Notable Properties of Specific Numbers
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247 is an Eulerian number. The Eulerian numbers, given by Sloane's A0295, begin: 0, 1, 4, 11, 26, 57, 120, 247, 502, 1013, 2036, ... and are simply the values of 2^{n}n1. These numbers usually manifest themselves as a permutation count. In this figure, each of the permutations has exactly one place where a larger number follows a smaller number (called an "ascent"):
A[2] = 1 12
A[3] = 4 132 213 231 312
A[4] = 11 1432 2143 2431 3142 3214 3241 3421 4132 4213 4231 4312
The Eulerian numbers show up as the first nontrivial diagonal of a number triangle (called Euler's triangle) that is like Pascal's triangle, but with rows that add up to successive factorials:
1 sum: 1 1 1 sum: 2 1 4 1 sum: 6 1 11 11 1 sum: 24 1 26 66 26 1 sum: 120 1 57 302 302 57 1 sum: 720 1 120 1191 2416 1191 120 1 sum: 5040 1 247 4293 15619 15619 4293 247 1 sum: 40320To compute numbers in this triangle, you add the two numbers above it, each multiplied by the index number of the diagonal shared by that number. For example, consider the first 26 in the 4^{th} row. The numbers 1 and 11 are above it. The 26 and the 11 share the 2^{nd} diagonal, counting from the left, so we take 11×2. The 26 and the 1 are in the 4^{th} diagonal, counting from the right, so we take 1×4. This gives 11×2 + 1×4, which is 26. For another example, 57×5 + 302×3 = 1191: the multipliers 5 and 3 are used because the number to be computed is in the 5^{th} and 3^{rd} diagonals.
The first few diagonals of the Euler triangle are A0295, A0460, and A0498, and the largest number in each row is given by A6551. All of these count permutations, but with different requirements for how many "ascents" there are.
2:47 is 13×19, but you're six minutes late.
251 is the first of a set of 4 consecutive primes that are spaced an equal distance apart: 251, 257, 263, and 269 are all prime, there are no primes in between, and the spacing between them is 6. 251 is the lowest number with this property; the next is 1741. See also 47, 9843019, 121174811 and 19252884016114523644357039386451.
252 = 10×9×8×7×6 / (5×4×3×2×1)
If you flip a coin 10 times in a row, there are exactly 252 ways in which it can turn out that you get exactly 5 heads and 5 tails (some examples are HTHTTTTHHH, THTHTHTHTH, HHHHHTTTTT, TTHHHHTHTT). Note the rather elegant way of expressing this using the numbers 1 through 10. This type of expression can be used to calculate any of the numbers down the center of Pascal's triangle. See also 2520.
This is 2^{8}, and is the number of possible values that can be stored in a byte, the most common unit of measurement for computer memory quantities. The number of values results from the fact that there are eight bits in a byte, and each bit can have two values (0 or 1). (Yes, during one period of history (16th century and earlier) there were also eight "bits" in a Spanish dollar, a fact that is related to the phrase "pieces of eight", and "two bits" (as in "shave and a haircut — two bits", slang for "25 cents"). The Spanish dollar was a gold coin with a value of eight reale, and was sometimes actually cut into eight wedgeshaped pieces to make change.)
Eight bits became the standard computer memory quantity because it is also a power of 2 (other smaller "bytes" were used in the early days when memory was expensive, but having the number of bits be a power of two makes many operations more efficient.) The next smaller powerof2 size would be 4 bits per "byte", but that only allows for 16 = 2^{4} possible byte values. The need to store and manipulate text, including a full alphabet (26 characters in the country where most of the computers were being developed) plus digits and many punctuation symbols meant that a byte that takes at least 100 or so values was necessary.
If computers had been developed in one of the many countries that have over 256 writing symbols (those that use the Korean, Ethiopic, Japanese, Chinese, and Vietnamese writing systems, among others) they might well have settled on a 16bit byte. In fact, the emerging world standard for text (Unicode) uses 16bit character codes, and allows distinct values for every symbol in every popular writing system.
257 is a Fermat number, a number of the form 2^{2N}+1. Fermat hypothesized that all of these were prime, but they are not. A discussion of them and their factors is here. See also 641.
