Notable Properties of Specific Numbers
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18980 = 2^{2}×5×13×73 = 52×365
Number of days in the roughly 52year cycle of the Mayan Calendar Round. This is the least common multiple of 260=13×20 and 365=5×73. See also 5126, 550420 and 1872000.
19683 = (3^{3})^{3} = 27×27×27
This is 3_{④}3 or 3^^3 by the lower (leftassociative) version of the hyper4 operator. It is also a "googol" in base 3: 10_{3}=3 and 100_{3}=9, so "10^{100}" in base 3 is 3^{9}=19683.
19683 is the largest "Dudeney number", defined as a perfect cube whose digits add up to its cube root: 1+9+6+8+3=27 and 27^{3}=19683. There are 5 others: 1^{3}, 8^{3}, 17^{3}, 18^{3}, and 26^{3}. It is easy to see^{104} that there is an upper limit for Dudeney numbers: for example, if it had 7 digits, its cube root would have to be at least 100 (because 100^{3}=1000000) but 7 digits can't add up to any more than 7×9=63.
See also 196883 and 4.434×10^{38}.
The number of days in the "Exeligmos" or triple saros cycle. 19756 days is very nearly equal to 726 draconic months or 669 synodic months.
The Minor planet number of Varuna, a fairly large object in the Kuiper belt originally known as "2000 WR_{106}". Varuna is about as large as the Dwarf planet Ceres. The notable round number 20000 was assigned to it early in 2001, commemorating the 200^{th} anniversary of the discovery of Ceres and the recognition of the fact that Varuna was significant because of its size:
The past six weeks have witnessed, not only the discovery, but also precovery observations back to 1954 and a size and albedo measurement, of the largest cubewano found so far. At 900 km, the transneptunian object 2000 WR_{106} has a diameter that is essentially the same as that of (1) Ceres, the largest member of the cisjovian belt. The coincidence of this recognition (cf. IAUC 7554, 2001 Jan. 2) with the Ceres bicentennial makes it particularly appropriate (cf. MPC 41901, 41911, MPO 7698) that 2000 WR_{106} should be numbered (20000).^{79}
An estimate (as of Oct 2014) of the number of proteincoding genes in the genome of the nematode worm C. elegans. See 959.
The number of days in a more accurate version of the metonic cycle suggested by Adam Goucher [214]). 20819 days is very nearly equal to 57 mean tropical years or 705 synodic months.
A "selfdescribing" number, like 2020 and 3211000; see 6210001000 for more.
This is the sum in the following "mirror image equality" by Benjamin Vitale^{120}:
13×13 + 42×42 + 53×53 + 57×57 + 68×68 + 97×97 = 79×79 + 86×86 + 75×75 + 35×35 + 24×24 + 31×31
If desired, 22024 can be placed in the centre, and will be symmetrical if expressed as the sum "22000 + 2 + 00022".
e^{10} is the highest value on the LL scale of most loglog slide rules that have such a scale. (However, a few go as high as 10^{10}.)
A number that occurs 6 times in Pascal's triangle: twice each in row 17 (because 17!/(8!×9!) = 24310), row 221 (because 221×220/2 = 24310), and of course row 24310. See 3003 for more.
The prisoner number of Jean Valjean, the prisoner who is freed in the prologue of Les Misérables. If this number is appears as a prisoner number in another work (examples include The Simpsons episodes "The Principal and the Pauper" and "Black Widower", and South Park's "Cartman's Silly Hate Crime 2000"), it is in tribute to the Victor Hugo novel. See also 525600.
This is the product in a multiplication problem involving all prime digits (2, 3, 5, and 7). The problem is called a "cryptarithm" and is posed this way: All digits are prime. Fill them in. __ __ __ x __ __ """""""""""""" __ __ __ __ __ __ __ __ """""""""""""" __ __ __ __ __
The solution puts 775 on the first line, 33 on the second, 2325 on each of the partialproduct lines, and 25575 on the final line. This puzzle goes back at least as far as W. E. Buker, American Mathematical Monthly 43 (Oct 1936), was reprinted in Martin Gardner's Mathematical Games column in Feb 1962 (crediting "25 years ago by Joseph Ellis Trevor"), and later in the book "Mathematical Quickies" by Charles W. Trigg (1985).
(precession of the Earth's axis)
As of 2003, the best known value of the Platonic Year, or precession cycle, in mean solar years, according to Fukushima [177]. The error is ±0.0015 years (less than a day!). This is the time it takes for the orientation of Earth's poles to go one full circuit in its circular path. The movement is caused by the tidal influence of the Moon and Sun (and to far lesser extent, other planets like Venus and Jupiter) on the Earth, which has a gyroscopic precession effect because of the Earth's bulging a bit around the equator. The value is approximated as 25600, 25800, 25920 or 26000, depending on the purpose and source. The raw data value from which this is derived is the "general precession in longitude" in computer models, 5028.7955 ± 0.0003 arcminutes per Julian Century^{8}. A Julian Century is 36525 mean solar days, so the precession period is computed by (360 × 3600 / 5028.7955) × (36525 / 365.242189670).
An oftenseen approximation to the precession cycle.
According to those ancient astronomers who knew of the "precession of the equinoxes" (now known to be movement of the Earth's axis of rotation, see 25772.1300), the ascending and descending nodes of the Sun move 1 degree through the zodiac every 72=6×12 years. This produces a cycle of 25920 years for the entire period. This isn't too far off from the modern value. An "age" of 2160 years, ^{1}/_{12} of the full period, is also given much significance. See also 2592.
An approximation to the precession cycle.
