Many of us are familiar with the everyday mathematical operators of addition, multiplication, and exponentiation. However, just around the corner, wilderness lurks. Each operation iterates the previous operation, so you can probably already imagine what the next operation can do:
Tetrational Numbers
We will begin with tetration, which is iterated exponentiation. a^^b is defined as a^(a^...(a^a)...) with b copies of a.
Just as with powers, 1 tetrated to any number is 1:
1^^2 = 11 = 1
1^^3 = 1^(1^1) = 11 = 1
1^^4 = 1^(1^(1^1)) = 1^(1^1) = 11 = 1
So it will always degenerate to 1 no matter how large the polyponent is. It turns out that tetration converges to a finite value if the base is between 1/(e^e) and e^(1/e) (between about 0.066 and 1.4447), which we will discuss later in the article. Next we will try base 2:
2^^1 = 2
2^^2 = 22 = 4
2^^3 = 2^(2^2) = 24 = 16
The third tetration of 2 has only two digits, and is one of only two 2-digit tetrational numbers. The next one is:
2^^4 = 2^(2^(2^2)) = 2^(2^4) = 216 = 65,536
2^^4 is big in terms of everyday numbers, but nothing surprising. Now we will consider 2^^5:
2^^5 = 2^(2^(2^(2^2))) = 2^(2^(2^4)) = 2^(2^16) = 265536
=
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2^^5 is a number with 19,729 digits! This would come as quite a shock, unless you notice that each tetration of 2 is 2 raised to the power of the previous. Next up is 2^^6:
2^^6 = 2^(2^(2^(2^(2^2)))) = 2^(2^(2^(2^4))) = 2^(2^(2^16)) = 2^(2^65536) = 220035299......19156736 (19729 digits)
With 2^^6, we have already reached a number larger than googolplex! The result of 2^(2^65536) is a number of about 6.031226*1019727 digits. One way to think of this number is: Take 2, square it once, and you get 4. Then square again, and we get 16. Then continue by squaring 2 a total of 65536 times! By comparison, we only need to square 2 334 times to get past a googolplex.
It can be easily shown that this number ends in a 6, and with a little more effort we can also prove that it ends in 36.
It turns out that the leading digits of this number can be computed, by taking the log10(2) to at least 19735 decimal places (19728 digits, plus a few additional digits to determine the digits just beyond the point), and multiplying it by the decimal expansion of 2^^5. Below are the first and last 40 digits of 2^^6.
2120038728808211984885164691662274630835...............8862693010305614986891826277507437428736
However, the leading digits of 2^^7 and beyond are inaccessible, as 2^^7 is equal to 2 to the power of 2^^6, and so to calculate is leading digits would require us to not only know log of 2 to 6.03122606*10^19727 decimal places, but also we would have to know the complete decimal expansion of 2^^6.
As you can probably guess, 2^^7 has about 30% of 2^^6 digits, and is greater than googolduplex.
You may notice that 2^^5 and 2^^6 both have the same last three digits. 2^^5 ends with ...5719156736, 2^^6 ends with ...7437428736, 2^^7 ends with ...9621748736, 2^^8 ends with ...9960948736, 2^^9 ends with ...7112948736, 2^^10 ends with ...4232948736, 2^^11 ends with ...1432948736, and 2^^12 and beyond end with ...3432948736. Indeed, the decimal expansions of the bigger 2-based numbers we will encounter later all end with ...3432948736.
Now we will try base 3.
3^^1 = 2
3^^2 = 3^3 = 27
3^^2 is the other tetrational number with 2 digits. Next is:
3^^3 = 3^(3^3) = 3^27 = 7,625,597,484,987
3^^3 is equal to about 7.6 trillion. This many miles would be about 1.4 light years, so this number is quite astronomical. This is a number that we will encounter many times when evaluating expressions using hyperoperators with the base of three.
3^^4 = 3^(3^(3^3)) = 3^(3^27) = 37,625,597,484,987
The result of 3^^4 is a number with 3,638,334,640,025 digits. This is approximately the 180,000,000th power of 2^^5, and so is MUCH larger than 2^^5. Here is how this number starts and ends:
1258014290627491317860390698203281215518.........4711077047886315075206738945776100739387
The next one is:
3^^5 = 3^(3^(3^(3^3))) = 3^(3^(3^27)) = 3^(3^7625597484987) = 312580142......00739387 (3638334640025 digits)
This number is much larger than 2^^6, in fact so much larger that even calculating its first leading digit would take 1000 times the amount of RAM on a typical PC (It might be possible to find the first digits of 3^^5 with more dedicated computational power).
