Notable Properties of Specific Numbers
First page . . . Back to page 7 . . . Forward to page 9 . . . Last page (page 25)
The centre number of the 3×3 prime magic square discussed here.
Euler asserted that there were solutions to the equation A^{4}+B^{4} = C^{4}+D^{4}, where A<C<D<B are all positive integers. He then found the first solution, A=59 (see 635318657).
The number of minutes in an hour, and the number of seconds in a minute. Degrees of angle are also divided into 60 minutes, 60×60 seconds, and the divisions can proceed further to 60×60×60 "thirds" (subdivisions of the third order)^{14}. These subdivisions into 60's are the most familiar relics of the Sumerian base60 number system (which was really closer to being a system of using base10 and base6 on alternating digits). It is widely acknowledged that 60 was used in a number system because it has lots of divisors; in fact it sets a record for having more divisors than any smaller number. Other popular ancient divisor recordsetters include 12 and 360; a less popular possibility is 1260, and another related to time divisions is 10080.
Hexagonal numbers
61 items can be arranged in a regular pattern to make a hexagon with 5 items on a side:
o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o oThus 61 is called the 5^{th} "hexagonal number". One can also make similar patterns that are almost hexagonal; see 27. I consider the halfway ones to be significant because they are often used in realworld applications such as cable bundles, etc. (and also because it makes my favourite number 27 be part of the sequence :)
The full sequence (including the halfway ones) runs: 1, 3, 7, 12, 19, 27, 37, 48, 61, 75, 91, 108, 127, 147, 169, 192, 217, 243, 271, ... (OEIS sequence A77043; my MCS1678). For integer sizes, the formula is 3N^{2}3N+1, and for halfinteger sizes, 3(N^{1}/_{2})^{2}.
N: 1 1.5 2 2.5 3 3.5 4 o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o 1 o o o o o o o o o o o o o o o o o o o o o o 3 o o o o o o o o o o o o o o o o o o o o 7 o o o o o o o o o o o o o o 12 o o o o o o o o o 19 o o o o 27 37Semiprimes
62 is the product of two primes. This sequence begins: 4, 6, 9, 10, 14, 15, 21, 22, 25, 26, 33, 34, 35, 38, 39, 46, 49, 51, 55, 57, 58, 62, 65, 69, 74, 77, ... (Sloane's A1358). Note that squares of primes are included, but not higher powers like 8 or 27. At first it seems that these outnumber the more composite numbers, but as you get higher, the opposite ends up being true. The same sequence omitting the squares of the primes is called distinct semiprimes. See also 1679.
In base 10, the square root of 62 has a remarkable decimal expansion; see 7.874007874... for more.
(a generalised Woodall number)
63 is a generalised Woodall number: 4×2^{4}1. Numbers of this type are discussed on a separate page. See also 648. The following table shows the Woodall numbers n×b^{n}1 for various bases b and multiplier/exponents n:
n=2 n=3 n=4 n=5 n=6 n=7 n=8 n=9 b=2 7 23 63 159 383 895 2047 4607 b=3 17 80 323 1214 4373 b=4 31 191 1023 5119 b=5 49 374 2499 b=6 71 647 5183 b=7 97 1028 9603 b=8 127 1535 b=9 161 2186 b=10 199 2999 b=11 241 3992 b=12 287 5183 b=13 337 6590 b=14 391 8231 b=15 449 b=16 511 b=17 577 b=18 647 b=19 721 b=20 799(compositions)
64 is the number of ways of distributing 7 indistinguishable balls into one or more distinguishable urns (with no empty urns). Here are examples of this idea for N=1 through 4:
1: (*) : 1
2: (**) (*)(*) : 10 11
4: (***) (**)(*) (*)(**) (*)(*)(*)
: 100 101 110 111
8: (****) (***)(*) (**)(**) (**)(*)(*)
(*)(***) (*)(**)(*) (*)(*)(**) (*)(*)(*)(*)
: 1000 1001 1010 1011 1100 1101 1110 1111
The sequence is just the powers of 2 (in this case, 2^{N1}). The correspondence is illustrated here: each distribution is matched with a binary number in which each digit is "1" if and only if the corresponding "*" is the first of its parenthesised group.
If the balls are distinguishable, we get instead the ordered Bell numbers. If the balls and urns are both indistinguishable, we get the partition numbers.
65 is the smallest number that can be expressed as the sum of two distinct squares in two different ways: 65 = 4^{2}+7^{2} = 8^{2}+1^{2}. 65 is the product of two primes, 5=2^{2}+1^{2} and 13=3^{2}+2^{2}, so Brahmagupta's identity shows that it is equal to (2×31×2)^{2} + (2×2+1×3)^{2} and also to (2×3+1×2)^{2} + (2×21×3)^{2}.
If the two squares do not need to be distinct, 50 is the smallest, and if 0^{2} counts as a square, 25 is the smallest. No such number can be prime; see 97. See 325 for more; see also 1729 and 635318657.
