Notable Properties of Specific Numbers
This is 2 to the power of the reciprocal fine-structure constant 137.0359... using the CODATA 2014 value of the latter. It is a simple use of the popular fine structure constant to produce a value close to the Dirac ratio 1040. See also 3.377×1038.
1.15868...×1042 = 64! / (32!×8!2×2!4×24)
This is (a corrected value for) the number of possible chess positions, originally given by Shannon in the 1950 article "Programming a Computer for Playing Chess." (Phil. Mag. 41, 256-275). The formula is based on the idea that you can theoretically arrange all 32 pieces in any position whatsoever (giving 64!/32!) but that all pawns of a given color are equivalent (8! for each color), as is each pair of rooks (22) and each pair of knights (another 22); the bishops are not interchangeable but each has only 32 squares to choose from (24). However, this is inaccurate for a number of reasons. First and most important, a pawn cannot switch columns (ranks), or move past the opposing pawn in its rank, unless it captures. The more captures take place, the more flexibility the pawns have, but that decreases the number of pieces which decreases the number of board positions. Also, the possibility of pawn promotion increases the number of combinations somewhat. A far better estimate is that by John Tromp.
20988936657440586486151264256610222593863921 = (2148+1)/17 ~= 2.098893665744×1043
In July 1951 Ferrier found this 44-digit prime using a mechanical desk calculator. It became the largest-known prime, breaking the record set by Lucas in 1876. This record did not stand long; it was broken by Miller and Wheeler in the same month. 34
The value of the number called zài in Chinese. See also 104096.
393050634124102232869567034555427371542904832 ~= 3.9305×1044
This is 141×2141+1, the smallest number of the form n2n+1 that is prime. Cullen (the same one after whom the Cullen numbers are named) investigated numbers of this form in 1905.
824792557184288824246737061810550733633916929 = 3×(7×392-1)/2 ≈ 8.247925...×1044
This is a lower bound found by Milton Green for the value of BB(8), where BB(n) is the busy beaver function.
7.4011968415649×1045 = 7!×36 × 24! × 24!/246 = 7401196841564901869874093974498574336000000000
(The 4x4x4 Rubik's cube)
The number of ways to arrange a 4×4×4 Rubik's Cube. The corner cubelets have the same number of combinations as the 2×2×2 cube (see 3674160). There are 24 edge pieces, which can be put in any of the 24!≈6.2×1023 permutations. There are 24 center pieces these would have 24! permutations, except for the fact that each of the four pieces of a given color are indistinguishable from each other; so there are 24!/246 combinations for those pieces.
Randall Munroe made this estimate of the number of "meaningfully different" 140-character Twitter messages in English, using Shannon's estimate of roughly 1.1 bits of information per letter.
An upper bound on the number of possible chess "diagrams" (a board configuration together with the facts of whose turn it is, who still has the option of castling, and any available en passant capture), computed by John Tromp. This estimate is better than that of Will Entriken and far better than mine.
2054221614063184107682218077003539824552559296000 = 29×35×53×72×112×132×172×19×23×29×31×37×41×43×47×53×59×61×79×83×89×97×101 ≈ 2.054×1048
An upper bound on the number of possible chess "diagrams" (a board configuration together with the facts of whose turn it is, who still has the option of castling, and any available en passant capture), computed by Will Entriken in 2006. This estimate is more carefully considered than mine; the estimate by John Tromp is better still.
(chess diagrams, by my estimate)
This is a simple upper bound on the number of possible chess diagrams ("positions" together with the knowledge of whose turn it is, for whom castling is still permitted, and where en passant might occur). It is computed in a similar manner to Shannon's estimate of 1.15868...×1042. It allows between 2 and 32 pieces in play, with no more than 16 of one color, including exactly one king of each color, and up to 8 pawns of each color (any of which might have been promoted to another piece). It is higher than Shannon's estimate because it allows pawn promotion, but is unrealistic because (among other reasons) one cannot promote all pawns without first capturing some other pieces.