I have also tried to find a series for the base-10 logarithm of 3 that may allow to calculate the initial digits of this number without calculating the entire integer part of its logarithm.
However, we do know that this number ends with ...6939489660355387. The number of digits in this number is approximately 6.00225*103638334640023. Notice how, again, the last 3 digits are the same as those of the previous number.
3^^6 = 3^(3^(3^(3^(3^3)))) = 3^(3^(3^(3^27))) = 3^(3^(3^7625597484987))
Similarly, the result of 3^^6 is considerably larger than 2^^7. The last digits of this number are ...5126595387. Just as with the tetrations of 2, the last digits appear to gradually stabilize as more 3s are added to the tower. 3^^7 ends with ...3200195387, 3^^8 ends with ...504195387, 3^^9 ends with ...8064195387, 3^^10 ends with ...6464195387, and 3^^11 and beyond end with ...2464195387.
Next we will try base 4:
4^^1 = 4
4^^2 = 44 = 256
4^^2 is equal to 2^8, the number of different values that can be represented with a byte.
4^^3 = 4^(4^4) = 4256 = 13,407,807,929,942,597,099,574,024,998,205,846,127,479,365,820,592,393,377,723,561,443,721,764,030,073,546,976,801,874,298,166,903,427,690,031,858,186,486,050,853,753,882,811,946,569,946,433,649,006,084,096
A 155-digit number. 4^^3 is already larger than a googol! Compare it with the previous two bases tetrated to 3.
4^^4 = 4^(4^(4^4)) = 4^(4^256) = 413407807...06084096 (155 digits)
4^^4 is a number that Jonathan Bowers used to call tritet, however, the tritet became the much larger 4^^^^4 (a number we will encounter later) when Bowers changed the definition of 3-entry linear arrays. Below are the first and last 40 digits of this number.
2361022671459731320687702749778179430946...…...5843019607448189676936860456095261392896
4^^5 = 4^(4^(4^(4^4))) = 423610226......61392896 w/ 8.0723*10^153 digits
4^^5 is larger than a googolduplex.
Now we will try base 5.
5^^1 = 5
5^^2 = 55 = 3,125
5^^2 is already quite sizable in everyday terms, but is still smaller than most of the smallest googolisms. Next up:
5^^3 = 5^(5^5) = 53125 = 1911012597945477520356404559703964599198081048990094337139512789246520530242615803012059386519739850265586440155794462235359212788673806972288410146915986602087961896757195701839281660338047611225975533626101001482651123413147768252411493094447176965282756285196737514395357542479093219206641883011787169122552421070050709064674382870851449950256586194461543183511379849133691779928127433840431549236855526783596374102105331546031353725325748636909159778690328266459182983815230286936572873691422648131291743762136325730321645282979486862576245362218017673224940567642819360078720713837072355305446356153946401185348493792719514594505508232749221605848912910945189959948686199543147666938013037176163592594479746164220050885079469804487133205133160739134230540198872570038329801246050197013467397175909027389493923817315786996845899794781068042822436093783946335265422815704302832442385515082316490967285712171708123232790481817268327510112746782317410985888683708522000711733492253913322300756147180429007527677793352306200618286012455254243061006894805446584704820650982664319360960388736258510747074340636286976576702699258649953557976318173902550891331223294743930343956161328334072831663498258145226862004307799084688103804187368324800903873596212919633602583120781673673742533322879296907205490595621406888825991244581842379597863476484315673760923625090371511798941424262270220066286486867868710182980872802560693101949280830825044198424796792058908817112327192301455582916746795197430548026404646854002733993860798594465961501752586965811447568510041568687730903712482535343839285397598749458497050038225012489284001826590056251286187629938044407340142347062055785305325034918189589707199305662188512963187501743535960282201038211616048545121039313312256332260766436236688296850208839496142830484739113991669622649948563685234712873294796680884509405893951104650944137909502276545653133018670633521323028460519434381399810561400652595300731790772711065783494174642684720956134647327748584238274899668755052504394218232191357223054066715373374248543645663782045701654593218154053548393614250664498585403307466468541890148134347714650315037954175778622811776585876941680908203125
5^^3 is a number of 2,185 digits! This number is approximately the 14th power of 4^^3, and is greater than the square of a googolchime. This number is greater than the number of years until all matter will become iron if protons are stable. Now we will consider 5^^4:
5^^4 = 5^(5^(5^5)) = 5^(5^3125) = 5^19110125...08203125 (2185 digits)
5^^4 is a number that has about 1.33574*102184 digits. Below are the first and last 40 digits of this number:
1111028808179997445286178274186057545167...…1798438687301313620992004871368408203125
5^^5 = 5^(5^(5^(5^5))) ~ 10^10^1.33574*10^2184
As expected, 5^^5 has nearly 5^^4 digits, and its first digits are beyond reach. The last 13 digits of 5^^5 are ...7618408203125.