The number of possible Tictactoe games, ignoring rotations and reflections.
See also 765.
27000 = 30×30×30 = 3^{3}×10^{3}
There is a kind of nice approximation to π: 84823/27000 = 271×313/(2×3×5)^{3} = 3.1415925925926... It's probably just a coincidence, but another larger one is 6109800817/1944810000 = 74203×82339/(2×3×5×7)^{4} = 3.1415926578946...
27000 is approximately the number of days in an expected lifetime for most readers of this page (and 75 "Babylonian" years of 360 days each). See also 405000).
This is 110110011100100 in base 2, and that sequence of 1's and 0's is the beginning of the "paperfolding sequence" A14577: 1, 1, 0, 1, 1, 0, 0, ... which defines the sequence of left and right turns in a Heighway dragon. The sequence can be generated by starting with a single 1, then repeatedly applying the steps "add a 1, then add the previous bits inverted and orderreversed". So from 1 we get 1 1 0 → 110; then 110 1 100 → 1101100; then 1101100 1 1100100 → 110110011100100, and so on. If at each step we convert the binary number to decimal we get: 1, 6, 108, 27876, 1826942052, 7846656369001524324, 144745261873314177475604083946266324068, 49254260310842419635956203183145610297351659359183114324190902443509341776996, ...
30240 = 2^{5}×3^{3}×5×7, and is a 4perfect number: Its divisors add up to exactly 4 times the number itself. Curiously, it also has the same digits, and most of the same divisors, as 40320. See also 120, 154345556085770649600 and 2.518504×10^{1906}.
A prime number, and one of many misspellings of the word "elite" ("eleet"; see table at leet) used by those who wished to hide their conversations from automatic detection on bulletinboard systems in the 1990's. This "language" possibly evolved partly from inverted calculatordisplay text (see 71077345).
See also 1337, 9001, 900901, and 5318008.
The largest perfect square of the form 2^{p}7. See 91.
This is a Leyland number because it is 15^{2}+2^{15}; and it is also prime, therefore a Leyland prime.
The width of the English Channel at its narrowest point, in meters.
See 3.1418708596056 and 137.035.
This is 187^{2}, and said to be the favourite number of Sesame Street's Count von Count.
This is 5!×4!×3!×2!×1!, or "5 superfactorial". See also 3456.
Number of days in a Julian century, see 365.25.
This is 8! (8 factorial), 8×7×6×5×4×3×2×1, and is the number of minutes in 4 weeks. It would also be the number of minutes in a month, if months were exactly 4 weeks long. I'm no revolutionary — I'm not suggesting we should change the calendar — but I do strongly believe February should be designated "International Factorial Appreciation Month", to be observed in all years except leapyears. Alas, all my pleas have fallen on deaf ears (: See also 10080,604800, 3628800 and 86400000.
In 1963 (July the 27^{th}) a set of bell ringers in Loughborough, Leicestershire, England became the first in history to ring eight tower bells in all possible permutations. Each bell was rung 40320 times, for a total of 8×8! = 322560 blows, with each different combination of the 8 notes occurring exactly once^{42}. The feat took almost 18 hours, and remains the longest performance of its type ever completed^{110}.
An approximate value (in kilometers per second) of the Earth's escape velocity^{109}. This will get you into orbit around the Sun, additional velocity is needed to escape the solar system altogether.
40585 = 4!+0!+5!+8!+5!, the largest number that is the sum of the factorials of its own digits^{103}. It is called a factorion and was discovered in 1964 by Leigh Janes.^{102} The only others in base 10 are 1, 2 and 145.
This is a square and also a triangular number: 41616 = 204^{2} = 288×289/2. Such numbers are fairly rare: 1, 36, 1225, 41616, 1413721, 48024900, 1631432881, ... (Sloane's A1110). See 204 for more.
Often related to 86400, 432000, etc. by people studying ancient and/or mystical things. Also, it has been noted that the great pyramid in Egypt has a perimeter of 3023.16 feet at its base, and estimated height (before erosion) of about 481.40 feet. If you multiply these by 43200 you get (almost exactly) the circumference of the Earth and its polar radius. See also 432, 432000, 4320000
The number of days Jeanne Calment lived; see 122.
45360 has 100 factors (including 1 and itself); no smaller number has as many. Its prime factorization is 2^{4}×3^{4}×5×7. To get one of the 100 factors of 45360, pick one number from each column in this table:

and multiply them together to get a factor. Since you have 5 choices in each of the first two columns and 2 choices in the other two columns, the total number of choices you can make is 5×5×2×2=100. Notice that these numbers (5,5,2,2) are one more than the exponents in the prime factorization of 45360 (2^{4}3^{4}5^{1}7^{1}). You can count the number of factors of any number just by taking the exponents in its prime factorization, adding one to each, and multiplying together. For 45360, the exponents are (4,4,1,1). Any other number with a prime factorization with two 4's and two 1's for exponents will have the same number of factors, including 2×3^{4}×5^{4}×13 = 1316250 and 2×5×7^{4}×11^{4} = 351530410 and so on. Based on this, you can see that:
 If N sets a record for most factors, no prime numbers are skipped in its prime factorization. It's easy to see why: if (for example) 2^{4}×3^{2}×5×11 were a recordsetter, then the smaller 2^{4}×3^{2}×5×7 would set the same record.
 Similarly, the exponents in the prime factorization of a recordsetter get smaller but don't get larger. For example, 2^{4}×3×5^{2}×7 definitely isn't a recordsetter because the smaller 2^{4}×3^{2}×5×7 would set the same record.