The length of a (Julian) year in Planck units. This is also the length of a light year in Planck length units. This can be used to convert the universe's age and size to Planck units. (Neither the age nor the size is known to sufficient precision for the discrepancy between the Julian year and other years, such as the tropical year, to make any difference.)
See also 5878625373183.6.
808017424794512875886459904961710757005754368000000000 = 246 × 320 × 59 × 76 × 112 × 133 × 17 × 19 × 23 × 29 × 31 × 41 × 47 × 59 × 71 ≈ 8.08...×1053
(the Monster group)
This is the "order" (number of elements) in the largest sporadic finite simple group, called the "Monster group" or the Fischer-Griess group.
(Some background: A "group" can be visualized as a set of transformations, e.g. rotations and reflections, that belong to an N-dimensional geometric structure such as a crystal lattice, or Rubik's Cube. A "simple" group has no "subgroups", which are subsets that themselves form a group; a "sporadic" group is one that does not fit into one of the infinite classes (cyclic, alternating, and Lie).)
See also 196883.
There are approximately 1057 neutrons in a neutron star. Neutron stars (by current estimates) range anywhere from 1.35 to 2.1 times the mass of the sun. The mass of the Sun is about 1.99×1030 kg, and the number of neutrons in a kilogram of neutrons is about 1000×6.02×1023, so the number of neutrons in a neutron star ranges between about 8.1×1056 and 1.26×1057. The upper and lower mass limits are a bit uncertain, so we can safely just call it "1057 plus or minus 20 percent".
Since a neutron star is made up (almost) entirely of neutrons, and about half of these were protons that have combined with electrons via a nuclear interaction similar to electron capture, one could think of a neutron star as being the nucleus of element number N, where N is anywhere in this range near 1057. Using systematic element names, we could whimsically say that a "nucleus" of
unniltripenthexhexoctuntriunquadunpentennbihexpenttrinilquad- hextritrisepttriseptoctniloctbiunseptununoctpentbihexquadhex- octbibihexnilseptseptennocthexenntrioctquadquadnilnilunium
is a neutron star.
Another large number that appears in the Lotus sutra texts of Mahayana Buddhism, where it appears as the word A-so-gi (あそぎ). See also 1011.
(age of the Universe in Planck units)
An approximate value for the age of the universe in Planck time units:
For various reasons, this number is not equal to the "radius", nor is it exactly 1/3 the radius of the visible universe. However, for rough calculations of things like the current volume and space-time volume, and particularly for larger derived values like the number of alternate universes, it is more than adequate.
See also 1040.
(radius of the visible Universe in Planck units)
An estimate of the radius of the "visible" universe in Planck length units. This is 47000000000 × 9.46×1015 / 1.616252×10-35. This accounts for changes in the rate of the universe's expansion, and the amount of its "curvature", according to the Lambda-CDM model.
Archimedes, in his writing psammites (better known as The Sand Reckoner), estimated the size of the universe according to the heliocentric model of Artistarchus, and how many grains of sand would fit in it. He arrived at a value equivalent to one vigintillion, or 1063. Even more impressive, he described a system of numbers extending as high as 108×1016. Curiously, the number of protons in those 1063 grains of sand is very nearly equal to the number of protons in the visible universe (the Eddington Number).
The word vigintillion is one of the number-names that had to be extrapolated by others based on the names established by Chuquet, and one of the few that appear in almost every dictionary; see this discussion, and see also 1033, 10303, 103003 and 103000003.
(Personal: For a while during 3rd grade this was the largest number I knew and on a few occasions I wrote it in the sand during recess: 1,000,000,000...,000,000. (counting out 21 sets of zeros). A mean kid would follow and wipe it out.)
The "alphabetically last prime" found by Knuth and Miller's fall 1980 CS 204 class at Stanford. Its name is American style English (without using the word "and" anywhere) is "two vigintillion two undecillion two trillion two thousand two hundred ninety three". With commas the number is 2,000,000,000,000,000,000,000,000,002,000,000,000,000,000,000,000,002,000,000,002,293.