Base 6:
6^^1 = 6
6^^2 = 66 = 46,656
6^^3 = 6^(6^6) = 646656 = 2659119772153226779682489404387918594905...…0672362313491705629432886056717863878656 (36,306 digits)
6^^4 = 6^(6^(6^6)) = 6^(6^46656) = 4446235190236934697525658222829200154057...…7815819279193580406457883859420138438656 (about 2.0692*10^36305 digits)
6^^6 is equal to 3$ (superfactorial under the Pickover definition, which takes n! tetrated to itself), and its last few digits are ...9127238656.
Below are a few base 7 examples.
7^^1 = 7
7^^2 = 77 = 823,543
7^^3 = 7^(7^7) = 7823543 = 3759823526783788538922130930895910817...2870132343 (695,975 digits)
7^^4 = 7^(7^(7^7)) = 7^(7^823543) = 7833005237480055565403854094754653082919...…8511366058254036038182879357182733172343 (about 3.1774*10^695974 digits)
Tetrations of 7 have the cool property that adding another 7 to the tower actually adds two convergent digits, not just one. This is because the last k digits of n actually determine the last k+2 digits of 7n, not just the last k+1 like with most bases, because 7^2 = 49 == -1 (mod 25).
8^^1 = 8
8^^2 = 88 = 16,777,216
8^^3 = 8^(8^8) = 60145207536513920379045068488......161150851025397555555421126656 (15,151,336 digits)
8^^4 = 8^(8^(8^8) = 64740329646697067997386625179........887920249826536946437619449856 (5.4*1015151335 digits)
9^^1 = 9
9^^2 = 99 = 387,420,489
9^^3 = 9^(9^9) = 9387,420,489 = 4281247731757470480369871159...2627177289 (369,693,100 digits)
9^^4 = 9^(9^(9^9)) = 9^(9^387420489) = 2141983294796805611333364373442480830147.........0163140828233401045865289 (about 4.085*10369693099 digits)
9^^4 is notable for being the first power tower of 9s greater than a googolplex, and also a number that appears frequently in the essay "Who can Name the Bigger Number".
10^^1 = 10
10^^2 = 10^10 = 10,000,000,000
10^^3 = 10^(10^10) = 10^10000000000 = 100000000000000000000000...…0000000000000000000000000 with 10,000,000,000 zeros
10^^4 = 10^(10^(10^10)) = 10^(10^10000000000) = 100000000000000000000000000...…00000000000000000000000000 with 10^10000000000 zeros
10^^5 = 10^(10^(10^(10^10))) = 10^(10^(10^10000000000)) = 100000000000000000000000000...…0000000000000000000000000000 with 10^^4 zeroes
If you would like to see the first and last digits of every tetrational number with a base less than 100 (I will add the rest of them eventually), click here.
If the base of tetration is between 1/e^e and e^(1/e) (between about 0.066 and 1.444), instead of blowing up to infinity, the result will actually converge to a certain, finite value as the height of the power tower increases. 1^^n is just 1 for any n, but other bases within the range where tetration converges produce more interesting results. Take, for instance, the square root of 2:
sqrt(2)^^1 = 1.414213562373095...
sqrt(2)^^2 = 1.632526919438152...
sqrt(2)^^3 = 1.760839555880028...
sqrt(2)^^4 = 1.840910...
sqrt(2)^^5 = 1.892712...
sqrt(2)^^6 = 1.926999...
