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4.281248...×10^369693099 = 9⁹⁹ = 9^387420489

The largest number you can express with just three base-10 digits and possibly some symbols and/or parentheses: 9⁹⁹, or 9^(9^9), etc. See also 387420489, 10460353203 and 10^{1.0979465941272×1019}.

8.808074...×10^646456992 = 2^2147483647-1 = 2^(231-1)-1

This is one of the largest Mersenne numbers ever tested by the Lucas-Lehmer test; it was found to be composite. If prime, it would have been a double Mersenne prime.

1.0621842147...×10^4990856845 = 3³²¹ = 3^10460353203

In high school, around the same time I was calculating large integers like this, I also made approximations of even larger numbers using logarithms on a calculator. This is the largest one I tried to actually write down in standard scientific notation. Due to the limited accuracy of my calculator, the closest estimate I could get was 9×10^4990856844. In my notebook I claimed that this was the value of 3_⑥2, where _⑥ represents the sixth function in the hyper series according to the lower "left-hand-associative" definition. But, due to an error in my formulas I thought 3_⑥2 was 3³²¹ when in fact it is 3³²⁰:

3_⑥2 = 3_⑤3
= (3_④3)_④3
= ((3³)³)_④3
= 19683_④3
= (19683¹⁹⁶⁸³)¹⁹⁶⁸³
= 3^{(9×19683×19683)}
= 3^3(2+9+9) = 3³²⁰

10¹⁰¹⁰

The largest finite number indirectly referred to in any published music (as far as I know). My Hero, Zero, the Schoolhouse Rock song about how the digit '0' is used to multiply any number by powers of 10, includes the lines:

Place a zero after one,
and you've got yourself a ten --
see how important that is!
When you run out of digits
you can start all over again --
see how convenient that is!
That's why with only ten digits, including zero,
you can count as high you could ever go --
forever, towards infinity.
No-one ever gets there, but you could try ...
with ten billion zeros.

It doesn't exactly say what is being done with those "ten billion zeros" (10¹⁰), but the picture on-screen during the lines "forever, towards infinity / no-one ever gets there, but you could try" shows a pyramid made up of the numbers 9, 80, 700, 6000, 50000, and so on — the screen ends up filled with small zeros — so I imagine they were implying the idea of writing some (nonzero) digit(s) followed by 10,000,000,000 zeros in a row — and then you'd get at least 10¹⁰¹⁰.

See also 10¹⁰, 100000000000 and 10¹⁰¹⁰⁰.

5.1843×10^22652507173

An upper limit of the volume of the universe, if one makes the following assumptions: 1) Not all of the universe can be observed directly (because of hyperinflation), 2) Life originated purely by the chance meeting of particles to form a single original cell, and 3) That event has happened only once, and all extant life is the result of it. A size of about 10^22000000000 is the size necessary to guarantee that each possible chance meeting of 75250000000 particles has occurred somewhere at least once; an additional factor of about 10⁵⁰ has to be added to ensure this happens in a hospitable environment (a habitable planet).

5.4093...×10^178485291567

This G₄₃, the first element in the Göbel sequence G_n that is not an integer, where G_n is given by:

G₀ = G₁ = 1
G_n+1 = 1/n × (SUM_i=0..nG_i²)     (for n>1)

the sequence starts: 1, 1, 2, 3, 5, 10, 28, 154, 3520, 1551880, 267593772160, 7160642690122633501504, ... (Sloane's A003504). The "1/n" in the formula makes it look like there should be fractional terms, but the sequence doesn't actually have any fractional terms until the 43^rd term.

10^8×1016

The highest value defined within the counting system set out by Archimedes in "The Sand Reckoner". See here and here. See also 10⁶³.

10¹⁰¹⁸

A very rough estimate of the number of possible life-experiences a person can have. This is based on a sensory bandwidth of 10¹⁰ bits per second.

10^{6000000000000000000} = 10^6×1018

In 2003 Y. Cheng showed that there is a prime between every pair of consecutive cubes N³ and (N+1)³ for all values of N less than 10^{2000000000000000000} (or N³ less than 10^{6000000000000000000}). Proving this for all integers seems like it ought to be easy, but it isn't. See also 1.3063778838.