1.0066961655...×1068 = 20!×319 × 30!×229 / 4
(the "Megaminx" puzzle)
This is the number of ways to arrange the pieces on a Megaminx, a dodecahedral puzzle similar in concept to Rubik's Cube. The reasons for the number are similar to those for the 3x3x3 Rubik's cube except for the parity factor, which is the division by 4. On the Megaminx, every turn performs an even permutation on the corners and an even permutation on the edges. Therefore, the total permutation of all the corners is always even, and likewise for the edges.
2.8287094227774×1074 = 8!×37 × 12!×210 × 24! × 24!/246 × 24!/246 = 282870942277741856536180333107150328293127731985672134721536000000000000000
(5x5x5 Rubik's cube)
The number of ways to arrange a 5×5×5 Rubik's Cube. Rotation of the center cubelets is ignored because it would be invisible. The 8 corners and the 12 central edge pieces together combine for the same number of combinations as the 3×3×3 Rubik's Cube (see 4.3252×1019). There are another 24 edge pieces, which can be freely placed into any of 24!≈6.2×1023 permutations. There are 48 movable center pieces, in two groups (the ones closer to the corners, and the ones closer to the edge-centers). Each of these two groups of 24 has 24!/246 arrangements for the same reason as the group of 24 center pieces of the 4×4×4 cube (see 7.4012×1045).
This is a prime, found by Miller and Wheeler in July 1951. This discovery has the distinction of being the first time the record for largest known prime was set by electronic computer. It broke the record set by Ferrier and was soon broken by Robinson. 34
31495448272550005155211307922363110936089435829054233418732462850152371262062592 = 2×136×2256 = 2×136×2223 = 17×2260 ≈ 3.149544...×1079
The Eddington Number
According to Arthur Eddington in his book *Mathematical Theory of Relativity* (1923, London, Cambridge University press), the number of charged particles in the universe is exactly 2×136×2256. In a 1939 lecture he stated this as follows:
I believe there are 15 747 724 136 275 002 577 605 653 961 181
555 468 044 717 914 527 116 709 366 231 425 076 185 631 031 296
protons in the universe and the same number of electrons.
(Quoted in Wikipedia, "Eddington number")
136×2256 is 1.574772...×1079, and twice this number is 3.149544...×1079. That would be the number of charged particles, but also close to the number of massive particles because there are many more protons and electrons than neutrons, (most of the matter in the universe is hydrogen). Note also that photons have no mass, and other charged particles like the muon were not yet known.
This number is notable for being the largest specific integer ever thought to have a unique and tangible relationship to the physical world. (All larger numbers in physics are estimates and approximations.)
Eddington was interested in showing that the various physical constants (the speed of light, the gravitational constant, the mass of the electron, etc.) were not accidental but were determined in some way that could be computed exactly. One of these constants was the fine-structure constant
In 1923 the fine-structure constant was known poorly enough that one could surmise that it is exactly 1/136. Eddington computed the number of particles in the universe from other measurements and observations and then found a simple mathematical formula based on integers that gave the same value. (When the fine-structure constant was later found to be closer to 1/137, Eddington repeated his work to make it fit that value. This hurt Eddington's reputation as a scientist, and he was jokingly called "Arthur Adding-one" by detractors.)
Eddington was shown to be wrong on other points. Many other estimates of the number of particles in the universe have been computed, all in the range from 1078 to 1080. Here is an example (which is way too simplistic for cosmologists but shows that the Eddington number was fairly close):
r = radius of visible universe
= age of universe × speed of light
= speed of light / Hubble constant
= 1.42 × 1026 meters
volume of universe = 4/3 π r3
= 1.2 × 1079 cubic meters
average density of universe
= 3 hydroden atoms per cubic meter
(from old models that give the minimum mass of a "closed" universe)
1 hydrogen atom = 4 particles (proton + electron, a proton is 3 quarks)
number of particles = 4.8 × 1079
If you include the various massless particles (photons, gravitons, other gauge particles, perhaps neutrinos) and virtual particles, the estimates become much greater. The only estimate I have been able to find gives the density of neutrinos in the cosmic background radiation as being 107 per cubic meter, which gives a value of 1.2×1086 neutrinos in the visible universe.
