Notable Properties of Specific Numbers
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(an oblong number)
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 A2378; my MCS1764) See also 210, 336, and 19958400.
For any whole number n, compute the square root of e^{n} and round up. For n=0, 1, 2, 3, ... through n=8 you get Fibonacci numbers: 1, 2, 3, 5, 8, 13, 21, 34, 55. But wait — when n=9 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. (See also 665.141633...).
Amateur mathematicians fall into this sort of trap all the time, and it frequently ensnares even the most professional and experienced. It is now known as "Guy's Strong Law of Small Numbers":
There aren't enough small numbers to meet the many demands made of them. - Richard K. Guy [153]
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.
Guy's paper [153] gives many examples of possible coincidences involving small numbers, presenting them as questions to answer "yes" or "no". His example 29 also mentions the number 91:
Example 29. Does each of the two diophantine equations
2x^{2}(x^{2}-1) = 3(y^{2}-1) and x(x-1)/2 = 2^{n}-1
have just the five positive solutions x = 1, 2, 3, 6, and 91?
and later gives the terse answer "True, but why the coincidence?" with no references. This has been an active problem for some time, and shows up in several different forms, some more clearly equivalent than others. It began, as have many other such curiosities (see for example 22 and 163), with Ramanujan.
In 1913 Ramanujan conjectured that the only values of 2^{N}-7 that are perfect squares are:
2^{3}-7 = 1 = 1^{2},
2^{4}-7 = 9 = 3^{2},
2^{5}-7 = 25 = 5^{2},
2^{7}-7 = 121 = 11^{2}, and
2^{15}-7 = 32761 = 181^{2}.
Ljunggren independently conjectured the same thing in 1943, and it was proven by Nagell in 1948 [137].
The problem of finding a Mersenne number that is triangular can be stated by the equation 2^{p}-1=z(z+1)/2. This problem turns out to be equivalent: starting with Ramanujan's 2^{n}-7=x^{2}, substitute x=2y+1 to get 2^{n}-7=4y^{2}+4y+1 which is the same as 2^{n}=4y^{2}+4y+8; then substitute n=m+2 to get 2^{m+2}=4y^{2}+4y+8 which is the same as 2^{m}=y^{2}+y+2 or 2^{m}-2=y^{2}+y=(y+1)y; then substitute m=p+1 to get 2^{p+1}-2=y(y+1) which is the same as 2^{p}-1=y(y+1)/2. So there are five solutions to this, which are:
2^{0}-1 = 0 = T_{0},
2^{1}-1 = 1 = T_{1},
2^{2}-1 = 3 = T_{2},
2^{4}-1 = 15 = T_{5}, and
2^{12}-1 = 4095 = T_{90}.
using T_{n} to refer to the n^{th} triangular number. Richard Guy states this as "x(x-1)/2 = 2^{n}-1" which makes x=91 be the largest solution.