For more about factors recordsetters, see 12, 60, 360, 840, 1260, 720720, 3603600, 245044800, 278914005382139703576000, 2054221614063184107682218077003539824552559296000 and 457936×10^{917}.
48625 = 4^{5}+8^{2}+6^{6}+2^{8}+5^{4}, the sum of its digits raised to the power of the same digits reversed in order. (from Matt Parker). See also 3435.
The Minor planet number of Quaoar, a fairly large object in the Kuiper belt originally known as "2002 LM_{60}". Quaoar is about as large as the Dwarf planet Ceres and is notable for having an orbit that is much more circular and much less inclined than that of Pluto, but of similar size (orbiting in 289 years, versus 248 for Pluto). The notable round number 50000 was assigned to it in late 2002, apparently for similar reasons as the assignment of 20000 to Varuna. See also 134340.
The highest value of k for which 2^{k1} + k is known to be prime, or is considered a "probable prime". The other lower values of k are 1, 3, 7, 237 and 1885. There ought to be about log(N) such k values for k < N, but since 2^{k} grows so quickly, it is very difficult to find more. It also happens to be difficult to prove whether there is or is not an infinite number of such k's.
This is 640320/12 and appears in the Chudnovsky series approximation of pi.
The number of astronomical units in a light year, calculated using the IAU value of the former.. In my old Macintosh space flight program Orion, I used 2^{16} as an approximation to this. The number of inches per mile (63360) is a closer approximation, giving folks in the United States an easy way to visualise the size of a light year. See the main light year entry and also 1.609344.
The number of inches in a mile: 12×5280. Its factorization is 2^{7}×3^{2}×5×11. (More about its origins at 5280.)
65504 = 2^{15} × ^{2047}/_{1024}
The largest finite value that can be represented in the 16bit floatingpoint format "s10e5", also called "half" or "fp16", used in professional computer graphics. This format uses all the IEEE754 rules with a 5bit exponent and 10bit mantissa. (More details here.)
This is 2^{16} = 2^{24} = 2^{222}, which is 2 ^^ 4 usingthe hyper4 operator. Also, since 2 ^^ 4 is 2 ^^ (2 ^^ 2), 65536 is 2 ^^^ 3 where ^^^ isthe hyper5 operator.
Appears in Zork and several later Infocom adventure games:
On the ground is a pile of leaves.
> count leaves
There are 69,105 leaves here.
The usage of the number probably originated from the MIT hacker culture^{74}; see 69.
In Through the LookingGlass by Lewis Carroll, the White Queen says that her age is 101 years, 5 months and a day. If the date is chosen carefully, this can work out to be 37,044 days, and if the Red Queen were born on the same day, their total would be 74,088 which is 42^{3}.
This number is all 1's and 0's when written in base 2, 3, 4, and 5: 82000_{10} = 10111000_{5} = 110001100_{4} = 11011111001_{3} = 10100000001010000_{2}. It has been known at least since 2008 (when Daniel Mondot noticed the number's properties and submitted sequence A146025 to the OEIS), and is discussed in a Numberphile video "why 82000 is extraordinary". Larger numbers with the property probably don't exist — it has been checked as far as 10^{1011}, and the huge number of digits needed makes the probability of another exceedingly low. That means there can't be such a number for bases 2 through 6 or beyond.
82944 = 2^{10}×3^{4}, (8×9×4)^{2}, or 8^{2}×9^{2}×4^{2}, the square of the product of its oddnumbered digits. Only 1 and 784=7^{2}×4^{2} share this property. If you multiply these three (1×784×82944) the answer is ((1×2×3×4×5×6×7×8×9)/(1+2+3+4+5+6+7+8+9))^{2}.
This led one D. G. Leahy to link 82944 and other cult numbers (including 37 and 666) to an old approximation of the fine structure constant.
The number of seconds in a day.
86400 = 1×2×3×4×5×6×5×4×3×2×1; see 3628800 (Source: Stephen Swanson).
In Biblical time, the number of weeks (approximately) in the time before the Flood. The Bible lists ten people during this period, of whom Adam and Noah are the 1^{st} and 10^{th}, and gives the ages of each when the next was born. According to the oldest known version of that account (which is in the Hebrew Torah and most Christian Bibles but differs from the Septuagint and Samaratan versions) the number of years before the flood was 1656 (Genesis5:3 through 5:29 and 7:11: Adam 130; Seth 105; Enos/Enosh 90; Cainon/Kenan 70; Mahalalel 65; Jared 162; Enoch 65; Methuselah 187; Lamech 182; Noah 600); this is almost exactly 86400 weeks (you have to fudge the math a bit: assume every 23 years has 5 leap years; then every 23 years would be exactly 1200 weeks; 1656=23×72 so there are 1200×72=86400 weeks.
There is also a link to 432000: since the Babylonian calendar had 5day weeks, 86400 weeks would be 432000 days in the Babylonian calendar. Under 432000 there is a description of the Babylonian flood story (Gilgamesh) and a period of 432000 years, which (as in the Biblical version) is expressed in terms of ten people who lived during the period prior to the flood.
The Babylonian story is older and it is generally acknowledged (by anthropologists, theologically neutral historians and such) that the Bible version is descended from it. It has been proposed that the sum was preserved and transformed, as the story traveled from the Babylonian to the Hebrew culture, in the following way: 432000 years became 432000 days, which then became 86400 5day weeks, which was then reinterpreted as 86400 7day weeks, which became 1656 years. This explanation works well because only the last step need be approximate, and the exact answer (86400 days = 1655.84 years) rounds up to 1656.