(Personal: the course number of my only Men's Weekend production)
The number of days in the tzol k'in, the almanac cycle Mayans used for divination (a system that originated at least 2600 years ago and is still used by native Americans in Yucatan, Guatemala, Belize, and Honduras). The days in the tzol k'in have successive names from a set of 20 (similar to the 20 days in the Mayan civil week) and numbers from a set of 13. There are thus 13×20 combinations of a name and a number and it takes this many days for the cycle to repeat, so the days were effectively named by a base {13,20} residue number system. It is not known why 260 was "chosen" although there are many theories. It is wellknown the Mayans recognized a period of 52×365 days (slightly less than 52 tropical years), which is exactly equal to 18980=73×260 days, with 73 being exactly a fifth of 365 — but this cycle does not line up with other important astronomical periods (including the eclipse seasons, the lunar month and the synodic period of Venus, each of which was wellknown and considered very important). See also 5126 and 1872000.
See 220.
288 = 4!×3!×2!×1! = 4^{4}+3^{3}+2^{2}+1^{1}
(4 superfactorial by the SloanePlouffe definition)
This is the value of "4 superfactorial" by the lower (Sloane and Plouffe 1995) definition of "superfactorial": 4!×3!×2!×1! = 24×6×2 = 288. By a rather nifty coincidence, it is also equal to 4^{4}+3^{3}+2^{2}+1^{1} = 256+27+8+1.
Barnes' Gfunction is to superfactorials as the Gamma function is to normal factorials. Barnes' Gfunction can be (very slowly) calculated by the formula:
G(z) = 2π^{z/2} e^{[z(z+1)+γ z2]/2} PRODUCT_{(n=1..inf)}[(1+z/n)^{n} e^{z+z2/(2n)}]
where γ is the EulerMascheroni constant. For sufficiently large values of z you can use the approximation:
G(n) ≈ (e^{1/12}/A) n^{n2/21/12}(2π)^{n/2}e^{3n2/4}
where A is the GlaisherKinkelin constant. See the MathWorld page ^{86} for more.
See also 10^{10102.0691973765×1036305}. There is also a hyperfactorial.
The reciprocal of the oblateness of the Earth according to the WGS 84 model. The diameter of the Earth measured through the equator exceeds the polar diameter by about 1/298.26 of its size. In other words, if P is the polar diameter and E is the equatorial diameter, then (EP)/E ≈ 1/298.26. WGS 84 is an approximation of the Geoid, which is a "gravitational equipotential surface".
See also 6378137.
The number of neurons in the nematode worm C. elegans. See 959.
I recently moved, and my house number changed from 27 to 304. I wanted to find relationships between these two numbers, and quickly found two rather straightforward ones:
304 = 3 × 10 × 10 + 4 and 27 = 3 + 10 + 10 + 4 (change × to +)
304 = 2^{4} × 19 and 27 = 2 × 4 + 19 (change ^ to × and × to +)
This sort of thing is very common.
323 is a Motzkin number; see 51.
323 = 17×19, a product of two primes that, together with 1001 and 2001, create the rather nice digit pattern of the primorial 6469693230.
325 = 18^{2}+1^{2} = 17^{2}+6^{2} = 15^{2}+10^{2}, and is the smallest number that can be expressed as the sum of two distinct positive squares in at least 3 different ways. The sequence of such numbers is: 5, 65, 325, 1105, 5525, 27625, 71825, ... (Sloane's A52199; see also A93195). An interesting thing about these numbers is that all of their factors are prime numbers of the form 4n+1 (5, 13, 17, 29, 37, ... the "Pythagorean primes" A2144), and their factorizations are similar to those of the highly composite numbers. See also 1729 and 635318657.