The other equation given by Richard Guy is 2x^{2}(x^{2}-1) = 3(y^{2}-1). It is fairly easy to see that since x and y are both integers, the left side is even and the right side is divisible by 3, and therefore both sides are divisible by 6; it follows that y is odd, so we can substitute y=2z+1 giving 2x^{2}(x^{2}-1) = 3(4z^{2}+4z+1-1) = 24z(z+1)/2 which is equivalent to x^{2}(x^{2}-1)/2 = 6z(z+1)/2. Once again using the abbreviation T_{n} for the n^{th} triangular number, the equation is T_{(x2-1)} = 6T_{z}, which is a pretty simple form, more likely to draw interest. The five solutions are:
2x^{2}(x^{2}-1) | = 3(y^{2}-1) | ; | T_{(x2-1)} | = 6T_{z} | |||
2×1^{2}(1^{2}-1) | = 0 | = 3(1^{2}-1) | ; | T_{(12-1)} | = T_{0} | = 0 | = 6T_{0} |
2×2^{2}(2^{2}-1) | = 24 | = 3(3^{2}-1) | ; | T_{(22-1)} | = T_{3} | = 6 | = 6T_{1} |
2×3^{2}(3^{2}-1) | = 144 | = 3(7^{2}-1) | ; | T_{(32-1)} | = T_{8} | = 36 | = 6T_{3} |
2×6^{2}(6^{2}-1) | = 2520 | = 3(29^{2}-1) | ; | T_{(62-1)} | = T_{35} | = 630 | = 6T_{14} |
2×91^{2}(91^{2}-1) | = 137133360 | = 3(6761^{2}-1) | ; | T_{(912-1)} | = T_{8280} | = 34283340 | = 6T_{3380} |
In 1995, Mignotte and Pethö [160] considered the paired system of equations x^{2}-6y^{2}=-5, x=2z^{2}-1, which can be expressed as a single diophantine equation by substituting the second equation into the first to get (2z^{2}-1)^{2}-6y^{2}=-5 (which clearly retains the requirement that the subexpression 2z^{2}-1 be an integer). This equation is equivalent to 4z^{4}-4z^{2}+1-6y^{2}=-5; 4z^{4}-4z^{2}=6y^{2}-6; 4z^{2}(z^{2}-1)=6(y^{2}-1); 2z^{2}(z^{2}-1)=3(y^{2}-1) which is thus the same as Guy's equation. It had long since been shown to be "effectively computable" (by Siegel in 1926) and the practical enumeration of the results was performed by computer in 1969 by Baker, but Mignotte and Pethö were the first to prove it. J.H.E. Cohn, who had failed in 1972 to prove it by elementary means, was then able to succeed [166]. In 2004 the result was generalised to other similar forms, such as (2z^{2}-1)^{2}-8y^{2}=-7, by Maohua Le [184] who used the generalised method to extract the above five solutions.
The number of muscles in the nematode worm C. elegans. See 959.
(a 3-smooth number)
96 = 2^{5}×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, ... This list is Sloane's A3586, and is created by multiplying all the powers of 2 by 3, 9, 27, and all higher powers of 3. The 3-smooth numbers can be used as the basis of a double base number system. See also 150.
96 is also of the form 3×2^{n}, which puts it in a more commonly seen subset of 3-smooth numbers; see 192, and 768.
Except for 4 itself, multiples of 4 ("doubly-even numbers") can be expressed as a difference of (positive) squares, and if they have enough even factors this can be done in multiple ways. This is done by expressing the number as a product of two even numbers and then changing that to (a+b)(a-b) = a^{2}-b^{2}. For example,
96 = 2×48 = (25-23)(25+23) = 25^{2}-23^{2}
= 4×24 = (14-10)(14+10) = 14^{2}-10^{2}
= 6×16 = (11-5)(11+5) = 11^{2}-5^{2}
= 8×12 = (10-2)(10+2) = 10^{2}-2^{2}
96 is the smallest even number for which this can be done in 4 different ways. The even numbers that set a new record in this way are: 8, 24, 48, 96, 192, 240, 480, 720, 960, 1440, 2880, 3360, 5040, 6720, 10080, 20160, 30240, 40320, 60480, 80640, 100800, 110880, 181440, 201600, 221760, 332640, 443520, 665280, 887040, ... Numbers that are even without being a multiple of 4 are not expressible as a difference between positive squares, because such a difference must be a sum of consecutive odd numbers, and all such sums are odd or are a multiple of 4. See also 105.
97 is a prime of the form 4n+1, and like all such primes, it can be expressed as a sum of two squares a^{2}+b^{2} in exactly one way (as can 2, the smallest prime). Such primes are called "Pythagorean primes" because √p is the length of the hypotenuse of a right triangle whose other two sides are a and b.
Fermat asserted that all primes 4n+1 can be expressed as such a sum, but a proof wasn't given until Euler. Once you know that there is one such sum a^{2}+b^{2} = 4n+1, and if you know about Gaussian primes, it is easy to see that there cannot be a second way to express it. That is because the product (a+bi)(a-bi) = a^{2}+b^{2} = 4n+1, and if there were another sum of squares c^{2}+d^{2} = 4n+1, then (c+di)(c-di) would also be 4n+1; but in the Gaussian integers factorisation is unique.