See also 31556952.
mean solar day
The number of SI seconds in a mean solar day is a little bit more than it should be (24×60×60 = 86400) because the speed of rotation of the Earth is gradually slowing down, and the SI second was calibrated at a time when Earth rotated faster. During the years 1995 to 2012, the number of SI seconds in the mean solar day varied between 86400.000 and 86400.002.
Over very long timescales, the two major causes of changes in the length of the day are Tidal acceleration and Postglacial rebound, buth described in the tropical year entry.
Recorded observations of occultations (when the moon passes in front of a star) enable us to know the precise speed of the Earth's rotation back to the early 1700's. Over that period, the length of the day has varied between 86399.997 and 86400.004 seconds (with the lowest value in 1867, and the highest value only 45 years later, in 1912). These may be caused by the "weather" in the Earth's core, as well as earthquakes, and longterm changes in ocean currents and surface weather. A related effect is the shifting of the poles (Chandler wobble).
During the past 50 years, atomic clocks allow us to measure much more shortterm effects^{49}. There are very fast (from one week or month to the next) somewhat randomlooking variations caused by air and water currents; these typically have a magnitude of less than a millisecond^{125}.
A unit of the (Asian) Indian number name system. It is called lakh when needed (primarily in Indian dialects of written English), a name that derives from Sanskrit laksha. See also 10000 and 10000000.
See 897612484786617600.
The number of days in a more accurate version of the metonic cycle described by William Williams [131]). 121991 days is very nearly equal to 334 mean tropical years or 4131 synodic months.
The number of days in 360 Babylonian "years" (see Prophetic Year); see also 453600.
The Minor planet number for Pluto. Like Ceres and many of the asteroids, Pluto was considered a planet for many years until it was discovered that there are many other bodies in similar orbits. After it was reclassified as a Dwarf planet, Pluto was given a number in the sequence used for the asteroids. Ceres' number is 1.
See also 20000, 50000 and 134340.
The number of days from a given day in the year 1601 to the same day in 1970. This is relevant to computer date conversion, see 11644473600.
This number has a property which exemplifies some of the many, obscure and somewhat arbitrary investigations into number theory that can be explored by anyone with the interest (and perhaps a personal computer). 140800 has the property that when expressed in each of 8 bases, the product of its digits equals the base times the sum of its prime factors. (The bases are all huge, and range from 198 to 2720.) There are many other similar (and some even more obscure) bits of number theory discussed on the Internet on Kevin Brown's math pages. See also 239.
This is the 6 digits after the decimal point in the fraction 1/7 (see 7). The decimal expansions of the fractions ^{1}/_{7}, ^{2}/_{7}, ^{3}/_{7}, ^{4}/_{7}, ^{5}/_{7}, and ^{6}/_{7} all consist of the digits 142857 repeated, and arranged in different ways. The multiples of 142857 are similar: 142857, 285714, 428571, 571428, 714285, and 857142 (see also 456790123).
Notice the factorization: add 7 and you get 3^{3}×7×11×13×37 = 999999, which is of course related to the facts that 3^{3}×37=999 and 7×11×13 = 1001.
A Kaprekar number: 142857^{2}=20408122449, and 20408+122449=142857.
The number of days in a more accurate version of the saros cycle found by Hipparchos (according to Ptolemy, as related by John Narrien [132]). 161178 days is very nearly equal to 5923 draconic months or 5458 synodic months.
175560 is a rather remarkable example of a product of two nonoverlapping sets of consecutive integers: 19×20×21×22 = 19×2^{2}×5×3×7×2×11 = 5×11×2^{3}×7×3×19 = 55×56×57. See 720 for other examples; see also 17297280 and Sequence A064224.
186282.39705122... = 299792458 / 1609.344
The speed of light in miles per second.
(the Leech lattice)
The number of vertices that define the Leech lattice in 24 dimensions. This lattice is the basis of a sphere packing in 24 dimensions which has been proven to be the best possible. It is also closely related to the Monster group, 26dimensional Lorentzian spacetime and bosonic string theory, and the fact that 1^{2}+2^{2}+3^{2}+...+24^{2}=70^{2} (see 4900). See also 8315553613086720000.
Richard Borcherds won the Fields Medal (the math equivalent of the Nobel prize) in 1998 for his proof of an amazing coincidence involving this number.
Some years after the Monster group was discovered, it was shown that the minimum number of dimensions of a crystal lattice whose symmetry rotations and reflections form the Monster group is 196883. It was then noticed that this is only one less than the number 196884 that occurs in the elliptic modular function responsible for the Ramanujan constant and connected to the proof of Fermat's Last Theorem.
There was much speculation about whether these two adjacent integers 196883 and 196884 appear for a reason (that is to say, is there a connection between the Monster group lattice symmetry and the FLTrelated elliptic functions?). It turns out the answer is yes — and for showing this Borcherds won the Fields Medal.^{82}
See also 19683.
A term in a polynomial that comes up in relation to the socalled Ramanujan constants. See also 196883.
This is 60^{3}+12^{3}, more artistically expressed as (3×4×5)^{3}+(3+4+5)^{3}. See also 134217728.
1260 years is approximately the lifetime of an eclipse in a saros series.