10^{1.0979465941272×1019} = 2³⁴¹

The largest number that can be formed from the digits 1, 2, 3 and 4 using the ordinary functions addition, multiplication and/or exponents. See also 163, 10460353203, 4.281248×10^369693099, 10^{8.0723047260281×10153} and 10^{(2.62086×106989)}.

10^{3.0056206947796095239×1029} = (27!)!

Very large factorials like this one can be computed with Stirling's series, a more accurate form of the better-known "Stirling's formula". The series gives a value for the logarithm of the Gamma function.

The Gamma function

The Gamma function comes up in lots of different places in mathematics, and is defined in terms of an integral¹⁰. For positive integers, the value of the Gamma function is equal to the factorial of the integer plus 1.

The Gamma function can be computed by the following series (which gives its logarithm)

ln gamma(z) = 1/2 ln(2 π) + (z + 1/2) ln z - z + SUM_n=1...inf [ B_2n / (2 n (2n-1) z^(2n-1) ) ]
   = 1/2 ln(2 π)
   + (z + 1/2) ln z
   - z
   + 1/(12 z)
   - 1/(360 z^3)
   + 1/(1260 z^5)
   - ...

where B_2n is a Bernoulli number.

The Barnes' G-function has a similar relationship to the superfactorials, as does the K-function to the hyperfactorials.

10¹⁰³⁶

The approximate odds against a person living at least 1000 years, as given by life insurance tables quoted by William Feller, in "Probability Theory and its Applications". (The tables don't actually go up that far; they simply give an extrapolation formula for ages above a certain point.)

10^{37218383881977644441306597687849648128} = 10^7×2122 ≅ 10^3.7218×1037

This number is described in the Mahayana Buddhist scripture Buddha-avatamsaka-nama-vaipulya-sutra (Flower Garland Sutra of Great Universal Buddha, in book 30, "the Incalculable"), which dates from about 420 CE. In Japanese its name is pronounced hukasetsuhukasetsuten (ふかせつふかせつてん). ⁵⁵,⁵⁶,⁵⁷.

See also 10⁴²¹.

5.45431...×10^{51217599719369681875006054625051616349} ≅ 10^{(5.1217599719369×1037)} = 2^{170141183460469231731687303715884105727}-1 = 2^(2127-1)-1

This is C₅ in Catalan's sequence and conjectured to be prime. It's a little too big to test.

2.604233075698...×10^{634704607339355474571695927232512278791} ≅ 10^{(6.3470460733936×1038)} = 27²⁷²⁷

This is 27²⁷²⁷ calculated to 50 significant digits with Hypercalc. It has over 10³⁸ digits, which is enough to pretty much guarantee that we will never find out, for example, whether its digits include a run of 40 consecutive 0's. Nevertheless, it is quite easy to figure out its first and last digits. The initial digits are found using logarithms: The logarithm to base 10 of 27²⁷²⁷ is log₁₀27×27²⁷, quite easily calculated to 50 decimal places as 634704607339355474571695927232512278791.41567985046... The integer part (to the left of the decimal point) tells us what power of 10 it has, and the fractional part (.41567985046...) tells us that the first few digits are 26042...

Perhaps more surprising, the last digits can be calculated by "modulo arithmetic". Modulo arithmetic exploits repeating patterns such as the alternating 125/625 in successive powers of 5. Modulo arithmetic shows that the last five digits of 27²⁷²⁷ are 03683: 27²⁷²⁷ mod 100000 = 27^{(2727 mod 5000)} mod 100000 = 27²⁸⁰³ mod 100000 = 03683.

By extending this method recursively (by the method described here) it can be shown that 27²⁷ ends in 9892803, 27²⁷²⁷ ends in 0403683, 27²⁷²⁷²⁷ ends in 7450083, 27^27272727 ends in 1242083, 27^2727272727 ends in 7002083, 27^272727272727 ends in 9802083, and all higher power towers of 27's end in 3802083. Each time you add another 27 to the power tower, another final digit becomes constant.