The number of legal positions in Go played on a 13×13 board (a popular size for beginners learning the game), computed by John Tromp (with Gunnar Farnebäck and Michal Koucký) in 2005 ; see OEIS sequence A94777.
2350988701644575015937473074444491355637331113544175043017503412556834518909454345703125 = 553 ≈ 2.3509887016443...×1087
In high school I wrote a special program (I think it was on a TRS-80) to calculate and display exact values of large exponents. Then I created tables of powerlogs, (which are integers of the form AAB where A and B are also integers), written by hand in a notebook. This is the largest of about 30 really big numbers in that table. See also 1.0621842147×104990856845.
(photons in the universe)
Rough estimate of the number of cosmic microwave background Photons in the visible universe. This is based on a figure of 3×108 photons per cubic meter that can be computed with Planck's law based on the cosmic background radiation temperature of 2.725 K. (See also Olbers' paradox).
The "alphabetically last power of two" found by Knuth and Miller's fall 1980 CS 204 class at Stanford. Its name is American style English (without using the word "and" anywhere) is "two untrigintillion one hundred thirty five trigintillion nine hundred eighty seven novemvigintillion thirty five octovigintillion nine hundred twenty septenvigintillion nine hundred ten sexvigintillion eighty two quinvigintillion three hundred ninety five quattuorvigintillion twenty one trevigintillion seven hundred six duovigintillion one hundred sixty nine unvigintillion five hundred fifty two vigintillion one hundred fourteen novemdecillion six hundred two octodecillion seven hundred four septendecillion five hundred twenty two sexdecillion three hundred fifty six quindecillion six hundred fifty two quattuordecillion seven hundred sixty nine tredecillion nine hundred forty seven duodecillion forty one undecillion six hundred seven decillion eight hundred twenty two nonillion two hundred nineteen octillion seven hundred twenty five septillion seven hundred eighty sextillion six hundred forty quintillion five hundred fifty quadrillion twenty two trillion nine hundred sixty two billion eighty six million nine hundred thirty six thousand five hundred seventy six". With commas the number is 2,135,987,035,920,910,082,395,021,706,169,552,114,602,704,522,356,652,769,947,041,607,822,219,725,780,640,550,022,962,086,936,576.
(fundamental particles in the universe)
Some of the larger estimates of the number of particles in the known visible universe are around this value, and result from including photons, neutrinos and other invisible particles, but not the dark matter or dark energy. The actual number of particles in the universe may be much larger for example, it might be that most of the universe is beyond our event horizon (redshift horizon). See also 7×1022 and 1.1×1089.
See also 3×1023.
Most pocket calculators max out at 9.9999999...×1099, which is just below a googol (10100, see next entry). See also 9.9999999...×10999 and the computer overflow values starting with 3.4028236692093×1038.
Main article: Googol and Googolplex
The dimensionless entropy of a black hole whose mass is about 3.087x1011 solar masses (roughly half the mass of the visible matter in our Milky Way galaxy) would be about 10100. See googolplex and 27.
An estimate of the number of subatomic particles that it would take to fill all the space in the universe. (From Straight Dope)
1.5715285840103×10116 = 7! × 36 × 24!2 × (24!/4!6)4 = 157152858401024063281013959519483771508510790313968742344694684829502629887168573442107637760000000000000000000000000
(6x6x6 Rubik's cube)
The number of ways to arrange a 6×6×6 Rubik's Cube. The corner cubelets have the same number of combinations as the 2×2×2 cube (see 3674160). There are two groups of 24 edge pieces, and 4 groups of 24 center pieces. Within each of these groups of 24 there are 24!≈6.2×1023 arrangements. The center piece groups each have 4!6 fewer combinations because they are in 6 different colors (with 4 of each color) and any pieces of the same color are indistinguishable from each other.