327 is a perfect totient number. This is like a perfect number, but adding together the iterates of the totient function instead of adding together the number's factors. The Euler totient function, Sloane's A0010, counts how many numbers smaller than N are relatively prime to N. For example, if N is 14, the numbers 1, 3, 5, 9, 11 and 13 are relatively prime (notice that 1 counts), so the totient function of 14 is 6. For a prime P we're counting all numbers smaller than P, which is P1, so the totient function always gives a smaller number except (by definition) when N=1. So, we can take a number like 327, and compute the Totient function (which gives 216), then compute the totient function of that number, and so on (giving 72, 24, 8, 4, 2 and 1), then add all these numbers together and we get 216 + 72 + 24 + 8 + 4 + 2 + 1 = 327. There is more about these numbers on a separate page here; see also 243.
327 also "combines" two of my favorite numbers, 3 and 27. When watching a baseball game I immediately notice when a batter's average or some similar statistic is .327, but it seems to happen more often then random chance would allow (.327 results from a record of 17 for 52, or a multiple like 34 for 104). This is the cult number effect in action.
A "3dimensional oblong number", the volume of a rectangular solid of dimensions 6×7×8. If you add the middle number (7) you get 7^{3}=343. Numbers of this type are always a multiple of 6, because in any three consecutive integers there is always exactly one multiple of 3 and at least one even number. The sequence starts: 6, 24, 60, 120, 210, 336, 504, 720, 990, 1320, 1716, ... (Sloane's A7531).
A "generalized oblong number" is the product of 2 or more consecutive positive integers. These include the regular oblong numbers, 3way products such as 336, and numbers like 120 formed by the product of 4 or more integers. The sequence starts: 2, 6, 12, 20, 24, 30, 42, 56, 60, 72, 90, 110, 120, 132, 156, 182, 210, 240, 272, 306, 336, 342, 360, 380, 420, 462, 504, 506, 552, 600, ... (Sloane's A45619). Each is the value of _{n}P_{r} for some values of n and r, n>r and r>1.
For all prime numbers p (except 2), 2^{p} modulo p is 2:
3 is prime, 2^{3} = 8 = 3×2+2
5 is prime, 2^{5} = 32 = 5×6+2
7 is prime, 2^{7} = 128 = 7×18+2
11 is prime, 2^{11} = 2048 = 11×186+2
13 is prime, 2^{13} = 8192 = 13×630+2
etc.
This is also true for some (but only a very few) composite numbers. The first composite number for which it is true is 341 = 11 × 31.
Checking a number p by this method is called a primality test because it allows you to quickly find out if a number is composite. The same method works with other bases in place of 2, as long as the base and p have no common factors. In the case of base 2, this means checking if p is odd or even. 341 "fools the test" for base 2 because it gives the same result a prime number would: 2^{341} is 2 modulo 341. However, it doesn't fool the test for base 3: 3^{341} is 168 modulo 341. See 561 for more.
341 is also 143 backwards.
343 equals 7^{3} = 3^{5} + 10^{2}
343 = (3+4)^{3}, a "Friedman number"
and (tragically) it is the number of New York City firefighters killed at the World Trade Center on 2001 Sep 11^{th}.
347 = 7^{3}+4, and is therefore able to be expressed in two different ways using the same 3 digits (3, 4 and 7). This is called a Friedman number.
360 can be evenly divided in 24 different ways — more than any other number of this size (see this entry for more about this).
360 is the number of degrees in a full circle. This size for angular division was chosen by the Babylonians because it is 60^{2}/_{10} and because it is close to the distance the sun moves through the zodiac each day (in other words, it is close to the length of the tropical year. In N days, the sun moves just about N degrees across the zodiac.
Conveniently for us Earthlings, the number of days per year happens to be pretty darn close to 360, a number with lots of factors. This means that, for lots of fairly common divisions (like 1/3 of a year, 1/4 of a year, etc.) it's pretty easy to remember how many days is involved because it's so close to what you'd get if the number were exactly 360. Thus: a quarteryear is just about 90 days; etc. For more accuracy you can easily add the fraction of the remaining 5 (e.g. adding 1.25 days to get 91.25 as a better approximation for the quarteryear).
If you assign consecutive values to the cards in a standard deck (Ace = 1, Jack = 11, Queen = 12; King = 13) the total of all 52 cards is 364. This is slightly more interesting considering that the number of cards (52) is close to the number of weeks in a year (52.1775 by the Gregorian calendar) and other less remarkable coincidences.
