See also 648.

74

This is currently the first number about which I have nothing (else) to say. See 39.

75

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, ... (Sloane's A000670). 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.

77 = 7×11

77 is the product of two consecutive primes. This sequence, a subset of the distinct semiprimes, begins: 6, 15, 35, 77, [143#lb143], 221, 323, 437, 667, 899, 1147, 1517, 1763, 2021, ... (Sloane's integer sequence A006094). See also 143, 323 and 1001.

77 is also the sum of the first 8 primes; see 58 for more.

77 = 4²+5²+6² — see also 25, 143, 216 and 8000.

78 = 2×3×13

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 A000977). However, after a while this type of composite begins to outnumber the semiprimes.

82

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 half-life, than nuclei with non-magic 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 (⁴He) and third (¹⁶O) most abundant isotopes in the universe. The other doubly magic nuclei are ⁴⁰Ca, ⁴⁸Ca, ⁴⁸Ni, ⁵⁶Ni, ¹⁰⁰Sn, ¹³²Sn and ²⁰⁸Pb (those shown in bold are stable). See also 111.

86

86 is a member of a sequence called the "Padovan" sequence, defined similarly to the Fibonacci and pell sequences: A₀ = 0; A₁ = 1; A₂ = 1; A_N+1 = A_N-1 + A_N-2. 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, ... (Sloane's A000931; my MCS197665) Because it has the same iteration rule as the Perrin numbers, both have the same asymtotic ratio, about 1.3247.

89

A Fibonacci number. The Fibonacci sequence starts:

F₀=0, F₁=1, F₂=1, F₃=2, F₄=3, F₅=5, F₆=8, F₇=13, F₈=21, F₉=34, F₁₀=55, F₁₁=89, F₁₂=144, F₁₃=233, F₁₄=377, F₁₅=610, F₁₆=987, F₁₇=1597, F₁₈=2584, F₁₉=4181, F₂₀=6765, F₂₁=10946, F₂₂=17711, F₂₃=28657, F₂₄=46368, F₂₅=75025, F₂₆=121393, F₂₇=196418, ...

(Sloane's sequence A000045; my MCS840). Each is the sum of the two before it.

The sequence is named after the pre-Renaissance Italian Leonardo Pisano (Leonardo of Pisa) whose original name was Leonardo Fibonacci. He wrote the book Liber abaci, a text on the use of Hindu-Arabic 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_mF_n-1 + F_m+1F_n.
• For any positive integer n, F_2n+1 = F_n² + F_n+1². It follows that F_n, F_n+1 and F_2n+1 have no common factors.
• Similarly, F_2n = F_n+1² - F_n-1², and F_n-1, F_n+1 and F_2n have no common factors.
• For any positive integer n, F_3n = F_n+1³ + F_n³ - F_n-1³.
• 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 F₁₂=144.)
• Apart from F₄=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 a²+b²=c² 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)/(2ⁿ√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 / 10ⁿ ]

which is the sum of

0.0
0.01
0.001
0.0002
0.00003
0.000005
0.0000008
0.00000013
0.000000021
(etc.)

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 non-obvious and surprising. (See 998 for another example).

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/(B²-B-1). In most bases B²-B-1 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 B²-B-1 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 F₄=3. The converse does not apply however; the first counterexample is 4181. The first few Fibonacci primes are: F₃=2, F₄=3, F₅=5, F₇=13, F₁₁=89, F₁₃=233, F₁₇=1597, F₂₃=28657, F₂₉=514229, F₄₃=433494437, F₄₇=2971215073, F₈₃=99194853094755497, ... (Sloane's A005478).

90

90 = 9×10, the product of two consecutive integers. Numbers of this type are called oblong numbers because they correspond to the area of an "oblong" rectangle whose length is one greater than its width. They are also called promic numbers; each is twice a triangular number. The sequence of oblong numbers starts: 2, 6, 12, 20, 30, 42, 56, 72, 90, 110, 132, 156, 182, 210, ... (Sloane's A002378; my MCS1764) See also 210, 336, and 19958400.

91

For each integer n, compute e^(n-2)/2 and round up. For the first several values of n you'll get the Fibonacci numbers: 1, 2, 3, 5, 8, 13, ... but wait! For n=10 you get 91, not the Fibonacci number 89. This is even a little more surprising to those who know that the Fibonacci sequence is an exponential sequence.

This is an example of Guy's Strong Law of Small Numbers:

There aren't enough small numbers to meet the many demands made of them. - Richard K. Guy

In other words, if you encounter a small number in two different phenomena, don't assume any actual cause-and-effect relationship between the two unless you already know (or can prove) that there is one.