65 is also the smallest number whose square can be expressed as the sum of two distinct squares in two different ways, if we require that there are no common factors; this is the same as saying that 65 is the hyponenuse of two different primitive Pythagorean triples: 33^{2}+56^{2}=65^{2} and 16^{2}+63^{2}=65^{2}. If common factors are allowed, 25 wins this distinction; but then for 65 we can add 39^{2}+52^{2}=65^{2} and 25^{2}+60^{2}=65^{2}. Also, 65 is the largest 2digit number that makes a Pythagorean triple with two larger twodigit numbers: 65^{2}+72^{2}=97^{2}.
65 is the magic constant of a 5×5 magic square. Here are two examples:
1 15 24 8 17 11 10 4 23 17 23 7 16 5 14 18 12 6 5 24 20 4 13 22 6 25 19 13 7 1 12 21 10 19 3 2 21 20 14 8 9 18 2 11 25 9 3 22 16 15All rows and columns add to 65, as do the two diagonals, and also any other set of 5 squares that is arranged in a pattern with 180degree rotational symmetry. (Magic squares with this property are called associative magic squares ^{89}). In addition, all the diagonals, including those that "wrap around" (like the five numbers 24, 5, 6, 12 and 18 in the left example) add to 65. This makes it a panmagic square ^{88}. 5×5 is the smallest size magic square that can be both associative and panmagic.
If it is an associative magic square, any pattern of N cells with 180degree rotational symmetry adds up to the magic constant. Most are rather nicelooking patterns. In the 5×5 case, there are 23 patterns, ignoring rotations and reflections. They are labeled here with unique numbers that I use to identify the pattern, and a 1, 2 or 4 telling how many more there would be if you count rotations and reflections:
10 4 11 4 12 4 13 1 15 2 16 4 X X . . . X . X . . X . . X . X . . . X X . . . . X . . . . . . . . . . . . . . . . . . . . . . . . . X . . . . . X . . . . X . . . . X . . . . X . . . . X . . . . X . . . . X . . . . . . . . . . . . . . . . . . . . . . . . . X . . . X . . . . . X X . . X . X . X . . X X . . . X . . . . X . . . . X 17 2 20 4 21 2 23 2 24 4 25 4 X . . . . . X X . . . X . X . . X . . . . X . . . . X . . . . . . X . . . . . . . . . . . X . . . . . X . . . . . X . . . . X . . . . X . . . . X . . . . X . . . . X . . . . X . . . X . . . . . . . . . . . . . . . . . X . . . X . . . X . . . . . . X . . X X . . X . X . . . . X . . . . X . . . . X . 26 4 27 2 28 4 29 4 33 4 34 2 . X . . . . X . . . . X . . . . X . . . . . X . . . . X . . . . . X . . . . . X . . . . . . . . . . . X . . . . . X . . . . X . . . . X . . X . X . X . X X X . . . X . . . . X . . . X . . . X . . . . . . . . . . . . . . . . . X . . . X . . . . . X . . . . X . . . . X . . . . X . . . X . . . . X . . 37 1 38 2 70 4 71 1 83 1 . . X . . . . X . . . . . . . . . . . . . . . . . . . . . . . . . . . . X X . . . X . X . . . X . . X . X . X . X X X . . . X . . . . X . . . X X X . . . . . . . . . . . . . X X . . X . X . . . X . . . . X . . . . X . . . . . . . . . . . . . . . . .More generally, one could refer to a "pattern" as any set of N numbers that adds up to the magic constant. In the 5×5 case, there are 1394 such sets. However, most of them do not have a nice geometrical layout. For example, consider: (3, 4, 15, 20, 23). These 5 numbers add up to 65, but if you find them in either of the squares above, you'll see they're in a sort of random pattern.
In the two sample magic squares, the left example is constructed by the following method. A similar method works for all N×N magic squares where N is odd:
1. Start in the centre square, with the value (N^{2}+1)/2.
2. Move one up; if this takes you off the top go to the bottom
of the same column.
3. Move two to the right; if this takes you off the right edge continue
at the left end of the same row.
4. Write the next number in sequence in this square (if the next number
is N^{2}+1, go back to 1)
5. Repeat steps 24 another N1 times.
6. Now move one down and one to the right; as before if this takes you
off the right continue at the left or if it takes you off the bottom
continue at the top of the same column.
7. Write the next number in sequence in this square.
8. Repeat steps 2 through 7 until you have filled all the squares.
The (11, 10, 4, 23, ...) example above is constructed by a similar method, but in steps 23 moving one down and one to the right and in step 6 moving one to the left.
66 is the lowest base with 'easy' divisibility tests for 5 different primes, assuming that the casting out 11's technique is not considered 'easy'.