This is over 10 million billion googol. Notice that it is only a little less than the number of ways to play a game of chess, and far greater than the number of valid chessboard positions. When making a move, you have 3×6=18 choices of what to turn and 3 choices of how far to turn it, for a total of 54 choices. This means that in order to get a reasonably good coverage of all the possible combinations, you have to make at least 67 moves to scramble the cube. By contrast, a normal 3×3×3 Rubik's Cube can be scrambled in about 15 moves.
(Chess games by Shannon's estimate)
This is the Shannon number, Claude Shannon's estimate of the number of chess games from his original 1950 paper on computer chess. It is calculated by the approximation 100040, based on the idea that at each move by White there are about √1000 ≈ 32 choices, to which Black has about 32 responses, and typical games last 40 moves. If the players aren't trying to win, however, a game can go much longer (see 5898, 8848, and 1012500).
Another example of innumeracy, this time in the opposite direction from the Rubik's cube example. On the TV program "Voyage to Pandora", and also seen in an NCSM newsletter (which, ironically, is an organization purporting to represent "Leadership in Mathematics Education"), I found two slightly-different wordings of the following passage:
A flight to Alpha Centauri (closest star to Earth, other than the sun) on a regular space shuttle would take 900 years and would require a mass of fuel greater than the mass found in the visible universe, 10137 kilograms. -- NCSM
To see how wrong this is, we can Do the MathTM. Assuming that the fuel is being burned at a constant rate over the 900 years, the burn rate would be:
10137 kg / 900 years = 4.37×10128 kg every 124 seconds
I chose 124 seconds for the final figure because this is the period of time during which the actual Space Shuttle solid rocket boosters fire. If we assume that all of the fuel in the Shuttle, tank and boosters is used during this time, and knowing that its total launch mass is 2.04×106 kg, we can make a comparison:
4.37×10128 / 2.04×106 = 2.14×10122
Consider how big the spaceship would need to be. Clearly it would be nearly all fuel. If the 10137 kg of fuel has a density like that of ordinary rocket fuel, then it would have a volume of 10137 liters, or 10134 cubic meters. A spherical fuel tank would have a radius of 2.88×1044 meters, which is about a decillion times the radius of the inner solar system (see astronomical unit) and 7×1027 times the distance of Alpha Centauri! Clearly, fitting all this fuel mass into a spaceship small enough to actually make the trip from Earth to Alpha Centauri would require a density so great that the entire thing would collapse into a black hole.
The TV program "Voyage to Pandora" also gives this factoid and credits it to a NASA study. Looking at that source, and knowing a bit about how multi-stage rockets work, the number makes a bit more sense. The number is an estimate of the weight of a Tsiolkovsky-style multistage rocket necessary to get to a speed of about 3 million miles per hour, which would be enough to get to Alpha Centauri in about 900 years. The additional assumptions are that: each stage of the multistage rocket weighs (say) twice as much as its payload (which comprises all of the stages that come after it); each stage is designed for a burn time of 500 seconds at an acceleration rate that can be endured by human passengers. At an acceleration rate of about 20 miles per hour each second, you need about 40 hours to reach a speed of 3 million miles per hour. Since each stage only lasts about 500 seconds, the multistage rocket ends up needing to have about 290 stages. If (as I stipulated before) each stage had about twice the mass of the stages above it, then the total mass of the rocket "at launch" would be about 3290 times the mass of its ultimate payload. That's where the unbelievable mass figure came from, and the reason it is so unbelievable is because the quotes (in the TV program and by NCSM) do not mention that the craft is not actually the size of "a regular space shuttle" but is a Tsiolkovsky-style exponentially-staged design with an incredible 290 stages.