The primes 4n+1 are: 5, 13, 17, 29, 37, 41, 53, 61, 73, 89, 97, 101, 109, 113, 137, 149, 157, 173, 181, 193, 197, ... (Sloane's A2144). See also 79.
98 is a multiple of a square (2 times 49), and is 2 less than a power of 10. Because of this, the digits of the square root of 2 start with "14" twice, then 21 then 35, all of which are multiples of 7 which is the square root of 49. All of this is not a coincidence and is explained in the entry for the (even more impressive) square root of 62.
See also 99.9998, 998, and 999998.
99 is one more than 2 times a square — 2×7^{2}+1=99. Add a 0 on the left and square the right, and the equation is still true: 2×70^{2}+1=99^{2}.
There is a sequence of numbers with the property that 2×N^{2}±1 is a square, each time involving a Pell number:
2×2^{2}+1=3^{2}
2×5^{2}-1=7^{2} and 2×7^{2}+1=99
2×12^{2}+1=17^{2}
2×29^{2}-1=41^{2} and 2×41^{2}+1=3363
2×70^{2}+1=99^{2}
2×169^{2}-1=239^{2} and 2×239^{2}+1=114243
2×408^{2}+1=577^{2}
2×985^{2}-1=1393^{2} and 2×1393^{2}+1=3880899
2×2378^{2}+1=3363^{2}
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 A1333; my MCS1724932). See also 841.
99.9998 = 2.54×39.37 = 2×31×127^{2}/10000
This is the approximation to 100 formed by the product of 2.54 (the number of centimeters per inch) and 39.37 (an approximation for the number of inches in a meter used in the United States prior to 1975). Because they are approximations, their product is not exactly 100 (the true number of centimeters in a meter) and because both approximations have a fairly small number of digits, the product ends up being off in just the final digit. This causes interesting digit patterns in the reciprocals of 2.54 and of 39.37, similar to the relationship between 1/27 and 1/37 (see 999). In this case there is also a doubling as seen in the reciprocal of 998:
1/2.54 = 0.3937 007874 015748 031496...
1/39.37 = 0.0254 000508 001016 002032...
The reciprocal of each contains the other as its starting digits because their product is just shy of a power of 10. Each subsequent block of 6 digits is multiplied by 2 because the product is a power of 10 times (10^{6}-2). For more of this type of decimal fraction, see my separate article, Fractions with Special Digit Sequences; see also 89, 998, 9801, 9899, and 997002999.
The same numbers in a slightly different form (254/2=127 and 3937×4/254=62), and the connections to 999998, cause the interesting repeating digits of the square root of 62.
Source: David Wilson, via math-fun^{114}
(a hundred)
100 = 1^{3} + 2^{3} + 3^{3} + 4^{3} = 10^{2}, a square and also a sum of consecutive cubes. If you add more cubes, the results are always squares: 100+5^{3} = 225 = 15^{2}; 225+6^{3} = 441 = 21^{2}; 441+7^{3} = 784 = 28^{2}, and so on. These numbers (1, 9, 36, 100, 225, 441, 784, ...) are "hyper-pyramidal" numbers (Sloane's A0537 or MCS3872). Their square roots (1, 3, 6, 10, 15, 21, 28, ...) are the triangular numbers.
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 bigger numbers are learned (perhaps thousand and then million) 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. See the table of big number names.
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.
101 is the smallest integer whose reciprocal has a 4-digit repeating decimal, 1/101 = 0.0099009900990099...
(area of a golden rectangle)
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 A1654; my MCS100945). You can generate them by multiplying Fibonacci numbers, or by the iterative definition: A_{0} = 0; A_{1} = 0; A_{2} = 1; A_{N+1} = 2 A_{N} + 2 A_{N-1} - A_{N-2}.
(7 double factorial)
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 A6882).