Also, because 27 is a factor of 999 we know that if we add the digits of 27²⁷²⁷ in groups of 3 the result will also be a multiple of 27.

Did I mention that I like the number 27?

10¹⁰⁵⁰

An estimate of the number of possible chess games, given by G. H. Hardy ("Ramanujan: Twelve Lectures on Subjects Suggested by His Life and Work", 1999). See also 1.15x10⁴² and 10¹²⁰.

1.000000...×10^{1.000000...×10100} = 10¹⁰¹⁰⁰

googolplex, for many people is the largest number with a name. Like googol it is a number that was "invented" just for the purpose of being large. As described under the googol entry, credit for the invention of the -plex suffix is indeterminate. See this great web page for more about this number.

10¹⁰¹⁰⁰ + 1

googolplex plus one. This number is known to not be prime, with the smallest known factor being 316912650057057350374175801344000001 = 2¹⁰⁴5⁶+1. Several other larger prime factors are known. Factors of many numbers of the form googolplex+n for small n are listed here.

Results like these are found using modulo arithmetic, similar to my description of how to find the last few digits of 27²⁷²⁷. See [Crandall 1997].

10^{(9.9565705518098×10101)}

Factorial of a googol. Notice how this appears to be only "a little larger" than googolplex.

10^{(3.4677786443...×10130)} = 27^86!

This is an example of a calculation that can be performed easily either with Maxima⁵ or Hypercalc. Both require a special command or syntax to get a full precision exponent. In Maxima:
(%i1) fpprec:200; (%o1) 200 (%i2) bfloat(27)^bfloat(86!); (%o2) 9.280229734930337461606281538723272714819546428444775230280\ 92521221347000632381451812127561138285074397038054986794948\ 96418628777914662492770386296874660782494654863382353322231\ 879854474444076252596714 b 34677786443012627135962232742326\ 49403369243699465867529793329082954277942846467082832165998\ 5543139375621253417161850434734823447802
In Hypercalc:
C1 = 27^(86!) R1 = 10 ^ ( 3.4677786443013 x 10 ^ 130 ) C2 = scale=150 Note: Factorial will only give 20 digits of accuracy. C2 = c1 C1: 27^(86!) C3 = format=1 4.2221364089683910967 x 10 ^ 3467778644301262713584883219782046054843086\ 208195414740688065133320263642461739090290922141022702407615373695372048\ 2181591037518137
Maxima, which is based on the original MIT MACSYMA, performs many of the same functions as the commercial programs Maple, Mathematica and MATLAB, but is free and open-source. It can do exact integer calculations up to about 10^1000000 and floating-point up to about 10^101000000.

10^{8.0723047260281×10153} = 4⁴⁴⁴

The highest value achievable in the "four 4's" puzzle. (This is the puzzle that asks you what numbers you can make using four 4's and the common operations on a calculator, for example, 1=⁴⁴/₄₄, 2=⁴/₄+⁴/₄, 3=⁴⁺⁴⁺⁴/₄, 4=⁴√(⁴/₄)×4, etc.) See also 10^{1.0979465941272×1019}.

10¹⁰¹⁶⁶

A somewhat lower estimate of the number of possible universe histories given by Dave L. Renfro and calculated by a different method (he estimates the Planck-unit volume of the universe at 10¹²³, the number of particles at 10⁸⁰ and the universe age at 10⁴¹ times an "interval" if 10^-24 seconds).³⁹ See also 10^3.67×10281 and 10^5.5×10405.

10^3.67×10281

A highly-simplified formula to compute the number of possible universes. N = e^{v n} where N is the number of possible universes, n is the number of fundamental particles in the universe and v is the number of particles that could be fit in the universe if it were packed full of particles.

10^5.5×10405

The factorial of the single-perturbation count, a highly theoretical estimate of the number of different ways all the particles in the known universe could be randomly shuffled at each moment in time since the universe's creation. In quantum mechanics, it is the number of universe timeline wave-functions that exist simultaneously from the viewpoint of an observer outside our universe. Another way of saying the same thing is that, if the universe is being created over and over again, it would take (on average) this many repetitions before one would expect to get an exact recurrance of "our" universe. See also 10¹⁰¹⁶⁶, 10^3.67×10281 and 10^{101.51×103883775501690}.