In India's ancient writings there are many references to large numbers with names; some are hard to attach to a specific value because of multiple conflicting or ambiguous uses. One of the larger numbers given a name in India is asankhyeya, commonly said to be 10140. In the Knuth -yllion naming system, 10140 is one myriad myllion quintyllion; in the more mainstream Conway-Wechsler system, it is one hundred quinquadragintillion. See also 10421 and 103.7218×1037.
This is the 13th Mersenne Prime and the first to be found by electronic computer. It was discovered in 1952 by Robinson and breaks the record set by Lucas in 1876, although that record was also broken by the non-Mersenne primes (2148+1)/17 and 180×(2127-1)2+1, which were found the year before. 34
1.95005511837..×10160 = 8!×37 × 12!×210 × 24!2 × (24!/246)6
(7x7x7 Rubik's cube)
The number of ways to arrange a 7×7×7 Rubik's Cube. Rotation of the center cubelets is ignored because it would be invisible. The 8 corners and the 12 central edge pieces together combine for the same number of combinations as the 3×3×3 Rubik's Cube (see 4.3252×1019). There are two additional sets of edge pieces with 24 in each set, which can be freely placed into any of 24!≈6.2×1023 permutations. There are 144 movable center pieces, in six groups (according to where they are in relation to the corners, and the edge-centers). Each of these six groups of 24 has 24!/246 arrangements for the same reason as the group of 24 center pieces of the 4×4×4 cube (see 7.4012×1045).
The number of legal positions in Go, computed by John Tromp (with Gunnar Farnebäck and Michal Koucký) in 2016 ; see OEIS sequence A94777. The exact number is 208168199381979984699478633344862770286522453884530548425639456820927419612738015378525648451698519643907259916015628128546089888314427129715319317557736620397247064840935.
There are 3(19×19) ≈ 1.74×10172 ways to place white and black stones on the board, and it is easy to test any given pattern to see if it is a legal Go position. Random sampling and heuristics had been used some time earlier to compute the estimates 2.089×10170 and 4.63×10170. Finding the exact value takes a lot more work.
An estimate of the number of legal positions in Go, computed by Achim Flammenkamp and posted to the rec.games.go USENET newsgroup on 1992 Sep 16:
I have made with Knuth's "heuristic sammple method", Math. of Comput. 1975, a new approch to get a more precise value for the number of go configurations. For a 19x19 board my best value is: 0.01197*3^361 with a varianz < 0.00001*3^361
The value 0.01197(1)×3361 is equivalent to 2.0838(17)×10170. This improved on the 1972 estimate of "10170" found in item 96 of HAKMEM72.
An estimate of the number of possible positions in Go, played on a 19×19 board. According to Wolfran Mathworld, this estimate came from "Beeler et al. 1972, Flammenkamp" but their citations are wrong. Beeler 1972 is a reference to item 96 of HAKMEM72, which indeed contains an estimate of "10170". The "Flammenkamp" reference points to this page, which as far as I know has never included any estimate of the number of legal Go positions, but it is most probably meant to refer to this value posted to rec.games.go by Achim Flammenkamp in 1992. I am presently unsure of the origin of this "4.63..." estimate.
The number of years in the longest time-period in the cosmology of Jainism, a religion and philosophy from India in the 6th century B.C. (From an article by J J O'Connor and E F Robertson)
(volume of the visible Universe in Planck units)
Brady Haran and the good folks over at Numberphile are always coming up with new challenges. One day they opened a tin of "Numberetti", a pasta product in which the bits of pasta are shaped like the digits 0 through 9. On the can is the enticing challenge, "What's the biggest number you can make?", and inside were the following 195 digestible digits:
As Landon Curt Noll's English name of a number site will readily tell you, this number in words is:
nine hundred ninety nine tresexagintillion,
nine hundred ninety nine duosexagintillion,
... (51 lines omitted) ...
one hundred eleven decillion,
one hundred nonillion
You can see Numberphile's video all about this here: spaghetti numbers.