It may not appear obvious from the values given above, but a power tower of sqrt(2)'s eventually converges to 2. With a base of 1.1, the value of the power tower converges to approximately 1.11178.
The limit of iterated exponentiation of x can be more easily found by finding the number y that satisfies xy = y, and iterated exponentiation of x increases without bound if there is no such y.
For a base just over 1.444667861..., tetration grows very slowly:
1.5^^2 = 1.837117307...
1.5^^3 = 2.106203352...
1.5^^4 = 2.349005318...
1.5^^5 = 2.592025704...
1.5^^6 = 2.860441497...
1.5^^7 = 3.189324761...
1.5^^8 = 3.644283987...
1.5^^9 = 4.382546732...
1.5^^10 = 5.911914873...
1.5^^11 = 10.990982932...
1.5^^12 = 86.181891743...
1.5^^13 = 1,499,263,005,586,576.357860319...
1.5^^14 ~ 1.232289776*10^264,007,110,309,345
1.5^^15 ~ 10^(2.1699*10^264,007,110,309,344)
The last digits of integer tetrations converge as the polyponent increases. The last d digits of ba in base 10 can be computed using the following recursive formula:
N1 = a
Nm = aN(m-1) mod 10d
Continue until Nb.
This works in base 10 for values of d ≥ 2 because the period of nk mod 10d evenly divides 10d itself. For instance, the last 3 decimal digits of powers always repeat after 100 powers if not sooner, thus 3…87 = …387 regardless of the digits before the …87 in the exponent. 2^100 also ends in ...376, and 8*376 = 3008.
While this works in base 10, this method fails in other bases. Watch what happens when we attempt to compute the last 3 digits of 7^^n in base 11 using this method:
7 ≡ 7 (mod 113)
27 = 823543 ≡ 985 (mod 113)
37 ≡ 7985 ≡ 857 (mod 113) (…70A11)
47 ≡ 7857 ≡ 378 (mod 113) (…31411)
57 = 7378 ≡ 64 (mod 113) (…05911) WRONG
However, if we use the correct method (which takes the exponent modulo φ(bd) where b is the numeral base and φ is Euler’s totient function), we get:
7 ≡ 7 (mod 113)
27 = 823543 ≡ 985 (mod 113)
37 = 7823543 ≡ 7743 (mod 1210) ≡ 332 (mod 113) (…28211)
47 = 77^823543 ≡ 7453 (mod 1210) ≡ 24 (mod 113) (…02211)
57 = 77^7^823543 ≡ 71113 (mod 1210) ≡ 266 (mod 113) (…22211)
67 = 77^7^7^823543 ≡ 71113 (mod 1210) ≡ 266 (mod 113) (…22211)
Thus, the last 3 base-11 digits of any power tower of at least 5 7s are …222.
The reason why the method works in base 10 is because the period of the last d digits of mn for integers m and n is always a divisor of 10d. The largest possible period for the last d digits is equal to lcm(phi(2^d), phi(5^d)) = lcm(2^(d-1), 4*5^(d-1)), which is 10^(d-1) for d greater than or equal to 3, and 20 for d = 2 (which is a divisor of 100, but greater than 10). However, in most other bases, the period of the last d digits has a divisor other than b that doesn’t evenly divide bd (for prime b, this is usually b-1). The method works only for at least 2 digits (trying it for 1 digit for x=2 would alternate between 4 and 6, even though the last digit of 2^^n is always 6 for n greater than or equal to 3). Additionally, it will not work if x is a multiple of 10 (for which the last few digits of tetrations will be trivially 0s), as the calculation will eventually return 0, which will result in 1 at the next step even though the correct result is 0.
The method will work in bases that are either powers of 2, or of the form 2^m*3^n, 2^m*5^n (this includes 10) when d*m is at least 2, 2^m*3^n*5^p when d*m is at least 2, or 2^m*3^n*7^p (where all exponents in each expression are at least 1). The reason for the condition that d*m be at least 2 in the bases divisible by 5 is because phi(5) = 4.
Thus, the method works in the following bases: 2, 4, 6, 8, 10, 12, 16, 20, 24, 30, 32, 36, 40, 42, 48, 50, 60, 64, ...