In base 10, there are wellknown tests for divisibility by the primes 2, 3, and 5, and by the composite numbers 9 and 10. (The tests for 2, 5 and 10 look at the last digit; the tests for 3 and 9 use the repeated digit addition technique.) Are there bases for which there are more numbers that can be easily tested? In base 9, you can test for divisibility by 3 and 9 by looking at the last digit, and by 2, 4, or 8 by using the digitaddition trick. So base 9 has just as many easy tests (5) as base 10. But the number of primes is less (just two: 2 and 3). In base 6, you can test for divisibility by the same three primes (2, 3, 5) as in base 10. Since division by composites can always be broken down into division by multiple primes, what really matters the most, for convenience and utility, is to have tests for as many primes as possible. Base 66 is the smallest base for which you can easily test for divisibility by 5 different primes: 2, 3, 5, 11 and 13. Notice they aren't the 5 smallest primes (7 is missing).
Here is a table of the recordsetters for this property. The primes in bold are the ones tested by looking at the last digit, the others require the digitaddition technique:

Note that if you want the recordsetters for just the lastdigit test technique, they are the same as the primorials.
If you count the casting out 11's method as an easy factorisation test, then you get the table shown under the 14 entry. If you want the recordsetters for all divisors (not just primes), see the entry for 21 (or 29 if the casting out 11 method is included).
(cousin primes)
67 and 71 are both prime, and differ by four. Because of this, they are called cousin primes (in an analogy to twin primes). The first few pairs of cousin primes are at: 3, 7, 13, 19, 37, 43, 67, 79, 97, 103, 109, 127, 163, 193, 223, 229, 277, ... (Sloane's A23200). The odd number in between (in this case 69) is always divisible by 3, except in the special case of the cousin primes 3 and 7.
smoot: 67 is also the number of inches in a "smoot", a nonstandard unit of length explained further here.
(a Perrin number)
68 is a member of an iterativelydefined sequence called the Perrin (or Ondrej Such) numbers. It is defined in a similar way to the Fibonacci and Pell numbers: A_{0} = 0; A_{1} = 2; A_{2} = 3; A_{N+1} = A_{N1} + A_{N2}. The sequence starts: 3, 0, 2, 3, 2, 5, 5, 7, 10, 12, 17, 22, 29, 39, 51, 68, 90, 119, 158, 209, 277, 367, ... (Sloane's A1608; my MCS197664). Because it has the same iteration rule as the Padovan numbers, both have the same asymptotic ratio, about 1.3247.
(a strobogrammatic number)
Another cult number, notable (among other reasons) for being the same when turned upside down. Such numbers are called strobogrammatic. The last three years with this property were 1691, 1881 and 1961; the next isn't until 6009. The complete sequence of invertable numbers starts: 0, 1, 8, 11, 69, 88, 96, 101, 111, 181, 609, 619, 689, 808, 818, 888, 906, 916, 986, 1001, 1111, 1691, 1881, 1961, 6009, 6119, ... (OEIS sequence A0787). To make an invertible number, start with any combination of the digits 0, 1, 6, 8 and 9; then add the same set of digits in reverse order and inverted. If the middle two digits are 00, 11 or 88, you can make another invertible number by removing one of the duplicated digits. For example, starting with 68 we can make the two invertable numbers 6889 and 689. In 1961 there was actually an article in a mainstream math journal titled "Strobogrammatic Years".
The Jargon File^{73} entry for 69 points out that 69_{16}=105_{10} and 69_{10}=105_{8}. 25 also has this property. See also 69105.
Many have found that 69 is the largest value of which you can calculate the factorial on a scientific calculator: 69! ≈ 1.711×10^{98}, and 70! overflows. Here's something you might find even cooler: Put in 69, then square it 5 times in a row: ((((69^{2})^{2})^{2})^{2})^{2} ≈ 6.969×10^{58}. See also 56.9612....
100 times the natural logarithm of 2, and the basis for the "rule of 72" (see 72).
(a Pell number)
The Pell numbers are similar to the Fibonacci numbers and are generated by the formula A_{n} = 2 A_{n1} + A_{n2}. The sequence runs: 1, 2, 5, 12, 29, 70, 169, 408, 985, 2378, 5741, 13860, 33461, 80782, 195025, 470832, 1136689, 2744210, 6625109, ... (Sloane's A0129; my MCS1684). The ratio of successive terms approaches 1 plus the square root of 2.
70^{2} is a square, and also a square pyramidal number, the only example of this kind apart from 0 and 1; this connects in rather nifty ways to 24 and some larger numbers (see 4900).
A "weird number" is a number whose proper factors add up to more than the number itself (it is "abundant"), but no subset of the factors does. In the case of 70, we have 1+2+5+7+10+14+35 = 74, and there is no way to take any of the numbers away from the left side to make the sum exactly 70. By contrast, the factors of 20 are 1+2+4+5+10 = 22, and you can take the 2 away to get 1+4+5+10 = 20.