Odd numbers can be expressed as a difference of squares, and if they're composite (but not the square of a prime) this can be done in multiple ways. This is done by expressing the odd number as a product of two odd numbers and then changing that to (a+b)(a-b) = a^{2}-b^{2}. For example,
105 = 1×105 = (53-52)(53+52) = 53^{2}-52^{2}
= 3×35 = (19-16)(19+16) = 19^{2}-16^{2}
= 5×21 = (13-8)(13+8) = 13^{2}-8^{2}
= 7×15 = (11-4)(11+4) = 11^{2}-4^{2}
105 is the smallest odd number for which this can be done in 4 different ways. The odd numbers that set a new record in this way are: 15, 45, 105, 315, 945, 1575, 2835, 3465, 10395, 17325, 31185, 45045, 121275, 135135, 225225, 405405, 675675, ... (related to Sloane's A53624). See also 96.
(the Lin-Rado busy beaver S() function)
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=2.56×10^{10} 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 2.5×10^{2879}. There is more on this topic here. See also 1.29149×10^{865}.
108 is 3^{3}×2^{2}×1^{1}, 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, ... (Sloane's A2109). 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^{1/3}Pn)^{1/12}n^{((n+1)!/(2(n-1)!))} × e^{[n^{2}/4 + 1/12 - B_{4}/(2×3×4×n^{2}) - B_{6}/(4×5×6×n^{4}) - B_{8}/(6×7×8×n^{6}) - ... ]}
where B_{N} are the Bernoulli numbers, P is 2^{2/3}π×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 the MathWorld entry ^{87} for more on the K-function.
See also this entry, and the lower and higher superfactorials.
108 is second in the "paperfolding sequence iteration interpreted as a growing sequence of binary numbers"; see 27876.
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.
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 organised 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.
(repunits)
111 is the magic constant of a 6×6 magic square; since the square has 6 rows, the sum of all numbers in the square is 666. See also 15, 34, 65, 153, 176, 177, 1665, 3051, 6561, 616617.
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, R_{4}=1111 can be divided by R_{2}=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 R_{5}=11111 = 41×271. It has been shown that R_{2}, R_{19}, R_{23}, R_{317} and R_{1031} are the only prime repunits smaller than 10^{10000}. That's about as high as we can go at present. In addition, R_{49081}, R_{86453} and R_{109297} are considered probable primes. See also 12345654321 and 2.25573...×10^{15599}.
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 later discovered), this element was first synthesised 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.
The hundredweight is a unit of weight equal to 112 pounds. I am told (by Chris Cotter) that sometime around the 1300's, this name was given to a unit of weight equal to 1/20 of a ton, which is called a "quintal" or a "zentner" in other languages. Since a ton (also called long ton for clarity) was 2240 pounds, the quintal is 112. There is also an "old" hundredweight of 108 pounds. The name has "hundred" in it because it is kind of close to a hundred; abbreviations for hundredweight include the letter C from the Roman numeral for 100.
113 is called a permutable prime because you can rearrange its digits in any order and get a prime: 113, 131 and 311 are all prime. 199 and 337 also have this property; it is thought that all larger ones are repunit primes (the largest known being 1.111111...×10^{1031}).
The aliquot sequence of 119 is: 119, 25, 6, 6, 6, ... Because its sequence ends in a perfect number, but the number itself is not perfect, 119 is called an "aspiring" number.
120 is a value of (3^{n}-3)/2, (Sloane's A29858; my MCS1979): 3, 12, 39, 120, 363, 1092, 3279, 9840, 29523, 88572, 265719, 797160, 2391483, 7174452, ... These numbers are sometimes involved in weighing problems (see 121) and are the denominators in the fractions discussed here.
120 is the smallest K-fold-perfect number for K=3, also called a 3-perfect or multiply perfect number. Its proper divisors add up to 2 times the number itself: 1+2+3+4+5+6+8+10+12+15+20+24+30+40+60=240. (By contrast, for an ordinary perfect number, the sum of proper divisors is equal to the number: 6 is perfect, and its proper divisors are 1, 2, and 3; 1+2+3=6).