All the information I have added here can be found on the Googology Wiki article on tetration.
For d of at least 2, the least common multiple of the totient of 2^d (2^(d-1)) and 5^d (4*5^(d-1)) is less than the totient of 10^d, and is in fact a divisor of 10^d. For d of at least 3, it is in fact equal to 10^(d-1).
Below are the first 20 "tetrational endings" (the last digits of x^^y for sufficiently large y) in base 10.
1: 1
2: ...17989855389270912121282841628624570145528696872600159853338098615075353432948736
3: ...62535796399618993967905496638003222348723967018485186439059104575627262464195387
4: ...50884638388641609659570660459260625312436575602139341359493622302555290411728896
5: ...55526928791141371573820176477133189339892294089651159083587117493152618408203125
6: ...95758144434105807618736420464074406497310819685060571069123767965593227447238656
7: ...87077568921945767774448256803562956628147659630680637333853643331265511565172343
8: ...288375102773892651193162095588510081521577035416895225856
9: ...37063161457310262991864868894914047889985007525482726503475610748087597392745289
10: ...0000000000000000000000000000000000000000000000000000000000000000000000000000000
11: ...0466162924608865061483429393549993495402015711896234397891720266703879172666611
12: ...4630928051457675352659769758401249332877791752396359135848584931714421454012416
13: ...9745827966741647334071395366566344986532376409686170296788842218195488255045053
14: ...3905084263199307138141621654341476432992350488936037184088267703282377567502336
15: ...4254258912210039409144134940393274456858774268397382911643944680690765380859375
16: ...1609507026662007387123763358711252899650083880264852268236678107504950290415616
17: ...720085777
18: ...315776
19: ...0609963179
20: ...08178710937600000000……………………000000000000000000000000000000000000000000000000
The tetrational ending for a is the 10-adic solution to ax = x.
Each time 1 is added to the polyponent, another terminating digit becomes constant. But with bases 5 and 7, two digits become constant each time 1 is added to the polyponent. With bases such as 15 and 57, three digits become constant when 1 is added to the polyponent.
If the base is of the form 20n - 2 or 20n + 2, the convergence of terminating digits gets off to a delayed start, because the last digit of powers of such a number will repeat after 4 powers but the number itself is not divisible by 4: 2222²² only has one convergent trailing digit, and already it has more than 1029 digits.
The last digits of any tetration yx where x is a multiple of 10 and has an even digit before the trailing zero(s) are …07743740081787109376 followed by zeroes.
Pentational Numbers
Pentation is the next operation after tetration, and is formed by iterated tetration. Below are the first few pentations of 2:
2^^^1 = 2
2^^^2 = 2^^2 = 2^2 = 4
2^^^3 = 2^^(2^^2) = 2^^4 = 65,536
2^^^2 is no bigger than 2^^2, and 2^^^3 is equal to just 2^^4. None of the above is anything surprising, but prepare to be mind-blown:
2^^^4 = 2^^(2^^(2^^2)) = 2^^(2^^4) = 2^^65,536 = 2^(2^(2^(2^(2^(2^(2^(...2^(2^(2^(2^(2^(2^(2^2))))))...))))))) with 65,536 2s
2^^^4 is already larger than any number in the previous section! If you wrote this out as a power tower with each 2 taking 2 centimeters, it would stretch for nearly a mile, and yet we saw that even just the topmost 12 centimeters would resolve to a number much greater than googolplex.
And yet we have only just begun with pentation:
2^^^5 = 2^^(2^^(2^^(2^^2))) = 2^^(2^^65536)
That's a power tower of 2s with as many 2s as a power tower of 2s with 65536 2s!
2^^^6 = 2^^(2^^(2^^65536))
2^^^7 = 2^^(2^^(2^^(2^^65536)))
Now we will try base 3.
3^^^1 = 3
3^^^2 = 3^^3 = 7,625,597,484,987
7625597484987 again. Next up is yet another number that completely transcends everything we covered in the last section:
3^^^3 = 3^^(3^^3) = 3^^7625597484987
This number is equal to a power tower of 3s 7,625,597,484,987 terms high. If you could write this power tower down with each 3 being two centimeters high, it would stretch all the way from the Earth to the Sun! And yet the topmost ten centimeters would already resolve to a number much larger than googolplex that we can barely even calculate the first digits of. Bowers calls this number tritri as it is equal to {3, 3, 3} in his array notation, which we will cover in part 3. It is also the third Ackermann number, the previous being 4.