10^{(6.6556570552...×10668)}

This is (10⁶⁶⁶)!, called the leviathan number. The word leviathan refers to a whale or sea-monster. Its Biblical references are all in the Old Testament, although 666 is more commonly associated with the last book of the New Testament.

10^{(4.64286499063×101102)} = 2^23657×203×10²³⁶⁵⁸

The number of lines of text one Adam Clarkson will owe his high school chemistry teacher on September the 22^nd 2008. The story runs as follows¹⁶:

In February (on the 26^th, to be precise, since it's a good day to celebrate, num' sein?) of 1998, a chemistry teacher gave a set of lines to one of his students, Adam Clarkson. The lines read, "I must always tuck my shirt in whilst participating in a Chemistry Lesson". He had to do a hundred of them. However, the clever part was that if he didn't do them by the next day, they would double, and if they weren't done by the day after that, they would double again. We pointed out that the lines would become too great to do pretty soon, but this didn't stop the teacher giving them. [On the 17^th] September 1998, 7 months late, due to a confused and old man apparently mishearing us (we tried to inform him of the vastness of the Lynz, but he didn't want to hear, and so he said "If they're not on my desk by tomorrow, they'll square!"), the lines were squared every day they weren't done. To this day, they still haven't been done.

Since the 17^th of September was 203 days after the 26^th of February, the lines had been doubled 203 times — on that day the assignment was 100×2²⁰³ ≅ 1.2855504354...×10⁶³ lines, a little over one vigintillion, and due the next day. Each day after that the number is squared, so the assignment on the 18^th of September was (100×2²⁰³)² = 100²×2^2×203, about 1.65×10¹²⁶. The day after that it was 100⁴×2^4×203, or about 2.73×10²⁵², and so forth:

So, on the 22^nd September 2008 the number of lines due will be 2^23657×203×10²³⁶⁵⁸, which is about 10^{(4.64286499063×101102)}.

The length of the assignment, as the author notes¹⁷, is "somewhat larger" than the just-mentioned count of the number of ways the universe's history could be shuffled. It is so large that if you tried to write the number of lines as a normal decimal number (that is, without using scientific notation) you'd be writing a number over 10¹¹⁰² digits long, a feat which could not be accomplished even if you could fit a googol digits on each particle in the observable universe.

This real-life story invokes a similar respect for the innumeracy of common people to that described in the ancient chess legend.

10^{(2.62086...×106989)} = .3^-(.2-(.1-4)) = (10/3)⁵¹⁰⁴

If decimal points are allowed in the digits 1 2 3 4 problem, this is the result. The subexpression .1^-4 is equivalent to 10⁴ = 10000; the subexpression .2^-(.1-4) is equivalent to .2^-10000 = 5¹⁰⁰⁰⁰ = 5.01237274958×10⁶⁹⁸⁹; similarly .3^-x is equivalent to 3.3333...^x. The idea for this was sent to me by Jim Denton (although his answer, 3^.2-(.1-4), was slightly smaller).

10^{1.55×104342944819032} = e^e1013

This is the value of the "inflation factor" in a model of the inflationary universe developed by Dr. Thanu Padmanabhan, resulting from the assumption that the cosmological constant lambda equals approximately 10^-8, a value arising from grand unification theories.³⁸

10^{3.5536897484442191...×108852142197543270606106100452735038} ≅ e^ee79 ≅ 10¹⁰¹⁰³⁴

The original (higher) value of the first (Riemann hypothesis true) Skewes number. It is normally written as "10¹⁰¹⁰³⁴". It was later reduced to e^e27/4, which is "merely" 8.1847946207224960623437×10³⁷⁰. In 2005 numerical techniques were used to determine the actual value of the crossover, 1.397162914×10³¹⁶.

10^1010100

Called "googolplexplex", one of many number names formed by extension of the -plex suffix. See also 200¹⁰⁰ and 10¹⁰¹⁰⁰.

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