This nomenclature contradicts the original classical distinctions of "deficient", "perfect" and "abundant" (see 496) because for the purposes of these "multiply perfect" numbers, the sum of all divisors (not just proper divisors) is taken. The original sense of "perfect number" is thus renamed "2-fold perfect", "2-perfect", or something similar. Thus, 120 counts as a divisor of itself, and the sum is computed as 1+2+3+4+5+6+8+10+12+15+20+24+30+40+60+120=360.
As a "3-fold perfect" number, 120 is followed by 672 and 523776 (Sloane's A5820). Multiply perfect numbers are known for N as high as 11. See also 30240 and 154345556085770649600.
120 = 2×3×4×5 = 4×5×6, the smallest number that can be expressed as a product of consecutive integers in two distinct ways. See 720 for more on this, also 210, 5040, 175560, 17297280 and 19958400.
Because 120 is a factorial, it counts the number of ways N distinguishable objects can be arranged in a row (where in this case N=5):
12345 12354 12435 12453 12534 12543
13245 13254 13425 13452 13524 13542
14235 14253 14325 14352 14523 14532
15234 15243 15324 15342 15423 15432
21345 21354 21435 21453 21534 21543
23145 23154 23415 23451 23514 23541
24135 24153 24315 24351 24513 24531
25134 25143 25314 25341 25413 25431
31245 31254 31425 31452 31524 31542
32145 32154 32415 32451 32514 32541
34125 34152 34215 34251 34512 34521
35124 35142 35214 35241 35412 35421
41235 41253 41325 41352 41523 41532
42135 42153 42315 42351 42513 42513
43125 43152 43215 43251 43512 43521
45123 45132 45213 45231 45312 45321
51234 51243 51324 51342 51423 51432
52134 52143 52314 52341 52413 52431
53124 53142 53214 53241 53412 53421
54123 54132 54213 54231 54312 54321
121 is 11^{2}, and also a sum of powers of 3: 81+27+9+3+1 = 121. All the numbers from 1 to 121 can be expressed by adding or subtracting these 5 powers of 3 — for example, 58 = 81-27+3+1. 121 is a member of the series (Sloane's A3462; my MCS979) that you get by adding the first n powers of 3: 1, 4, 13, 40, 121, 364, 1093, 3280, 9841, 29524, 88573, 265720, 797161, 2391484, 7174453, 21523360, 64570081, ... In general each is (3^{n}-1)/2 for some value of n.
There are a couple old math problems involving weighing that involve numbers of the form (3^{n}-1)/2 (or in a few cases, (3^{n}-3)/2 or (3^{n}+1)/2). One such problem states, "how many weights do you need to be able to weigh things up to 40 ounces on a two-pan balance?" (The answer is 4 weights, of 1, 3 9 and 27 ounces). Another states, "how many weighings do you need to determine which one of 13 coins is of a different weight, if you know for certain that only one is different?" (The answer is three weighings; you start by comparing two groups of 4.^{41})
121 is a Friedman number.
121 is an illustration of the binomial theorem. The digits of 121 are the same as the 3^{rd} row of Pascal's triangle. And "11^{2}=121" in any base larger than base 2 (where you get "1001"). (See also 81)
The number of classification-systems for 7 object types. This value for N=7 was known in 1979, but determining the number for N=8 was so difficult it was not computed until thirty years later, in 2009. See 1015 and my A005646 page.
122 is the oldest age reached by any living person in modern recorded history. Jeanne Calment was born in 1875 and lived until 1997. Her age at death was 44724 days.
This is 5^{3}. The powers of 5 are: 1, 5, 25, 125, 625, 3125, 15625, 78125, 390625, ... . If you learn these you also learn the powers of 1/2, because 1/2^{N} = 5^{N}/10^{N}: 1/2 is 0.5, 1/4=0.25, 1/8=0.125, 1/16=0.0625, and so on. Starting with 25 they all end in "25", and the alternation of 125/625 continues indefinitely. The last 4 digits repeat in a cycle of 4: 0625, 3125, 5625, 8125. The last 5 digits repeat in a cycle of 8, the last 6 repeat in a cycle of 16, and so on. Patterns like this can be used to determine the trailing digits of certain huge numbers (see for example 27^{2727}).