Numerology knows no limits. While researching 217728 I found the following false claim: "In Revelation Chapter 12 there are five numbers; 12, 24, 3, 1296, and 2, which when multiplied together total 217728." 217728 is actually 12×24×3×126×2, so at first it looks like the numerologist missed the 9 key on his calculator as he was entering 1296. (Not hard to believe, as he goes on to say that "217728 also equals 81 times 2268"; it is actually 81×2688.) But, if you actually read Rev 12, you find that the numbers mentioned there are (in order) 12, 7, 10, 7, ^{1}/_{3}, 1260, 2; there is also a "half" in Rev 12:14. You may or may not choose to count the fractions; our numerologist seems to have counted "a third" in Rev 12:4 as a 3. Apparently he decided to combine the 7+10+7 into 24 (perhaps because they occur in the same verse), and dropped the 0 on 1260 to make 126. That would at least make the product equal 217728, but the basic claim still isn't satisfied. If he had done it "right" his product would have to be 12×7×10×7×(^{1}/_{3})×1260×2 = 4939200, or 14817600 if you don't count the ^{1}/_{3} as a number, or 44452800 if you count it as a 3, and various other non217728 values if you add a ^{1}/_{2} or 3^{1}/_{2} from Rev 12:14 (see 1260 for more on Rev 12:14).
For more about numerology, see 666.
246924 = 19^{3}+(2×19)^{3}+(3×19)^{3}
Someone recently searched for this number and found this page — even though I did not write anything about it (until now). We aim to please here at Munafo Robotics, so I added this entry.
The question as encountered by the reader was:
If a, b, and c are in the ratio 1:2:3, and a^{3}+b^{3}+c^{3}=246924, what is the value of c?
We can solve this type of problem directly by noting that (for example) if you double a, the values of b and c have to double as well, and then all the cubes go up by a factor of 8 so the sum of cubes also goes up by 8. In other words, we can backsolve from the sum of cubes and the known ratios:
a^{3}+b^{3}+c^{3} = 246924
a^{3}+(2a)^{3}+(3a)^{3} = 246924
36 a^{3} = 246924
a^{3} = 246924/36 = 6859
a = 19
c = 3a = 57
(4 hyperfactorial)
This is another way to define an exponentbased factorial function (the bestknown version is the hyperfactorial). This "powertower factorial" grows much faster than hyperfactorials because the height of the "tower of exponents" keeps growing: 1, 2^{1}=2, 3^{21}=9, 4^{321}=262144, 5^{4321} = 6.206069882×10^{183230}, 6^{54321} = 10^{(4.829261049×10183230)}, ... (however, Pickover's superfactorial grows a lot faster).
Expressed as "4^{32}", this number is a pair with 2^{34} in two different ways: using the digits 2, 3 and 4; and using the digits 1, 2, and 8 (because 4^{32} = 2^{18} and 2^{34} = 2^{81}). See also 2.417851...×10^{24}, 10^{1.0979×1019}, and 6pt1.86×10^{3148880079}.
The mass of the Sun divided by that of the Earth, according to NASA in 2013. See also 354710, 1304000, and 1.32712442099(10)×10^{20}.
The solar mass ratio (ratio between the mass of the Sun and that of the Earth) in the 1809 calculations of Carl Friedrich Gauss; see 0.01720209895 and 365.2568983.
9 factorial, 1×2×3×4×5×6×7×8×9, is the number of ways to arrange the 9 digits 1 through 9. This is the number of solutions to a single 3x3 section of a Sudoku puzzle (see 3546146300288 and 6.6709×10^{21}).
The first of a set of 111 consecutive composite numbers: Every number from 370262 through 370372 is composite. This is the first time there are 100 or more composite numbers in a row. The preceding prime, 370261, and the following prime 370373, together set a record for largest gap between primes. For every number of digits, there are about 4 primes that set a record like this. The sequence is: 2, 3, 7, 23, 89, 113, 523, 887, 1129, 1327, 9551, 15683, 19609, 31397, 155921, 360653, 370261, 492113, 1349533, 1357201, 2010733, 4652353, 17051707, ... (Sloane's A2386). See also 1693182318746372.
The ratio between broad daylight and a clear night under a full moon. It is a rather impressive feat of nature that the human eye can handle this range so well. The expression "2.512^{14}" refers to the differences between the Moon's and Sun's apparent visual magnitudes (which is 14) and the ratio represented by one unit on the logarithmic magnitude scale.
(base30 Hindu time divisions)
The nimesha, 1/405000 of a day, is a division of time in one (of several) Hindu system of measurement. This one is notable for involving consecutive powers of 30. (this one uses powers of 60).

The nimesha breaks the pattern of powers of 30, but sometimes a unit of "halfnimesha" is used (see 2202). As I have shown, one can also extend the system in the other direction by adding the 30day month, which is the more commonlyused month in Hindu calendars. Continuing further you get 900 days (roughly 2.5 years), 27000 days (73.92 years, which is curiously just about a lifetime), and 810000 days, 2217.7 years or a little longer than the time for precession to go from one zodiac sign to the next. See also 26244000000, 4665600000000.
432000 years is also 120 sars (a sars is 3600 years), the length of the kaliyuga, which in Hindu mythology is the length of the age in which the early writers placed themselves. There are several longer lengths of time in the same mythology; see 4320000.
A Babylonian version of the flood story (analogous to that in Genesis) tells of ten kings who ruled during the period prior to the flood; the total length of that period is 432000 years. (See 86400 for a link to the Biblical version.)
In preChristian Germanic mythology, there is sort of a reference to the number 432000: 800 men at each of 540 gates of Wodan's palace.
Similarly, in the Viking doomsday tale of the Day of Ragnorook, 800 divine warriors emerge from each of 540 doors of Valhalla.
Since 120=5×24, 432000 seconds is exactly 5 days (see also 86400).
See also 432, 43200, and 4320000000.
Number of days in a Biblical prophetic period: three and a half prophetic "years" of 129600 days (see 3.5 and 1260). It is also 10 times the highlycomposite number 45360.