70 appears in many places throughout the Hebrew Bible (Gen 46:27, 50:3; Exo 1:5, 15:27, 24:1; Num 7:13 etc., 11:16 etc., 33:9; Deu 10:22; Jud 1:7, 8:30 etc.; 2Ki 10:1; and so on)
The first of a pair of twin primes. Twin primes, like 3 and 5, or 71 and 73, are primes that differ by 2. The first few pairs of twin primes are at: 3, 5, 11, 17, 29, 41, 59, 71, 101, 107, 137, 149, 179, 191, 197, 227, 239, 269, 281, 311, 347, 419, 431, ... (Sloane's A1359). See also 2003663613×2^{195000}±1
The "Rule of 72" is used for interest calculation estimates. As it is usually stated, the "number of periods" needed for a quantity to double at a given interest rate per period is 72 divided by the interest rate. For example, at an interest rate of 4.5%, a quantity will double in 16 periods because 72/4.5 = 16.
To anyone familiar with the actual formulas used to calculate such things (which use logarithms and exponents), 72 is clearly an approximation. The number it approximates is 69.3147..., which is 100 times ln(2). The 100 comes from the fact that interest rates are in percentage units.
However, a "rule of 69.3" only works well for very small interest rates. This is a consequence of the compounding effect seen in any exponential function (for example: 1.1^{2} exceeds 1.2, and 1.1^{3} exceeds 1.3 by an even greater amount, and so on).
The approximation is most exact when the interest rate is near 8% (for which the number of periods is 9). The further from this (in either direction), the less well the approximation is. This table shows several different interest rates, expressed as a percentage and as a compounding (multiplication) factor, the number of periods (which usually includes a fraction) needed for exact doubling, and the answer given by the "rule of 72":

The table also shows the best "rule of" number for each interest rate. For smaller interest rates like 2% or 4%, a "rule of 70" works better, and for very small interest rates (or for daily and continuous compounding), a "rule of 69" or "rule of 69.3" is even better.
The pixel size "72 pixels per inch", used on Apple systems back to the original Macintosh, is based on the "pixel" being one point, with 12 points per pica and 6 pica per inch. It corresponds to 2834.64566929... pixels per meter.
72 appears in many spiritual traditions, inclusing Judaism (Cabalists in particular); Christianity (Jesus is said to have had 72 disciples, the 12 apostles plus 60 more); Confucianism (like Jesus, Confucius had 72 disciples); and Islam (72 martyrs of Karbala, 72 sects or 72 plus the saved community, etc. and also appears often in the Arabian Nights).
The number of typographical point per inch advocated by Donald Knuth in connection with his ^{T}_{E}^{X} computerised typsetting system, which derives from a definition of the "American pica" as being 0.013837 feet. There are 12 points per pica and 12 inches per foot, so the precise value would be 1/0.013837 = 72.2700007227...
There are other similar values. The comp.fonts FAQ (from 1996) states, "[...] the United States Typefounders Association in 1886 [...] realised that the inch was an archaic measure [and defined] the point [to be] 1/12 of a pica, and an 83pica distance was made equal to 35 centimeters [...]". This gives (2.54×83×12)/35 = 72.281142... pica/inch. A glossary by digital typefoundry Foam Train states, "In traditional printersâ€™ measure, the pica is 4.22 mm or 0.166 inch: close to, but not exactly, one sixth of an inch." This gives either 12/0.166 = 72.289156... or 25.4×12/4.22 = 72.227488... It seems likely that "72.27" and these alternate values derive from the same source, converted and rounded in different ways.
73 is a prime of the form 4n+1, and all such primes are expressible as a sum of two squares.
This was claimed by Fermat (and called "Fermat's Two Squares Theorem" or simply "Fermat's theorem" though Fermat never actually provided a proof) and first proven by Euler. That proof, along with later proofs by Lagrange and Dedekind, and a proof based on Minkowski's theorem are all pretty long, hard to follow, and/or require more advanced concepts than simple arithmetic and algebra. But Zagier's "onesentence proof", based on HeathBrown in turn based on Liouville, is very easy to see if converted into a geometric representation.
We start by noting that since the prime p is odd, if it is to be the sum of two squares, one will be an odd number squared. Call that x^{2}. The remaining part will be a multiple of 4 (because all odd squares are of the form 4m+1 and because our total p is also 4n+1). Therefore, for each odd integer x less than √p, there will be one or more ways to choose integers y and z so that p=x^{2}+4yz.
Zagier showed that the easy way to prove that there is a sum of two squares is to show two different ways of pairing up the different choices of (x,y,z) that satisfy p=x^{2}+4yz.