Friedman numbers
126 is a number that can be formed from an expression made from its own digits: 6×21=126. Other such numbers include: 25=5^{2}, 121=11^{2}, 125=5^{1+2}, and so on. They are described in [171], OEIS sequence A36057, and the Wikipedia article on Friedman number.
(a Mersenne number)
127 is 2^{7} - 1, an example of a number of the form 2^{P} - 1 where P is prime. All such numbers are called Mersenne numbers, but most are not prime. For example, the next Mersenne number after 127 is 2^{11} - 1, which is 2047, but 2047 = 23 × 89, not a prime.
If a Mersenne number is prime (as is the case for 127) it is called a Mersenne prime. Here is a list of all known Mersenne primes, and this is the current record.
127 also has the special property that 2^{127}-1 is also prime, see that entry for more.
Given a sequence S, you can create another sequence T by starting with T_{0} as the first term of S, then T_{1} = the T_{0}^{th} term of S, T_{2} = the T_{1}^{th} term of S, and so on. The Ackermann function consists of sequences of this type.
127 is a member of the sequence you get when you do this using the prime numbers for S: The 1^{st} prime is 2, the 2^{nd} prime is 3, the 3^{rd} prime is 5, the 5^{th} prime is 11, and the 11^{th} prime is 31 — Continuing this process, you get: 127, 709, 5381, 52711, 648391, 9737333, 174440041, 3657500101, ... (Sloane's A7097) Each number T_{n} is approximately equal to ln(T_{n-1}) times T_{n-1}. This sequence grows slower than any exponential sequence A_{n} = K^{n} for K>1, but faster than any fixed-power sequence B_{n} = n^{K} for K>1.
See also 8127.
All numbers larger than 128 can be expressed as the sum of two or more distinct squares.
There no larger powers of two for which all of the digits are also powers of two.
128 is a power of two, and therefore vaguely related to computers; also the number of a state highway in the region of Boston, USA where a lot of computer companies got their start.
Gratuitous connection to 27: Make the 7 an exponent: 27 → 2^{7} = 128.
130 equals 1^{2}+2^{2}+5^{2}+10^{2}, the sum of the squares of its four smallest divisors 1, 2, 5, and 10. A member of the website dxdy.ru presented this problem and it was quickly answered.
Let X be a positive integer with at least 4 divisors, and assume that the sum of the squares of the smallest four divisors of X is equal to X. Fairly quickly one can see that X cannot be an odd number — because if so, all the divisors would be odd, and the sum of the 4 squares would be even. And the number cannot be a multiple of 4 either, because then either 1 or 2 of the 4 smallest divisors would be odd (1 is a divisor of every number, 2 and 4 would be divisors, and the other one of the 4 smallest divisors is either an odd prime or 8). Since the square of any odd number is 1 greater than a multiple of 4, the sum of the 4 squares would exceed a multiple of 4 by either 1 or 2, and thus the sum cannot be the same as the original number.
Thus any solution must be of the form 4N+2 for some N. That means 1 and 2 are divisors but 4 is not. The four smallest divisors would be either be (1, 2, P, 2P) for some prime P; or (1, 2, P, Q) for two different primes P and Q. The latter case is again easily ruled out because the sum of squares would be odd. Thus the sum needs to be X = 1^{2} + 2^{2} + P^{2} + (2P)^{2} = 5×(1+P^{2}). Since P divides into the unknown X, it must also divide into the expression 5×(1+P^{2}); and since (1+P^{2}) is not divisible by P, P has to be 5. Thus there is only one solution: X=5×(1+5^{2})=130.
(Contributed by Evegny "прлопр рпопро")
See also 50, 100, 128, 239, 325, 666, 841 and 4900.
132=11×12 is a multiple of 11 with 3 digits that uses each digit from 1 to 3 exactly once. 231=11×21 also satisfies this condition. With four digits, there are four solutions: 1243=11×113, 1342=11×122, 2431=11×221 and 3421=11×311. There is more about these numbers and their relation to an abstract type of "molecule" on this page.
132 is also a Catalan number.