A prime number; the next prime is 492227. The gap between these two (492227492113 = 114) is the largest gap between two primes less than 1000000. (from numberphile)
A number with a specific name, crore, in Iranian usage until recently ^{67}. This is the largest numbername I know of that is not a power of 10 (a word like mebibyte, meaning 2^{20}=1048576 bytes, refers to a quantity of something specific, not the number itself. Huge number like the Mega and the GrahamRothschild number are not used with other words to make a composite name). In India, crore is still in current usage but means 10000000. See also 144, 1728, 10000 and 100000.
510510 is a primorial: 510510 = 2×3×5×7×11×13×17. It is also the product of four consecutive Fibonacci numbers: 510510 = 13×21×34×55. The digits repeat because 1001 is the product of the 3 consecutive prime numbers 7×11×13.
Number of minutes in a year, if you take a year to be exactly 365 days. This number is repeated several times in the song "Seasons of Love" in the stage musical Rent. At 11 syllables, it is the longest number name that I know of that has been set to music. It beats out 8675309 (which is sung "eight six seven five three oh nine") by three syllables. See also 24601, 10000000000, 0118 999 881 999 119 725 3, and 10^{1010}.
Number of days in the approximate Sothic cycle, 1460 Julian years of 365.25 days each, that was adopted by the Romans for the Julian calendar (see 1508.0833 for more).
This is 365×1508, and is very nearly equal to 1507 tropical years. It is also the number of days in 29 full *calendar rounds of the Mayan/mesoamerican people (see 18980). There is written evidence that the Mesoamericans knew this and thus, that their estimate of the length of the tropical year was 365.242203...
See also 365, 5126, 1872000 and 23040000000.
604800 = 60×60×24×7, the number of seconds in a week. It is also 10!/6. See also 10080, 40320, 86400 and 86400000.
Magic constant of this 7×7 magic square found by Sebastien Miquel, in which all the numbers are perfect cubes:

For this feat Miquel won €500 in 2015.
The number of positions that can be reached on an ordinary 3×3×3
Rubik's Cube using only 180^{o} turns. In group theory terms, the
order of the
In the Thistlethwaite cubesolving algorithm from 1981 [148] this subgroup is the last stage in a series that progressively descends from largerindex to smallerindex subgroups of the whole Rubik's Cube group; once reaching an element of this group the cube is solved using only halfturns. See also 19508428800, 21119142223872000, and 43252003274489856000.
640320 = 2001×320 = 2^{6}×3×5×23×29
One of the numbers that comes up in relation to the Ramanujan constant, also notable for being easy to remember. It appears in the Chudnovsky series approximation of pi.
A quadruple factorial and a highlycomposite number.
It takes this many Hebrew years for the Hebrew calendar to repeat itself. The factor of 19 is the metonic cycle, 7 is the number of days in the week, and the other factors (a total of 2^{6}×3^{4} = 5184) come from the units of division used to approximate the length of the synodic month (see 29.5305941358)
The number of atoms in a poliovirus, as modeled by the simulation program VirusX. See also 7.525×10^{10}, 2.165×10^{25}, and 7×10^{27}.
The first of about 20 successive factorization recordsetters that have a repeating digit pattern. 720720 has 240 divisors; the next recordsetters are 1081080, 1441440, 2162160, 2882880, 3603600, and so on. The digits repeat because the numbers are multiples of 1001, and multiples of 1001 occur as divisibility recordsetters because 1001 is the product of 3 consecutive prime numbers.
The allnumeral leetspeak spelling of googol.
This is the number of days from the date given in Ezekiel 31:1 (as generally accepted: Sivan 1, 3174; see Book of Ezekiel) until September 11, 2001. The sum 1290 + 1335 is numerically significant, see 1260; also 360 is the number of days in the Prophetic Year (see 129600). A symbolic connection between Ezekiel 31 and 9/11 is seen in passages such as "...because it was proud of its height, I handed it over to the ruler of the nations, for him to deal with according to its wickedness. I cast it aside, and the most ruthless of foreign nations cut it down and left it. ..." (Exekiel 31:1012, New International Version).
This is 2 less than a power of 10, and more significantly the square of 127 times 62. This causes an exceptional pattern of digits in the square root of 62.
The factors of 999998 consist of a 2, a 31, and two 127's; all but the 2 are Mersenne primes. David Wilson, via mathfun^{114}, has suggested there might be a connection between this and the square root of 62 phenomenon, based roughly on the following:
127 is a little less than 2^{7},
√62 is a little less than 2^{3},
so 127√62 is a little less than 2^{10};
which coincides with 1000 being a little less than
1024.
This looks promising, since 127√62=999.99899... is indeed very close to 1000. However, there is no Mersenne connection to any of the other special square root digit patterns, including the many examples in other bases shown in the 7.874007874... entry.
999998 is also closely related to the properties of the metric conversion constants 2.54 and 39.37, described at 99.9998.
See also 998.
Fermat's Little Theorem states that if p is a prime number and not a factor of n, then n^{p1}  1 is divisible by p. When n = 10 and p = 7 we get 10^{71}  1 = 10^{6}  1 = 999999 which is divisible by 7. This gives us the first "interesting" repeating decimal fraction, 1/7.
If p is 2 the formula is trivial; in all other cases p is odd and we can express n^{p1}  1 as:
n^{p1}  1 = (n^{(p1)/2} + 1) (n^{(p1)/2}  1)
which gives at least two other factors; in the case of 999999 these are 999 and 1001. By Fermat's Little Theorem, at least one of these must be divisible by 7 (in this case it's 1001).