96

96 = 2⁵×3. It has no prime factors larger than 3, and this makes it a 3-smooth number. The 3-smooth numbers are: 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, 48, 54, 64, 72, 81, 96, 108, 128, 144, 162, 192, 216, 243, 256, 288, 324, 384, 432, 486, 512, 576, 648, 729, 768, 864, 972, ... (Sloane's A003586) 3-smooth numbers can be used as the basis of a double base number system. See also 150.

96 is also of the form 3×2ⁿ, which puts it in a more commonly seen subset of 3-smooth numbers; see 192, and 768.

99

99 is one more than 2 times a square — 2×7²+1=99. Add a 0 on the left and square the right, and the equation is still true: 2×70²+1=99². There is a sequence of numbers with that latter property, and each time the other number is a Pell number:

2×2²+1=3²
2×5²-1=7²   and 2×7²+1=99
2×12²+1=17²
2×29²-1=41²   and 2×41²+1=3363
2×70²+1=99²
2×169²-1=239²   and 2×239²+1=114243
2×408²+1=577²
2×985²-1=1393²
2×2378²+1=3363²

These are the same numbers shown in the description of the square root of 2 — the numerators and denominators of the fractions (³/₂, ⁷/₅, ¹⁷/₁₂, etc.) that approximate it.

Both sequences are generated the same way: each term is 2 times the previous term plus the term before that. The only difference is that this sequence begins with (1, 1) and the Pell numbers start with (0, 1). This sequence starts out: 3, 7, 17, 41, 99, 239, 577, 1393, 3363, 8119, 19601, 47321, 114243, 275807, 665857, 1607521, 3880899, 9369319, ... (Sloane's A001333; my MCS1724932). See also 841.

100

For many people, 100 is the first number they ever thought of as being "really really big". It probably also held (if only for a brief period) the title of "biggest number I've ever heard of". As you get older, you learn bigger numbers (perhaps thousand and then million) and 100 becomes simply "big". After 100, 1000 and 1000000, the answer to "what's the biggest number you've heard of?" will usually be one or more of: billion, trillion, Avagadro's number, googol and googolplex.

101

The first 3-digit prime number.

To test for divisibility by 101, take the digits of the number in groups of 4 starting from the right, and add the resulting numbers together. If the result is more than 4 digits, repeat the process. If the resulting number is of the form ABAB or A0A (like 2727, 4242 or 808) the original number is divisible by 101. (This test works because of the casting out 9's principle and because 101 is a factor of 9999.) See also 1001.

103

The 27^th prime.

104

104 is 8×13, the product of two consecutive Fibonacci numbers. These numbers are called "golden rectangle numbers"; a rectangle with two consecutive Fibonacci numbers as sides approximates the ideal Golden Rectangle whose aspect ratio is exactly phi. The golden rectangle numbers are: 1, 2, 6, 15, 40, 104, 273, 714, 1870, 4895, 12816, 33552, 87841, 229970, 602070, ... (Sloane's A001654; my MCS100945). You can generate them by multiplying Fibonacci numbers, or by the iterative definition: A₀ = 0; A₁ = 0; A₂ = 1; A_N+1 = 2 A_N + 2 A_N-1 - A_N-2.

105

If you calculate a factorial but leave half the numbers out, you get a (somewhat misnamed) "double factorial". 105 is 7 double factorial (sometimes written "7!!") because 105=7×5×3×1. The double factorials are: 1, 2, 3, 8, 15, 48, 105, 384, 945, 3840, 10395, 46080, 135135, 645120, 2027025, ... (Sloane's A006882).

107

107 is the maximum number of steps that a 4-state, 5-tuple Turing Machine can make, on an initially blank tape, before halting. The busy beaver problem was introduced by Lin and Rado in 1965, who showed that the maximum number of steps for 1, 2, and 3-state machines is 1, 6 and 21 respectively. The maximum for 4 states was shown in 1983 by Allan Brady. This is not a trivial task as the number of possible Turing machines increases exponentially — there are 25600000000=2.56×10¹⁰ different 4-state machines. Brady had to thoroughly investigate many of them using sophisticated pattern-matching techniques to prove that they would never halt. The answers for higher numbers of states are not known. For 5 states it is at least 47176870, and for 6 states an astounding 2.5×10²⁸⁷⁹. There is more on this topic here. See also 1.29149×10⁸⁶⁵.