Gratuitous connection to 27: 132=10^{2}+2^{5} and 27=10×2+2+5.
The moon's phases occur on the same day of the year and the same day of the week every 133 years. 133 is 7 × 19, which is the number of days in the week multiplied by the number of years in the metonic cycle.
Gratuitous connections to 27: 1×3^{3} = 27.
135 = 1^{1} + 3^{2} + 5^{3}, a cute arithmetical coincidence. Another example is 89 = 8^{1} + 9^{2}. After the single-digit numbers, such numbers are fairly rare: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 89, 135, 175, 518, 598, 1306, 1676, 2427, 2646798, 12157692622039623539, ... (Sloane's A32799). See also 126, 153, and 3435.
Gratuitous connection to 27: 13.5+13.5=27.
137 is ... uh, let's see ... the prime number closest to the average of 127 and 143! But perhaps it is more notable for being almost the reciprocal of the Sommerfeld fine-structure constant.
(the Sommerfeld fine-structure constant)
Main article: The number 137.035...
This is a recent (CODATA 2014) value for the fine-structure constant (or more precisely, its reciprocal), originally called the "Sommerfeld fine-structure constant" and often referred to by the Greek letter α (alpha). It is a dimensionless constant in physics.
An example of its use is seen in the Bohr Model of the atom, and specifically in the formula for the velocity of the electron in the innermost orbital (or shell with lowest energy state) according to a "classical" analysis where one considers only the effects of electric charge and ignores relativity and quantum mechanics. The electron is considered to be "orbiting" the nucleus like a planet orbiting the Sun, and the velocity of the innermost electron in its orbit turns out to be Zαc, where Z is the Atomic number and c is the speed of light. Since α is about 1/137, that means that if the nucleus had 137 protons the innermost electron would be travelling at the speed of light. When applying relativity and quantum mechanics, the formulas become more complex and the limit ends up being higher, close to atomic number 173. Above that, a bare nucleus would grab an electron out of the vacuum (see Quantum foam) causing its pair positron to be given off as radiation. In nuclear synthesis experiments that make very heavy nuclei, such positrons have been detected.
The fine-structure constant has a long-time cult following among the scientific community and quasi-scientific followers, beginning with Arthur Stanley Eddington and including some interest from more recent scientists such as Richard Feynman. The fact that it is unitless, like π, and that it is connected with the fundamental laws of subatomic physics, seems to make people think that it should have some precise (perhaps Divinely determined) mathematical value. (More details here: The Cult of 137).
See also 2.0023193043768, 1836.15267245 and 1838.6836605.
137.0359990744319454736519206830802368980220378291196412187098...
This is (ln(2+√4-sqrt(e)) - ln(3))^{-e}, and is one of many wonderful approximations that Fine Structure cultists can find using RIES. See the number 137.035... for more.
After learning that 3^{2}+4^{2}=5^{2} and 3^{3}+4^{3}+5^{3}=6^{3}, one might wonder if the pattern continues in a similar way to 4th powers. But it doesn't: 3^{4}+4^{4}+5^{4}+6^{4}=7^{4}-143. See 8000.
143 is 11×13, a product of twin primes. Numbers with this property are: 15, 35, 143, 323, 899, 1763, 3599, 5183, ... (Sloane's integer sequence A37074). This is a more specific case of a number that is a product of consecutive primes.
The last 143 digits of 143^{143} (or, in other words, 143^{143} mod 10^{143}) is a prime number. (That's not really such abig deal, unless you consider how much effort it took someone to figure it out!)
The Minot lighthouse in Massachusetts blinks in a 1-4-3 pattern: once, then four times after a pause, then three times after another pause, then a longer pause before repeating. Seamen (or their wives on shore) would see the light and imagine it was a message from their beloved: "i love you" = 1 letter, 4 letters, 3 letters.
The legend of Aladdin and the other stories of the Arabian Nights might not be as well-known if they had been called "143 Arabian Weeks".
Gratuitous connection to 27: 10^{1.43143} ≈ 27.004. See also MCS01. ralimon ラリモン
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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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