Length of the Harvard Bridge in Boston (which actually bridges Boston to M.I.T. and is nowhere near Harvard University), measured in "smoots". In October 1958, as a fraternity prank, thenfreshman Oliver Smoot (MIT class of 1962) was used as a unit of measurement to measure the length of the bridge (which MIT students use to walk to and from Boston.) Appropriately, Smoot went on in life to eventually serve as President of the International Organization for Standardization (ISO), and as chairman of the American National Standards Institute (ANSI).
His height at the time, 5 feet 7 inches, is a fairly well known "nonstandard unit" of length measurement. It is available in Google's builtin calculator (type "27 feet in smoots" in any Google search box) and as an optional unit in Google Earth (in the Ruler tool). The markings themselves are so wellaccepted by the public that when the bridge was replaced in the 1980s, the Cambridge Police department requested that the markings be maintained, since they had become useful for identifying the location of accidents on the bridge. The construction crew went one better, by actually scoring the concrete surface of the sidewalk on the bridge at 1smoot intervals instead of something more conventional, like 4 or 6 feet (see here).
Often used in quick and sloppy calculations as the number of days in a year. 365 is 52 weeks plus one day.
The ancient Mayan and Egyptian cultures each (independently) came up with calendars that divided most of the year into equalsized months (18×20 or 12×30 days respectively) and then added a 5 or 6day intercalary period.
The Egyptian calendar is still in use within the Coptic Orthodox church in Egypt; the 56 day period is called Pi Kogi Enavot and falls in the second week of September.
The Mayan 365day count (called Haab') is still kept by indigenous peoples in Guatemala and nearby parts of Mexico. It was known to the early Mesoamericans that this was not an accurate approximation; see 550420.
The tropical year is the amount of time for the Earth to complete one orbit around the Sun, measured from equinox to equinox or from solstice to solstice. It differs from the sidereal year because of the Earth's axial precession and the effects of other planets.
This is the most important definition of "year" for most people because it is the ratio between the frequency of day and night (the average solar day) and the frequency of the seasons. It is the value that solar and lunisolar calendars try to approximate.
A mean tropical year is an average over a very long period of time (such as a 25800year precession cycle). However, there are many types of mean tropical year. Normally a tropical year is measured from one solstice or equinox to the next. However, this presents two accuracy problems:
 The Earth orbits the Solar System's Barycenter, and the Sun also orbits this in a different orbit mainly determined by the gas giants. This makes the EarthSun angle change at a rate that is different from what you would expect from simply the Earth going in an elliptical Kepler orbit.
Length of Time Between Successive March Equinoxes
(from [151], page 41)

 The positions of the equinoxes depend on the Earth's axis, which moves slowly (axial precession). The Earth's orbit is an ellipse, the orientation of which also changes slowly (apsidal precession). Since the Earth moves more slowly near aphelion, the position of each solstice and equinox moves at different rates. This makes for several different mean tropical years:
Mean Tropical Years Between Each Pair of Solstice or Equinox Type
(for the year 2000, from [151], page 42)
The "days" here are 1900 mean solar days ("ephemeris days")

On geologic timescales the Earth's rotational period (length of the solar day) increases by about 2.3 milliseconds per day per century. Longer days means fewer days per year, so the length of the tropical year decreases every year by about 2.7×10^{10} days. This is mainly due to the tidal interaction with the Moon^{126} slowing Earth's rotation (see 9192631770 for more about this). In addition, there is an effect called postglacial rebound that operates on shorter timescales (the period since the last ice age) causing the Earth's rotation to speed up by about 0.7 milliseconds per day per century, making the tropical year increase by 8.1×10^{11} days per year. This adds to the lunar effect, making for an overall increase of the day length of +1.7 milliseconds per century, and an overall decrease of the number of solar days per tropical year of 1.96×10^{10} every year.
There are also shorterterm variations that fluctuate so much that it doesn't even make sense to try to average them into a year length figure; see mean solar day.
The following numbers, close to 365.242, are approximations to the tropical year; compare to the sidereal year constants which are higher.