The first and very obvious pairing method involves swapping y and z. Here we list all the x^{2}+4yz for p=73, pairing the ones that are equivalent by swapping y and z:
73 = 1^2 + 4*1*18 73 = 1^2 + 4*18*1 73 = 1^2 + 4*2*9 73 = 1^2 + 4*9*2 73 = 1^2 + 4*3*6 73 = 1^2 + 4*6*3 73 = 3^2 + 4*1*16 73 = 3^2 + 4*16*1 73 = 3^2 + 4*2*8 73 = 3^2 + 4*8*2 73 = 3^2 + 4*4*4 = 3^2 + 8^2 73 = 5^2 + 4*1*12 73 = 5^2 + 4*12*1 73 = 5^2 + 4*2*6 73 = 5^2 + 4*6*2 73 = 5^2 + 4*3*4 73 = 5^2 + 4*4*3 73 = 7^2 + 4*1*6 73 = 7^2 + 4*6*1 73 = 7^2 + 4*2*3 73 = 7^2 + 4*3*2As you can see, every way of expressing 73=x^{2}+4yz can be grouped in pairs, except the single solution where y=z which demonstrates that 73 is the sum of two squares.
If we could prove that there is always an odd number of ways to express a prime p as x^{2}+4yz, then we would know that there is always at least one that cannot be paired up as shown in this example, meaning at least one case with y=z and therefore a way to express the prime as a sum of two squares.
This turns out to be easy to prove! The way to do it involves using a different pairing of the x^{2}+4yz forms, which Zagier expressed this way:
(x,y,z) → (x+2z, z, yxz) if x < yz
(1)
(x,y,z) → (2yx, y, xy+z) if yz < x < 2y
(2)
(x,y,z) → (x2y, xy+z, y) if x > 2y
(3)
These conversions between (x,y,z) triples can be illustrated clearly by showing that the pairings correspond to windmilllike shapes such as these:
b b a a b b a a a a a a a a a a b b a a a a a a a a a a a a a a a a a a b b d d d * * * * * * * a a a a a a a a a a a a a a a a b b d d d * * * * * * * a a a a b b b b d d d * * * * * d d * * * b b * * * * * * * d d d d * * * b b b b d d d * * * * * b b b d d * * * b b * * * * * * * d d d d * * * b b b b d d d * * * * * b b b d d * * * b b * * * * * * * d d d d * * * b b b b d d d * * * * * b b b d d c c c c c c c c * * * * * * * b b b d d d d c c c c * * * * * b b b d d c c c c c c c c * * * * * * * b b b c c c c c c c c d d c c c c c c c c c c d d c c c c c c c c c c d d c c 73 = 3^2 + 4*4*4 = 3^2 + 8^2 73 = 5^2 + 4*4*3 73 = 3^2 + 4*8*2 73 = 7^2 + 4*2*3 (3,4,4) > (2*43, 4, 34+4) (3,8,2) > (3+2*2, 2, 832) by conversion (2) by conversion (1) a b a b a b a b a b a a b a d d d d d d * * * * * * * a a a a a a a a a a a a b : * * * * * * * d * * * * * b a * * * * * * * d * * * * * b a * * * * * * * d * * * * * b d d...d d * b b...b b * * * * * * * d * * * * * b c * * * * * * * d * * * * * b c (18 letters * * * * * * * b b b b b b d c c c c c c c c c c c c : on each "arm") c d c c d c c d c d 1^2+4*1*18 c d (1,1,18) > (1,1,18) c d by conversion (2) 73 = 7^2 + 4*1*6 73 = 5^2 + 4*12*1 (7,1,6) > (72*1, 71+6, 1) by conversion (3)The rules for making one of these shapes are: x^{2} is a square of *'s in the centre; four blocks of yz are added as rectangles of a, b, c, d in clockwise order; in the rectangle of as the width is y and the height is z; and choose the arrangement that has the lowerright a directly above the upperright *.
The way to pair up the shapes should be clear from the examples illustrated, but notice you often have to flip one over (mirrorimage) for the shapes to match exactly.
Except for the plain cross shape shown last, where x=y=1, each shape occurs exactly twice, once with the inner square of *'s completely surrounded by letters a, b, c, d, and the other not. This is fairly easy to see when you consider the task of taking the whole shape and dissecting it into a central square and four congruent rectangles. (It may be useful to remember that y must be different from x because x^{2}+4yz is prime, except in the special case of x=y=1; and similarly for z.)
Because there is only one case possible with x=y=1, and all the others occur in pairs, we can see that there will always be an odd number of ways to express p=x^{2}+4yz. So if we go back to the first method of listing choices of x^{2}+4yz, the number of those must be odd too, so there is at least one with y=z, which proves that p is the sum of two squares.
A good description of this proof, with better illustrations, is at MathOverflow here. Mathologer also made this video about it.
A reader wrote to inform me (thanks to the writers of The Big Bang Theory) that 73 is more notable than I had thought. It and its "reversal" 37 are prime (this is called an emirp, see Sloane's sequence A6567), and that 73 is the 21^{st} prime while 37 is the 12^{th} prime: 12 and 21 are also "reversals". Also 73 is a palindrome in base 2 and a repunit in base 8.
There was also a magazine called 73, named after the Morse code abbreviation for "best regards", part of the Western Union 92 Code.