999999 is divisible by 99 and by 999 precisely because 6 is divisible by 2 and by 3. That's true because you can break it up into identical groups of digits, and therefore you can factor it: 999999 = 990000 + 9900 + 99 = 99×(10000 + 100 + 1) = 99×10101; and (as already stated) 999999 = 999000 + 999 = 999×(1000+1) = 999×1001.
(a million)
One million. This number is probably associated with the concept "really big number" by more people than any other number (see also thousand and billion).
One million is at about the limit of direct physical perception. You can just barely put 1,000,000 dots on a large piece of paper and stand at a distance such that you can perceive each individual dot as a distinct dot, and at the same time be within viewing distance of the other 999,999 dots. (I have actually done this, just for fun!). Because it is near the limit of physical perception, I use 1000000 as the boundary between the class 1 and class 2 numbers.
The English name million comes from French (and Old French) milion, probably from Old Italian milione, which is a colloquial way of saying "big thousand".^{54}
In the classical Chinese 3×3 magic square, the rows can be treated as 3digit numbers, and their squares sum to the same value as the reversed form of the same numbers: 492^{2}+357^{2}+816^{2} = 294^{2}+753^{2}+618^{2} = 1035369. If we do the same thing with the columns we get 1172421.
See also 1665.
In the classical Chinese 3×3 magic square, the columns can be treated as 3digit numbers, and their squares sum to the same value as the reversed form of the same numbers: 834^{2}+159^{2}+672^{2} = 438^{2}+951^{2}+276^{2} = 1172421. If we do the same thing with the rows we get 1035369.
See also 1665.
The volume of the Sun divided by that of the Earth, according to NASA in 2013. See also 333000.
(the SI Mebi prefix)
The square of 1024, and another power of 2 that is close to a power of 10. There is an SI prefix for 1048576, mebi (abbreviated Mi). (There are more prefixes in this table: standard names and SI prefixes.)
A near miss to Fermat's Last Theorem, 1419869 = 13^{5}+16^{5} = 17^{5}+12. (And, you can remove the exponents — 13+16 = 17+12)
Fermat was reading the Arithmetica by Diophantus, and he conjectured that there are no solutions to the equation
A^{n} + B^{n} = C^{n}
for positive integers A, B and C and positive exponent n>2. He then wrote, in the margin of his copy of Arithmetica, his famous (or infamous!) comment:
Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et generaliter nullam in infinitum ultra quadratum potestatem in duos eiusdem nominis fas est dividere: Cuius rei demonstrationem mirabilem sane detexi hanc marginis exiguitas non caperet. (Latin: [For] a cube [to be the sum of] two cubes, or a 4^{th} power [to be the sum of] two 4^{th} powers, or in general [for] any [number that is a power] greater than a square [to be] divided into [a sum of] two like [powers], is not possible: I have discovered a truly remarkable proof which this margin is too small to contain.)
Thus began almost 400 years of speculation and intensive effort by the mathematics community to discover what, if any, idea Fermat might have had when he made such a claim, and more importantly, whether the "theorem" (actually a conjecture) was in fact true.
Certain exponents (values of n) are easy to prove. For example, if the exponent n was a product of primes n=pq, and if there was a solution A^{n}+B^{n}=C^{n}, then it could be rewritten (A^{p})^{q}+(B^{p})^{q}=(C^{p})^{q}, or D^{q}+E^{q}=F^{q} where D=A^{p}, E=B^{p}, and F=C^{p}. Thus, if the "theorem" is proven for any exponent q, all multiples of q also get proven.
And Fermat himself wrote a proof that if x^{2}+y^{2}=z^{2} and x,y,z are positive integers, xy/2 cannot be a square. From this it follows that A^{4}+B^{4}=C^{4} has no solution in positive integers A,B,C. That leaves only the odd prime exponents. In the late 1700's, Euler (almost) proved the case n=3, but his proof had flaws which had to be fixed posthumously. The case n=5 was proven next, but it took until 1825 and the work of several leading mathematicians.
Work got progressively more complex; by 1847 all odd primes up to 31 had been proven; many more were proven by hand and then computer techniques were used to prove all primes up to 4000000. The actual complete proof came in 1993 and 1994 and was based on the study of elliptic curves, something seemingly unrelated to Fermat's Last Theorem.
There are of course many "near misses" to the theorem. The smallest and bestknown is 10^{3}+9^{3}=12^{3}+1=1729.
The number of millimeters in a mile (an exact integer, see 1609.344). Reader Neil Croll pointed out to me that this is also equal to 9×11×127×128.
1658880 = 5^{20}×4^{21}×3^{22}×2^{23}×1^{24}×0^{25}. It is a member of Somos' sequence A52129 associated with the constant 1.66168.... See also 576.
1969920 = 24×1080×76 = 2^{8}×3^{4}×5×19
(used in the Hebrew calendar)
The number of divisions of the day in the Talmudic (traditional Jewish law) measurement system used for astronomy, including the Hebrew calendar (see 29.5305941358):

Another, more symmetrical system is used in the Talmud for nonastronomical periods of time:

See 1001, 2001 and 6469693230.
This is 128^{3} and is also equal to 125^{3}+50^{3}+25^{3}+15^{3}+3^{3}. That's not particularly special, but if you multiply everything by 4 you get the sum mentioned here.