One value of the tropical year in mean solar days. This figure was in the Wikipedia Year article in 2008; later shortened to "365.24219". Both are intended to be for the year 2000. The mean solar day and other factors vary too much to get more than about 9 digits of accuracy; see that entry and the tropical year article for details.
The length of the mean tropical year in "SI days" of 86400 SI seconds, for the year 2000 based on [154] ("The constants P_{1} and e_{0}", p. 664 and "Mean elements of the Earth", p. 675):
365.25×1000 × 360×60×60 / (1295977422.83429 + 50288.200) = 365.242190402
This calculation is derived the same way as that for the sidereal year value 365.256363004 (see that entry for details), but adding Simon's rate of movement of the solstices, 50288.200 arcseconds per millennium.
This number differs from other values of the tropical year for reasons discussed under tropical year and mean solar day.
Length of the tropical year, in "SI days" of 86400 SI seconds, by the "Besselian year" definition; see also 2415020.31352.
The approximation to the tropical year known to the Mayan/mesoamerican people, who knew that 29 calendar rounds was very nearly equal to 1507 tropical years. See 550420.
Length of the tropical year (in mean solar days) under the "Revised Julian calendar" of the Eastern Orthodox Church. In it, years divisible by 4 are leap years, except that years divisible by 100 are only leap years if dividing the year by 900 leaves a remainder of 200 or 600. This gives the approximation 365 + 1/4  1/100 + 2/900 = 365.242222222..
An approximation to the tropical year from measuring between vernal equinoxes. As of 2012, this is given in the Wikipedia article Leap year, calling it the "vernal equinox year" (earlier versions gave 365.242375).
The length of time from one equinox to the next of the same type is not exactly the same as the tropical year, because the Earth's axial precession (see 25772.1300) moves at a different rate from the precession of the perihelion of the Earth's orbit (see anomalistic year). In modern times the Earth is furthest from the Sun (and thus moves slowest) in early July.
An approximation to the tropical year from measuring between vernal equinoxes. This is seen in some discussions, most getting the number from Wikipedia Leap year, which cited this value as the "vernal equinox year"; later it switched to 365.242374.
This is the approximation that the Gregorian calendar uses for the number of days in a tropical year. Note the series of approximations:
365, the Egyptian civil year
365 + 1/4 = 365.25, the Julian year
365 + 1/4  1/100 = 365.24, a slightly better approximation (but not used
in any calendar)
365 + 1/4  1/100 + 2/900 = 365.242222222.., the
revised Julian calendar year
365 + 1/4  1/100 + 1/400 = 365.2425, the Gregorian year
This is the approximation that the Hebrew lunisolar calendar uses for the number of days in a tropical year.
The Julian year was the length (in days, or more accurately mean solar days) of the tropical year under the Julian calendar.
This approximation was probably first known to the Egyptians, who used a civil calendar of 365day years (with no leapyears) for a staggeringly long period of time, and were therefore able to measure how inaccurate it was through observation of star risings (effectively measuring the difference between their 365day civil year and the sidereal year). The discrepancy between this and the more accurate modern values of the tropical year results from the Egyptians' use of Sirius, a star far from the Ecliptic, to determine the length of a sidereal year and assuming that the tropical and sidereal years were the same. See the Sothic cycle and 1508.0833.
The Coptic calendar and Ethiopian calendar still use this approximation, because they have a leap year every 4 years without exception. There is further discussion of this and the Maya calendar (also still in use) in the entry for 365.
A sidereal year is the amount of time it takes Earth to complete one orbit around the Sun when viewed from a nonrotating reference frame (which is usually approximated by averaging together the positions of many distant stars). It differs from the tropical year because of the Earth's axial precession.
There are different values for the sidereal year because the rate of Earth's rotation changes over time, and it has been measured in different ways; see the tropical year and mean solar day entries for details.
The following numbers, close to 365.256, are approximations to the sidereal year; compare to the tropical year constants which are lower.
Simon Newcomb's value for the length of a sidereal year in mean solar days. In Newcomb's work it appears as a parametrized formula:
t = 365.25636042 + 0.00000011T
where T is the number of centuries that have elapsed since 1900. Thus, for the year 2012, the sidereal year would be about 365.25636054 mean solar days.
