41150205802468724281893

4115205761358024650205781893

411522427983580246909465020781893

41152261316872469135802057613170781893

4115226316872428024691357983539094670781893

411522633539094650246913580242798353909670781893

41152263372427983539135802469135390946502059670781893

4115226337427983539094691358024691316872427983559670781893

411522633744650205761316913580246913576131687242798559670781893

41152263374483539094650205802469135802468724279835390948559670781893

Below is the sequence of repunit cubes. The digits converge to 1/729, similar to the repunit squares converging to 1/81.

1331

1367631

1371330631

1371700960631

1371737997260631

1371741700960260631

1371742071330590260631

1371742108367626890260631

1371742112071330589890260631

1371742112441700960219890260631

1371742112478737997256519890260631

1371742112482441700960219519890260631

1371742112482812071330589849519890260631

1371742112482849108367626886149519890260631

1371742112482852812071330589849149519890260631

1371742112482853182441700960219479149519890260631

1371742112482853219478737997256515779149519890260631

1371742112482853223182441700960219478779149519890260631

1371742112482853223552812071330589849108779149519890260631

1371742112482853223589849108367626886145408779149519890260631

1371742112482853223593552812071330589849108408779149519890260631

1371742112482853223593923182441700960219478738408779149519890260631

1371742112482853223593960219478737997256515775038408779149519890260631

1371742112482853223593963923182441700960219478738038408779149519890260631

1371742112482853223593964293552812071330589849108368038408779149519890260631

1371742112482853223593964330589849108367626886145404668038408779149519890260631

1371742112482853223593964334293552812071330589849108367668038408779149519890260631

This time, the left side converges to the digits of 1/729. And the right side converges to the last repeating digits of 1/729.

The first repunit cube with full convergence is

1371742112482853223593964334705075445816186556927297668038408779149519890260631000960219478737997256515775034293552812071330589849108367626886145404663923182441701001371742112482853223593964334705075445816186556927297668038408779149519890260631

which is R823.

The first repunit cube that is greater than googolplex is R33333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333353, which is approximately 1.37174211248285322359396433470507544581618655692729766803840877914951989026x10(googol+2)

Below is the value of Rgoogol3. The middle digits have the pattern ...29355281207133058984910836762688614540466...

1371742112482853223593964334705075445816186556927297668038408779149519890260631001371742112482853223593964334705...503429355281207133058984910836762688614540466392...9149519890260631001371742112482853223593964334705075445816186556927297668038408779149519890260631

1371742108367626890260631 is the only repunit cube with this property. If you multiply it by 9, the answer begins with all the digits except zero in sequence. It is the pandigital number 12,345,678,975,308,642,012,345,679.

The sequence of pandigital repunit cubes begins as 1371742112071330589890260631, 1371742112478737997256519890260631... Soon all repunit cubes are pandigital.

Below is another sequence, the fourth powers of repunits.

14641

151807041

1523548331041

15240969373571041

152415180613625971041

1524157293095656149971041

15241578417924106081389971041

152415789666209420210333789971041

1524157902149062642981252857789971041

15241579026977594878838591678097789971041

152415790275262917238226794695930497789971041

1524157902758116140832190306355738454497789971041

Note the fourth power of 111,111,111. It has the pattern ...420210... in it, where a block of three is twice the next.

Repunit fifth powers converge to yet another pattern. It begins as 169350878084302867110365967.... These are the first digits of the reciprocal of 59,049.

161051

16850581551

1692662195786551

169342410709747836551

16935003133160595268336551

1693507934088807600450473336551

169350869616759132246099002523336551

16935087723754847838236041084523023336551

The last digits also converge to ...767278023336551, which are the last digits of the 10th term in 3, 1, 37, 12345679, 4115226337448559670781893...

In a repunit sixth power, the digits converge to 18816764231589....

1771561

1870414552161

1880547699518858161

1881563525396008211918161

1881665133128606900860142518161

1881675294153349101744125879448518161

1881676310258338178650085334026072508518161

1881676411868862234942354805142998028003108518161

For repunit seventh powers, the main interest is the first digit becoming 2. This happens because 97 = 4782969, and 4782969 is less than 5000000.

19487171

207616015289871

2089288494165451416871

20906052330675047242622686871

209073694607052641361471295335386871

2090750117762021873788017450039922462386871

And yet another sequence is created, where each term is the nth power of Rn.

121

1367631

1523548331041

169342410709747836551

1881665133128606900860142518161

2090750117762021873788017450039922462386871

232305712669729896500117628007183171975956078054818623681

2581174768482624149493549155062886015819364354785066837288515125486819591