(the "uninteresting number paradox")
This is currently the first number about which I have nothing (else) to say.
Note that such things are highly objective — for example, one reader told me that the 8^{th} time that the digit 8 occurs in π is in the 74^{th} digit (which of course depends on whether or not you count the 3 as the first digit of π). I don't consider that interesting enough to talk about (except as an example of why such things aren't interesting) but the writer and many other people clearly do.
One can also question my choices of which number to use for a given attribute. For example 129 is used for the Cullen numbers, but that description could have gone under 51, 73, 99, etc. As described in the introduction, I try to place such descriptions under the first number that does not have some other (somehow "better") property. For quite a while there was nothing else under 129, and one could say that I was "cheating" in order to have an entry for 129.
(ordered Bell numbers)
The "ordered Bell numbers" count the number of ways of placing N distinguishable balls into one or more distinguishable urns (but with no empty urns). The sequence runs: 1, 3, 13, 75, 541, 4683, 47293, 545835, 7087261, ... (OEIS sequence A0670). For example:
1: (1)
3: (12) (2)(1) (1)(2)
13: (123)
(23)(1) (1)(23)
(13)(2) (2)(13)
(12)(3) (3)(12)
(3)(2)(1) (3)(1)(2)
(2)(1)(3) (2)(3)(1)
(1)(2)(3) (1)(3)(2)
75: (1234)
(1)(234) (234)(1) (a total of 2×4 cases like this)
(12)(34) (34)(12) (a total of 6 cases like this)
(1)(2)(34) (2)(1)(34) (2)(34)(1)
(1)(34)(2) (34)(1)(2) (34)(2)(1) (a total of 6×6 cases like this)
(1)(2)(3)(4) (a total of 24 cases like this)
1+8+6+36+24=75
If the urns are indistinguishable, the sequence is the normal Bell numbers. If the balls are indistinguishable, then we get the powers of two.
(the Lucas sequence)
A member of the Lucas number sequence (not to be confused with the LucasLehmer sequence), a Fibonaccilike sequence that begins with 2, 1 rather than with 1, 1. The sequence, Sloane's A0032, begins: 2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843, 1364, ... (more terms here). It can be computed the same way as the Fibonacci numbers: Start with 2, 1, and each time you add two consecutive terms to get the next term. The sequence grows at the same rate as the Fibonacci numbers, asymptotically approaching a ratio of phi=1.618033... between consecutive terms.
A Lucas number squared, plus or minus two, gives an evenindexed Lucas number:

More similar formulae are in the OEIS sequence A0032.
(a product of consecutive primes)
77 is the product of two consecutive primes: 7 times 11. This sequence, a subset of the distinct semiprimes, begins: 6, 15, 35, 77, 143, 221, 323, 437, 667, 899, 1147, 1517, 1763, 2021, ... (Sloane's integer sequence A6094). See also 143, 323 and 1001.
77 is also the sum of the first 8 primes; see 58 for more.
77 = 4^{2}+5^{2}+6^{2} — see also 25, 143, 216 and 8000.
The smallest number with a persistence of 4: 7×7 = 49; 4×9 = 36; 3×6 = 18; 1×8 = 8.
77 is a Grafting number; see that for more.
(3prime composites)
78 is divisible by three distinct primes. This property is rare at first: the sequence starts 30, 42, 60, 66, 70, 78, 84, 90, 102, ... (Sloane's A0977). However, after a while this type of composite begins to outnumber the semiprimes.
79 is a prime of the form 4n+3, and like all such primes, there is no way to express it s a sum of two squares. This is quite easy to show: all squares are of the form 4n or 4n+1, so any sum of two squares can only be of the forms 4n, 4n+1, or 4n+2.
Since it is not a sum of two squares, a number of the form 4n+3 cannot be expressed as a product of a conjugate pair of Gaussian integers (a+bi)(abi). This means that any prime of the form 4n+3 is also a Gaussian prime.
The primes 4n+3 are: 3, 7, 11, 19, 23, 31, 43, 47, 59, 67, 71, 79, 83, 103, 107, 127, 131, 139, 151, 163, 167, 179, 191, 199, ... (Sloane's A2145). See also 97.
81 = (8+1)^{2}
The largest 2digit square in any base (such as 81 in the case of base 10) always has the property that the sum of its digits is equal to its square root. For example, consider base 7:
51_{7} = 5×7+1 = 36 = 6^{2} = (5+1)^{2}
Algebraically, x^{2} = (x+1)(x1)+1.