108

108 is 3³×2²×1¹, or 27×4. Numbers of this form are sometimes called hyperfactorials.

Hyperfactorials and the K-function

The hyperfactorials are: 1, 4, 108, 27648, 86400000, 4031078400000, 3319766398771200000, 55696437941726556979200000, ... . The hyperfactorials can be extended to the real numbers, the result is the K-function, which is related to Barnes' G-function, the Gamma function and the Riemann Zeta function. n hyperfactorial is equivalent to K(n+1). There is an infinite series:

K(n+1) = (2^1/3Pn)^1/12n^{((n+1)!/(2(n-1)!))} × e^{[n²/4 + 1/12 - B₄/(2×3×4×n²) - B₆/(4×5×6×n⁴) - B₈/(6×7×8×n⁶) - ... ]}

where B_N are the Bernoulli numbers, P is 2^2/3π×e^{gamma-1-Z'(2)/Z(2)}, gamma is the Euler-Mascheroni constant, Z is the Riemann Zeta function and Z' is its derivative.

For sufficiently large values, an approximation is given by:

K(n+1) ≅ A n^{n2/2+n/2+1/12}e^-n2/4

where A is the Glaisher-Kinkelin constant. See here for more on the K-function.

See also this entry, and the lower and higher superfactorials.

108 occurs frequently in eastern religions. The Hindu Krishna dances with 108 gopis (cowgirls). In Buddhism, there are 108 arhat (perfected saints); the Tibetan scriptures Tanjur and Kanjur have 108 parts; at Japanese Buddhist temples, on New Year's Eve at midnight, a special bell is rung 108 times (once for each type of evil in the world). See also 1080.

110

A number I have occasionally heard referred to as "eleventy". Most well-known is the "eleventy-first" birthday party for Bilbo in Tolkien's Lord of the Rings (By the way, there were several organized events honoring Tolkien's own eleventy-first birthday on Jan 3rd, 2003). I have also heard "eleventy billion" used as a jocular term for zillion.

111

111 is the next repunit after 11. A repunit is simply a number that consists of the digit 1 repeated. The term repdigit is used to refer to these plus numbers like 666 and 999999 that consist of some other digit repeated. These numbers were first discussed in Albert Beiler's 1966 book, Recreations in the Theory of Numbers (appropriately, in chapter 11). The shorthand R_n is used to refer to the repunit with n 1's. Repunits are the subject of a few conjectures and much recreational investigation. For example, which ones are prime? It is easy to show that R_n is prime only if n is prime — just divide by another repunit R_f where f is a factor of n. For example, R₄=1111 can be divided by R₂=11: 1111=101×11. The opposite need not be true — just because n is prime, does not mean that R_n is prime. For example, 111=3×37, and R₅=11111 = 41×271. It has been shown that R₂, R₁₉, R₂₃, R₃₁₇ and R₁₀₃₁ are the only prime repunits smaller than 10¹⁰⁰⁰⁰. That's about as high as we can go at present. In addition, R₄₉₀₈₁, R₈₆₄₅₃ and R₁₀₉₂₉₇ are considered probable primes. See also 12345654321 and 2.25573...×10¹⁵⁵⁹⁹.

111 is the atomic number of the element below gold on the periodic table. Given the nickname eka-gold (after Mendeleev's names for elements he predicted and were lated discovered), this element was first synthesized in 1994. Its most stable isotope has a half-life of 3.6 seconds. In 2004 it was officially given the name Roentgenium, after Roentgen, the physicist who discovered X-rays and received the first Nobel prize in physics. It is now possible to continue Hofstatder's amusing recursive story (pp. 103-126 of Gödel, Escher, Bach) one level deeper:

Meta-Meta-Genie: I'll have to send it through Channels, of course. One quarter of a moment, please... (And, twice as quickly as the Meta-Genie did, this Meta-Meta-Genie removes from the folds of his robe an object which looks just like the gold Meta-Meta-Lamp, except that it is made of Roentgenium, and it has "MMML" etched on it in even smaller letters, so as to cover the same area.)

Voice-of-Achilles: (another octave higher than before) And what is that?

Meta-Meta-Genie: This is my Meta-Meta-Meta-Lamp... (He rubs the Meta-Meta-Meta-Lamp, and a huge puff of smoke appears. In the billows of smoke, they can all make out a ghostly form towering above them.)

Meta-Meta-Meta-Genie: I am the Meta-Meta-Meta-Genie. You summoned me, O Meta-Meta-Genie? What is your wish?

[and so on...]

See also 82.

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