See also 365.2568983.
Length of the sidereal year in "SI days" of 86400 SI seconds, for the year 2000 based on [154] ("Mean elements of the Earth", page 675). It gives the rate of Earth's motion around its orbit as 1295977422.83429 arcseconds per 1000 Julian years. A Julian year is 365.25×86400 SI seconds, so the rate of movement would be
R = 1295977422.83429 / 1000×365.25×86400 ≈ 0.0410670463797
arcseconds per SI second. The number of arcseconds in a full circle is 360×60×60, so the length of the sidereal year in "SI days" would be:
Y_{s} = 360×60×60 / 86400 R
To compute directly from the original number:
365.25×1000 × 360×60×60 / 1295977422.83429 = 365.256363004
This number differs from other values of the tropical year for reasons discussed under tropical year and mean solar day.
See also 365.242190402 and 365.2568983.
An approximate value of the sidereal year used in Carl Friedrich Gauss' Theoria Motus, and cited by [133]. See also 365.2568983.
The value of the sidereal year corresponding to the following definition of the astronomical unit:
[An Astronomical Unit is] the radius of an unperturbed circular orbit a massless body would revolve about the sun in 2π/k days (i.e., 365.2568983.... days), where k is defined as the Gaussian constant exactly equal to 0.01720209895.
See also 365.256363004 and 365.2563835.
An approximate value (for the year 2011) of the number of days between successive passages of the Earth through its Perihelion. This is the longest version of the "length of a year".
When viewed from the north ecliptical pole in a nonrotating position (i.e. "from the fixed stars"), the ellipse of the Earth's orbit moves "counterclockwise" making one full turn in about 112,000 years. This is Apsidal precession; most of this effect comes from Earth getting "pulled forward" by Jupiter while it orbits the Sun.
Meanwhile, the direction of the Earth's rotational axis moves "clockwise", making a cycle in about 25,800 years. Since the Earth itself moves "counterclockwise" in its orbit, the tropical year is shorter than the sidereal year, while the anomalistic year is longer.
Number of days in a leap year. See 365 and 365.25 for a discussion of 365day calendars that are still in use.
A reader asked (roughly) the following:
What are the odds of seeing the digits 3, 6 and 8 (in any order, but contiguous) on an automobile license plate? If I see 1000 plates during my drive to work, it is unusual to see 10 cars that feature one of the combinations 368, 386, 638, 683, 836 or 863?
Of course, the answer depends a lot on how many license plates have digits at all, and whether there are lots of plates with 4 or more digits in them, etc. You have to consider how many cars there are with each different number of digits, and figure the odds for each.
For plates with 3digit numbers, the answer is easy: 6 out of 1000. If all the plates had 3digit numbers, then seeing 10 out of 1000 would be a bit higher than expected, but not much.
In general, the odds of seeing your N special digits in an Ndigit number are N!/10^{N}. (This is assuming the special digits are different from each other. If some of them are the same, the odds go down.)
For plates with more than N digits, the odds are a little more complicated to compute. For example, for 3 digits in a 4digit number, the odds are 114 out of 10000. This is 6 less than 120/10000, which is what you get if you take the 6/1000 odds for the 3digit case and double it. The reason is that you get:
 60 cases with the 3 digits at the beginning of the 4digit number
 60 cases with the 3 digits at the end
 minus 6 cases that got counted twice because the first three are a match and the last three are also a match. These are the 6 cases that consist entirely of the 3 special digits, and have the first and last digit the same: 3683, 3863, 6386, 6836, 8368 and 8638.
114 out of 10000 is equivalent to 11.4 out of 1000. So if all the plates had 4digit numbers, seeing 10 out of 1000 on your way to work is just about what you'd expect.
The general formula for the N+1 case is (2×101)N!/10^{(N+1)}
As you would expect, the formula keeps getting more complicated. For the N+2 case it's ((3×102)×101)N!/10^{(N+2)}. So the odds of seeing 368, 386, etc. in 5digit license plate numbers is 1674 out of 100000, or 16.74 out of 1000.
See also 0.543643....
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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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