See also 9, 89, and 121. (This item was suggested by Ivan Zlatarski)
In physics, 82 is one of the nuclear magic numbers. These numbers arise in the study of nuclear stability. Nuclear isotopes that have a "magic number" of protons or of neutrons are more likely to be stable, or if unstable, have a longer decay halflife, than nuclei with nonmagic numbers of protons and neutrons. The explanation for this (involving the Pauli exclusion principle) is similar to the numbers governing chemical stability of various elements and chemical bonds, based on the number of electrons. The sequence of magic numbers (as far as is known so far) is: 2, 8, 20, 28, 50, 82, and 126. 40 is also "semimagic" for protons, and there is speculation that 114 is magic for protons. Nuclei for which both the proton and neutron count is magic are "doubly magic"; these include the second (^{4}He) and third (^{16}O) most abundant isotopes in the universe. The other doubly magic nuclei are ^{40}Ca, ^{48}Ca, ^{48}Ni, ^{56}Ni, ^{100}Sn, ^{132}Sn and ^{208}Pb (those shown in bold are stable). See also 111.
The Tetrahedral Numbers
84 is the 7^{th} tetrahedral number, which is the sum of the first 7 triangular numbers. These numbers count how many identical round objects can be stacked in a triangular pyramid with N layers (see 20 for an example).
The sequence of tetrahedral numbers starts: 1, 4, 10, 20, 35, 56, 84, 120, 165, 220, 286, 364, ... (OEIS sequence A0292). You get the next tetrahedral number by adding the appropriate triangular number: 15 is the 5^{th} triangular number, and 20+15=35, so 35 is the 5^{th} tetrahedral number. Perhaps less obvious is the fact that you can add the nexthigher square to get the following tetrahedral number: 6^{2}=36, and 20+36=56, so 56 is the 6^{th} tetrahedral number. This relation holds because each square is the sum of two consecutive triangular numbers (In this case, 36 is the sum of the 5^{th} and 6^{th} triangular numbers).
A tetrahedral number can also be expressed as a nice sum of products, illustrated by the example: 4×1 + 3×2 + 2×3 + 1×4 = 4+6+6+4 = 20. The reason for this is more clear if you imagine taking the tetrahedral stack of oranges and balancing it along one edge. Then the 4 layers look like this:
o . . o o o . . . o o o . . o o . . . . o o o o o . . . o o o . . o o (top) o (bottom)84 ("eightyfour") is the only number whose name consists of exactly ten letters, all of them different. There are rather few numbers whose names have no duplicated letters: 0, 1, 2, 4, 5, 6, 8, 10, 40, 46, 60, 61, 64, 80, 81, 84, and 5000. See also 34, 1005, 1025, 1084, and 1000000000008020.
(the Padovan sequence)
86 is a member of a sequence called the "Padovan" sequence, defined similarly to the Fibonacci and pell sequences: A_{0} = 0; A_{1} = 1; A_{2} = 1; A_{N+1} = A_{N1} + A_{N2}. The sequence runs: 1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28, 37, 49, 65, 86, 114, 151, 200, 265, 351, ... (OEIS sequence A0931; my MCS197665) Because it has the same iteration rule as the Perrin numbers, both have the same asymptotic ratio, about 1.3247.
Narayana's cows
88 is a member of a sequence called "Narayana's cows", defined similarly to the Fibonacci sequence: A_{0} = 0; A_{1} = 1; A_{2} = 1; A_{3} = 1; A_{N+1} = A_{N1} + A_{N3}. The sequence runs: 1, 1, 1, 2, 3, 4, 6, 9, 13, 19, 28, 41, 60, 88, 129, 189, 277, 406, 595, 872, 1278, 1873, 2745, 4023, ... (OEIS sequence A0930; my MCS401473). It has an asymptotic ratio of about 1.46557.
(a Fibonacci number)
89 is a Fibonacci number. The Fibonacci sequence starts:
F_{0}=0, F_{1}=1, F_{2}=1, F_{3}=2, F_{4}=3, F_{5}=5, F_{6}=8, F_{7}=13, F_{8}=21, F_{9}=34, F_{10}=55, F_{11}=89, F_{12}=144, F_{13}=233, F_{14}=377, F_{15}=610, F_{16}=987, F_{17}=1597, F_{18}=2584, F_{19}=4181, F_{20}=6765, F_{21}=10946, F_{22}=17711, F_{23}=28657, F_{24}=46368, F_{25}=75025, F_{26}=121393, F_{27}=196418, ...
(Sloane's sequence A0045; my MCS840). Each is the sum of the two before it.
The sequence is named after the preRenaissance Italian Leonardo Pisano (Leonardo of Pisa) whose original name was Leonardo Fibonacci. He wrote the book Liber abaci, a text on the use of HinduArabic numerals and calculation techniques. The book includes this problem:
A man put a pair of rabbits in a place surrounded on all sides by a wall, to find out how many pairs of rabbits will be produced there after a year, if it is assumed that every month a pair of rabbits produces a new pair, and that rabbits begin to bear young two months after their own birth.