As the video mentions, you can make a far greater number by making the digits into a big tower of exponents, like my 2345... example at 6pt1.86×103148880079. By the power-tower method, you'd be able to make the monster 161pt1017 (which has 222... at the bottom and ...1010100000000000000000 at the top). This is "a bit" shy of Moser's "Mega".
This is 27140, the number of possible 140-character Twitter messages when you're allowed just 26 letters and a blank space.
5.7324701932...×10207 = 75600000000000 × 840000028
Another number from measurement of time in Jainism. A purvi is 22×33×7×1011 = 7.56×1013 years, and a shirsha prahelika is 840000028 purvis, which works out to 5.7324701932...×10207 years.
Value of "googoc" from the Googology page. This page introduces a lot of novel number names and also summarizes a lot of other people's names for special large numbers.
The basic idea is to derive word endings, prefixes and infix parts by reverse etymology from existing names like "googol". According to that page, André Joyce noticed that googol is "googo" followed by the Roman numeral L (representing 50) and that googol=10100 is equal to 10050. He extrapolated this with the generalization that "googo" followed by any Roman numeral(s) z is (2z)z, which he chose to express as ack-h(2,ack-h(1,2,l),l) using the Herbert version of the Ackermann function. This means that "googoc" would be 200^100, "googod" is 1000500=101500, etc. In another similar generalization, anything followed by -ple- and Roman numeral(s) such as x is the result of raising x (or whatever) to the power of the thing to which -plex was added (perhaps he imagined "ple" stood for something like "placé dans l'exposant de"). Thus, for example, twoplex = two, 2, followed by -ple- and x representing 10, is 102; threepleiii is 33=27; and googodplem = 1000googod = 10001000500 = 103×101500.
(space-time volume of the visible Universe in Planck units)
The four-dimensional volume (in space + time) of the known universe using the formula for the volume of a hypercone (with spherical cross-section) and the universe's age, expressed in Planck units. The hypercone has a 4-dimensional volume given by
V = (1/4 h) (4/3 π r3)
where h is the height of the hypercone and r is the radius of the sphere that forms the hypercone's base. h is the age of the universe in Planck time units and r is its current radius in Planck length units. Due to complexities of relativity and the way the universe expands, h and r are not the same. This gives 1.75×10245 for the 4-D volume.
Real models of the universe used by astrophysicists and cosmologists are much more complex and do not admit to such a simple calculation of volume, but most models would arrive at a figure close to this one.
See also 1.41×10408.
9.1197528826...×10262 = 20!×319 × 30!×229 × 60!/(5!12) × 60!/(5!12) × 60! / 25
(the "Gigaminx" puzzle)
This is the number of ways to arrange the pieces on a Gigaminx, a dodecahedron-shaped combinatorial puzzle with 230 movable pieces. The Gigaminx is to the Megaminx as the 5x5x5 cube is to the normal 3x3x3 Rubik's Cube. This video shows one in action, and here is another..
(base of Hypercalc level-index representation)
This is the base of the level-index representation used by Hypercalc. I chose it because it is close to the limit of the IEEE 754 double type. The bit of extra space above 10300 makes addition and roundoff algorithms much simpler.
It is also the basis of my ASCII-lexicographic ordering of arbitrarily high positive numbers, used internally by the source file of this web page. Here is an illustration of the system through examples:
These ASCII strings can be used as labels, variables or function names in popular programming languages (including C, Perl and PHP) and are valid search keywords (for example, a Google search for "e023_602" finds this page).
Quick index: if you're looking for a specific number, start with whichever of these is closest: 0.065988... 1 1.618033... 3.141592... 4 12 16 21 24 29 39 46 52 64 68 89 107 137.03599... 158 231 256 365 616 714 1024 1729 4181 10080 45360 262144 1969920 73939133 4294967297 5×1011 1018 5.4×1027 1040 5.21...×1078 1.29...×10865 1040000 109152051 101036 101010100 footnotes Also, check out my large numbers and integer sequences pages.