The number of days in the "long count" cycle of the Mayan calendar (also subsequently used by the Aztec and Toltec cultures), a bit less than 5126 years. This cycle is 13 bak'tun, where a bak'tun is a period of 20 k'atun, a k'atun is 20 tun, a tun is 18 unial (pronounced wee nal), and a unial is 20 k'in or days. The unial is the same length as the Mayan "week", a period of 20 days each with a distinct name; 18 such periods plus the 5day uayeb period of prayer and mourning made up the haab, or vague civil year. When this is combined with the 260day tzol k'in, a larger cycle of 18980 days (52 haab) results, this is the least common multiple of 365 and 260, and thus is the number of days that elapses before the tzol k'in and civil year repeat with the same alignment. The tzol k'in itself was comprised of a 13day cycle overlaid with the 20day unial; see Residue Number System. Mayans were also aware of the need for leap years. See also 260, 2012, 550420, 23040000000 and 10331233010526315789473684112000.
The Julian date of the astronomical epoch used to define the start of the "Besselian year". It corresponds to approximately 7:31 PM on December 31, 1899 (GMT). See also 365.242198781.
The Julian Date of midnight in the early morning of January 1, 1958. This is the epoch (similar to an astronomical epoch or calendar epoch) of the timescale variously called T_{m}, AM, A3, and TA(BIH). This was the first international atomic time scale, defined by the BIH (InternationalTimeBureau), and capable of being used by observatories around the world to coordinate astronomical observations.
See also 2436204.5.
The Julian Date of the moment that is 32.184 past midnight in the early morning of January 1, 1977. This is the epoch (similar to an astronomical epoch or calendar epoch) of the time standards TCB (Barycentric Coordinate Time), TGC (Geocentric Coordinate Time), and TT (Terrestrial Time). All of these measure time in "seconds" (but not minutes, hours, days or any larger unit), and they all use the SI second, but they tick at different rates due to general relativity (most significantly, gravitational time dilation). TT measures time as experienced by an atomic clock on the Geoid (the equipotential surface equivalent to mean sea level), and is the slowest of these because it is in the Earth's and all larger gravity wells.
As compared to TT, TGC (which is outside the Earth's gravity and rotation but remains at rest relative to the Earth, and therefore is still experiencing frame acceleration) runs faster by 22 milliseconds per year; and TCB (which does not move with the Earth, but moves with the Solar System's barycentre) runs faster by about 490 milliseconds per year. When TDB and TCG are plotted against each other at sufficiently high precision (see this image: recent differences in time scales one sees a sinusoidal variation with amplitude ± 2 milliseconds in TCB. (As noted at ^{127}, "Relativistic variation of TCB and TDB exaggerated by factor of 1000". TDB (Barycentric Dynamical Time) is a scaled version of TCB designed to run at the same speed as TT, and thus useful for astronomical purposes like coordinated observation of Solar System events by telescopes at different locations on Earth.)
This is the Julian day number of January 1, 2000. It is sometimes seen in astronomical ephimerides, referred to as "J2000".
See the 137.035 article (Michael Wales conjecture).
2646798 = 2^{1} + 6^{2} + 4^{3} + 6^{4} + 7^{5} + 9^{6} + 8^{7}, a sum of its digits raised to consecutive powers^{51}. See 135.
Last denominator in the greedy Egyptian fraction expansion of ^{4}/_{17} = ^{1}/_{5}+^{1}/_{29}+^{1}/_{1233}+^{1}/_{3039345}. This is an example of why the greedy algorithm doesn't work too well.
A "selfdescribing" number, like 21200 and 42101000; see 6210001000 for more.
This divisibility recordsetter is really easy to remember because it has 360 factors (360 itself is a recordsetter), and includes two 360's in its digits. The repetition of 360 relates to 1001. And its prime factorization is reasonably easy to remember if you learn it as 3^{2}×4^{2}×5^{2} × 7×11×13; that should also help if you're trying to find its divisors without using trialanderror.
10! = 10×9×8×7×6×5×4×3×2×1 = 3628800 can also be expressed as 6!×7×6! because 6! = 720 = 10×9×8. (Source: Gary Rosys) See also 5040. 10! seconds is exactly 42 days; see also 10080, 604800 and 86400000.
There are no other factorials that can be broken up in this way, beyond the trivial case of 2! = 1×2×1, but see 86400.
Call the middle number x (so, in the 10! case, x=7). Then the product of interest is (x1)!×x×(x1)!. The latter (x1)! is also expressed of the form (x+1)×(x+2)×(x+3)×..., so we have:
1×2×3×...×(x2)×(x1) = (x+1)×(x+2)×(x+3)×...
The numbers (x+1), (x+2), etc. on the right side of the equation have to be composite, because if any of them is prime then they would not be a prime factor of the product on the left. The formula works for 7 because the numbers right after 7 (8, 9, 10) are not prime.
Also, all of the prime factors on the left side have to occur on the right hand side. Since the left side includes all the primes up to (x1), all of those primes have to occur on the right. In the case of 7, the needed primes are 2, 3 and 5, which is not very many. Let p be the largest prime on the left. For example, if x is 13, p is 11. A multiple of p has to occur on the right, and the first multiple of p is 2p. This requires a fairly big gap in the distribution of primes. When x is bigger than 7, there will always be some other prime on the right side, that is greater than x and less than 2p. Proving this part is a little tough.
3628800 is also the number of seconds in 6 weeks:
6 × 7 × 24 × 60 × 60
= 6 × 7 × (8×3) × (3×4×5) × (10×2×3)
= 6 × 7 × 8 × 3×3 × 4 × 5 × 10 × 2 × 3
= 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 × 10.
Here is video by Numberphile: 10 factorial.
See also 86400000.
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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 5×10^{11} 10^{18} 5.4×10^{27} 10^{40} 5.21...×10^{78} 1.29...×10^{865} 10^{40000} 10^{9152051} 10^{1036} 10^{1010100} — — footnotes Also, check out my large numbers and integer sequences pages.
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