During the first month there is the original pair of rabbits. A new pair is born at the beginning of the second month, so during the second month there are two pairs. The first pair produces another pair at the beginning of the third month, but the second pair is not old enough to reproduce yet, so during the third month there are three pairs. At the beginning of the fourth month two pairs are born, one to the first pair of rabbits and another to the pair that was born at the beginning of the second month, so during the fourth month there are five pairs. Continuing in this manner, we see that the number of pairs of rabbits during each month are:

Since the story starts at the beginning of the first month, "after 1 year" would have to be the beginning of the 13th month, so Fibonacci's answer is 377 pairs of rabbits. It has also been argued that "end of the 12th month" is more appropriate, giving the answer 233. And if the original pair isn't counted ("how many pairs of rabbits are born?"), the answer is 232 or 376.
The Fibonacci sequence is the first recursive sequence studied in Europe; the relationship to phi was noticed in the 18th century and the occurrences of Fibonacci numbers in nature (such as flower petals) in the 19th century.
You can generate the Fibonacci numbers on most "cheap" calculators by pressing: 1 + = + = + = + =... Every time you hit + = you'll get another Fibonacci number.
There are a lot of surprising properties of the Fibonacci sequence. Here are some examples:
 Any set of three consecutive Fibonacci numbers is pairwise relatively prime — in other words, for any n, F_{n}, F_{n+1} and F_{n+2} share no common factors except 1.
 For any two positive integers m and n, F_{m+n} = F_{m}F_{n+1} + F_{m1}F_{n}. (Example: m=4, n=3, F_{m}=3, F_{n}=2, F_{m+n}=F_{7}=13, F_{m}F_{n+1}=3×3=9, F_{m1}F_{n}=2×2=4)
 For any positive integer n, F_{2n+1} = F_{n}^{2} + F_{n+1}^{2}. It follows that F_{n}, F_{n+1} and F_{2n+1} have no common factors.
 Similarly, F_{2n} = F_{n+1}^{2}  F_{n1}^{2}, and F_{n1}, F_{n+1} and F_{2n} have no common factors.
 For any positive integer n, F_{3n} = F_{n+1}^{3} + F_{n}^{3}  F_{n1}^{3}.
 For any two positive integers m and n, F_{mn} is divisible by F_{m} and by F_{n}. An equivalent statement is: if p is divisible by n, then F_{p} is divisible by F_{n}. This makes the fibonomial triangle possible. (However, as a general rule F_{n} is not divisible by n, except for a few notable exceptions like F_{12}=144.)
 Apart from F_{4}=3, all prime Fibonacci numbers F_{n} have a prime subscript n. (The reverse is not true, see 4181.)
 For any positive integer n, there are integers a, b and c such that a^{2}+b^{2}=c^{2} and F_{n}×F_{n+1}×F_{n+2}×F_{n+3}=a×b/2 — in other words, the product of any four consecutive Fibonacci numbers is the area of a Pythagorean right triangle. (See also 510510.)
 For any positive integer n, F_{n} = ((1+√5)^{n}(1√5)^{n})/(2^{n}√5).
 For all n greater than 6, F_{n}1 and F_{n}+1 are both composite.
89 is a particularly special Fibonacci number in base 10 because 1/89 is
0.01123595505617977528089887640449438202247191011235...
a repeating decimal of 44 digits which is also equal to the sum
SUM [ F_{n} / 10^{n} ]
which is the sum of
0.0 0.01 0.001 0.0002 0.00003 0.000005 0.0000008 0.00000013 0.000000021 + ... ...  0.01123595505...adding all the Fibonacci numbers while shifting each one one place further to the right. The fact that this sum creates a repeating decimal (and furthermore, the reciprocal of a Fibonacci number) seems at first to be nonobvious and surprising. 9899 is similar; see also my article Fractions with Special Digit Sequences.
We can take the same sort of sum for any base. For base 2 the sum is exactly 1.0, and for base 3 the sum is 1/5 = 0.01210121..._{3}. In the general case for base B this sum is 1/(B^{2}B1). In most bases B^{2}B1 isn't a Fibonacci number. The only bases less than base 1000 for which it is are bases 2, 3, 8, and 10. The Fibonacci numbers grow exponentially and as we go to higher Fibonacci numbers the odds of any given Fibonacci number F being of the form B^{2}B1 are about 1 in √F, so statistically we don't expect to find any more, and it's actually quite surprising that there are as many as 4. (See 454539357304421 for another example of this type of phenomenon.)
89 is also a Fibonacci prime, a number that is both Fibonacci and prime. Based on the fact (stated above) that F_{ab} is divisible by both F_{a} and F_{b}, it is easy to show that for F_{n} to be prime, n must be prime except for the one special case F_{4}=3. The converse does not apply however; the first counterexample is 4181. The first few Fibonacci primes are: F_{3}=2, F_{4}=3, F_{5}=5, F_{7}=13, F_{11}=89, F_{13}=233, F_{17}=1597, F_{23}=28657, F_{29}=514229, F_{43}=433494437, F_{47}=2971215073, F_{83}=99194853094755497, ... (Sloane's A5478).
First page . . . Back to page 7 . . . Forward to page 9 . . . Last page (page 25)
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.
s.11