3^^^4 = 3^^(3^^(3^^3)) = 3^^(3^^7625597484987)
This number is equal to a power tower of 3s with as many 3s as a power tower of 3s 7625597484987 terms high! Below are examples of pentation with the next few bases.
4^^^1 = 4
4^^^2 = 4^^4 = 2361022671.........5261392896 (about 8.0723*10^153 digits)
4^^^3 = 4^^(4^^4) = 4^^2361022671.........5261392896
4^^^4 = 4^^(4^^(4^^4))
5^^^1 = 5
5^^^2 = 5^^5 = 5^5^5^5^5 ~ 10^10^1.33574*10^2184
5^^^3 = 5^^(5^^5) = 5^^(5^5^5^5^5)
5^^^5 is sometimes referred to as boogafive, because it is 5 pentated to 5.
The next hyperoperator is even more insane...
Hexational Numbers
Hexation is the next operation after pentation, and is formed by iteration of pentation.
2^^^^2 = 4
2^^^^2 is the same as 2^^^2 and 2^^2. In general, any arrowed expression with two 2's degenerates to 4. Also 2^^^^3 = 2^^^4 = 2^^65,536 (a number we already encountered in the last section), so we will begin with:
2^^^^4 = 2^^^(2^^65,536)
To express 2^^^^4 in terms of power towers, we would have a row of power towers of 2's where the number of power towers (including the lone 2) is 2^^65,536. This can be expressed more compactly as a "tetration tower" with 2^^^4 terms:
2^^^^4 = 2...22 where the number of terms is 2^^^4
3^^^^2 = 3^^^3 = 3^^7,625,597,484,987
3^^^^3 = 3^^^(3^^^3) = 3^^^(3^^7,625,597,484,987)
That's 3 pentated to a power tower of 7.625 trillion 3s! This number is referred to as G1 because it is the first term in the sequence defining Graham's number, which we will cover in a later article. To express this number in power towers, we would have a row of 3^^^3 power towers of 3s (including the lone 3):
3^3^3^3^3^...^3^3^3^3^3 } 3^3^3^3^3^...^3^3^3^3^3 } ... } 3^3^3^3^...^3^3^3^3 } 3^3^3 } 3
(where the number of towers is 3^^^3)
4^^^^4 is a number that has been dubbed the tritet, because it is equal to {4, 4, 4} in Bowers' array notation (which we will cover in a later article), and the sixth booga number is equal to 6^^^^6 (6 hexated to 6).
We can continue with heptation, octation, enneation, decation, etc.
In general, any arrowed expression with two 2s is a degenerate case, as each step will just decrement the number of arrows until we reach 2^2 = 4. An arrowed expression with a base of 2 and a polyponent of 3 is a semi-degenerate case, as even though it is (usually) a lot larger than 4, it resolves to a number in the class of the previous operator as 2^^^...(n arrows)...^^^3 = 2^^^...(n-1 arrows)...^^4.
For instance, if you only knew 1^^3 = 1 and 2^^3 = 16, you would likely expect 3^^3 to be a few thousand, but it is actually about 7.625 trillion as we saw. Similarly, knowing 1^^^3 = 1 and 2^^^3 = 65536, one would expect 3^^^3 to maybe have a few dozen digits or so, but in reality it is indescribably larger than that.
A curious observation is that 2 ^m n seems to be the 2-based equivalent of 3 ^m n-1. For tetration, the comparison is not completely accurate (7,625,597,484,987 is considerably larger than 65,536). However, 2^^^^4 is very much the 2-based equivalent of 3^^^^3 (their visualizations in terms of power towers are very similar).
The sequence n^^^...n arrows...^^^n is known as the Ackermann numbers. The first few Ackermann numbers are 1, 4, 3^^^3, 4^^^^4, ... However, if we use the n-th operation instead of n arrows, we can better capture how quickly the operators grow: 2, 4, 27, 4^4^4^4, 5^^^5, ...
NEXT >> Extensions to Arrow Notation