Notable Properties of Specific Numbers

First page . . . Forward to page 3 . . . Last page (page 25)

Many properties, mostly for trivial reasons (see 3.) Perhaps its most profound³ property is that 1³=1⁷=1²⁷=...

This is the ratio between 12 "perfect fifths" (an interval in music, a pitch ratio of precisely 3/2) and 7 perfect octaves (a ratio of precisely 2/1).[200] It is called the Pythagorean comma because of the practice in Pythagoras' time of defining all musical intervals (including all the notes on the major scale) as ratios between some power of 2 and some power of 3.

The value is close to the 51^st root of 2, however, dividing the octave divided into 51 parts with a note-to-note ratio of 2^1/51 = 1.0136839... does not make a better musical scale. In the 53 equal temperament scale the "semitone" closely approximates the ratio 3¹²/2¹⁹.

See also 0.0833333... and 1.202056....

1.66168794963359...

This is the Somos quadratic recurrence constant, √(1√(2√(3√4...))) = (2/1)^1/2 (3/2)^1/4 (4/3)^1/8 (5/4)^1/16 ... and related to Sloane's sequence A52129. See also 576 and 1658880.

1.7632228343519...

The reciprocal of the Lambert W function of e, which has the special property 1.7632228343519^{1.7632228343519}=e. Thus x^x² = (x^x)^x = x_④3 = e^x when x is this number. See 2.506184... for more.

1.7724538509055... = √π

The square root of π. This is the area under a bell curve of unit height and unit standard deviation. It is also equal to Gamma(1/2), and is sometimes also called (-1/2)!, the factorial of -1/2. See also 7.188082....

1.85540442929...

This is the 3^rd hyper-root of 7, using the "higher" definition of the hyper4 operator: if x=1.855404429290030564371386293833..., then x^{x^x} = x^④3 = x↑↑3 = 7.

See also 1.763222..., 1.855404..., 2.506184..., and 56.9612....

Many properties, mostly for trivial reasons (see 3.) See prime, powers of 2, doubling, Fibonacci.

2.00231930436153(53)

The "gyromagnetic ratio of the electron", a dimensionless constant of physics notable for being known very precisely: within 12 significant figures or better. The "(53)" represents the magnitude of uncertainty in the last few digits: instead of ...153, those last 3 digits might be as low as 153-53=100, or as high as 153+53=206. This number is a favourite of amateur theorists in much the same way as the fine-structure constant.

2.263348517438173216473

The thickness in millimeters of MAD magazine number 26, used as the basis of the Potrzebie system of weights and measures by Donald Knuth and published in MAD #33 (June 1957).

2.3130367364335829063839516026417824763...

This is the number whose continued fraction expansion gives all the prime numbers: It is 2+1/(3+1/(5+1/(7+1/(11+1/(13+1/(17+1/(19+1/...))))))). The digits of this constant are Sloane's sequence A064442. Compare to 2.56654383217....

2.5029078750...

(the second Feigenbaum constant)

The second Feigenbaum constant, commonly designated by the Greek letter alpha. (The other Feigenbaum constant is 4.6692016091....)

2.50618414558876925629294092237784727177139605213321283014...

The solution of x^x=10, and notable for being close to √2π.

It is appropriate to note here that this value is transcendental — and there is no "simple" (closed-form) way to turn x^x=y into an expression for x in terms of y. Instead, the function is solved by Newton's method. For example, the value of x for x^x=10 was calculated this way:

step	action	notes
1.	T = 10.0	target value
2.	X = 2.0	first approximation of answer
3.	Y = X^X	calculate the function
4.	dY = Y (1 + ln(X))	derivative with respect to X
5.	new X = old X + (T-Y)/dY	new approximation by Newton's method
6.	go back to step 3	repeat until accurate enough

Despite the rather poor initial approximation X=2, it only takes 8 repetitions of steps 3 4 and 5 to get the first 14 digits correct and 10 repetitions to get 57 digits.

In general, x^x=y can be solved for x in terms of y using the Lambert W function:

x = e^W(ln(y)), or

x = ln(y)/W(ln(y)).

These are special cases of the more general solution to the inverse lower hyper₄ function with the complete derivation given on my hyper4 page.

For instructions on how to calculate the Lambert W function, see the wikipedia page. Alternately, just go to Wolfram Alpha [228] and put in "x^x=10". I use Maxima; here are examples showing the calculation of 1.559610... and 1.763222...:

(%i1) exp(lambert_w(log(2.0))); (%o1) 1.559610469462369 (%i2) log(2.0)/lambert_w(log(2.0)); (%o2) 1.559610469462369 (%i3) % ^ %; (%o3) 2.0 (%i4) exp(lambert_w(log(float(%e)))); (%o4) 1.763222834351897 (%i5) 1.0/lambert_w(1.0); (%o5) 1.763222834351897 (%i6) % ^ %; (%o6) 2.718281828459046 (%i7) exp(lambert_w(log(10.0))); (%o7) 2.506184145588769 (%i8) log(10.0)/lambert_w(log(10.0)); (%o8) 2.506184145588769 (%i9) % ^ %; (%o9) 9.999999999999998

Although x^x=y cannot be solved precisely with "normal" functions like ln(x) and e^x, there is an iterative solution for x^1/x=y (see 1.444667...) that can be adapted to solve x^x=y for certain values of y:

given: x^x=y, y is known, find x.
1. Substitute x=1/z: (1/z)^(1/z) = y
2. Substitute y=1/w: (1/z)^(1/z) = 1/w
3. Take the reciprocal of both sides: z^(1/z) = w
4. If w is not within the range (e^-e,e^(1/e)) — approximately (0.065988, 1.444667) — then either there is no solution to x^x=y or else the solution cannot be found this way.
5. Otherwise, find the limit of the series w, w^w, w^{w^w}, ...: this is z (as shown by Euler).
6. Take the reciprocal to get x.

Example: starting with x^x=10, we get w=1/10. Iterating, we get w^w = 0.7943282347..., w^{w^w} = 0.1605727204..., w^{w^{w^w}} = 0.6909192287..., and so on, converging after about 350 iterations to 0.39901297826...; the reciprocal of 0.39901297826... is 2.50618414558... . (This solution, based on work of Euler, was pointed out to me by Fantini and Kloepfer.)

The equation x^y=y^x is much easier to solve, as is x^y=xy.

2.50662827463... = √2π

Notable for being kinda close to the solution of x^x=10.

2.54

(length of the inch in centimeters)

Since July 1 1959, by international agreement, the inch has been defined to be exactly 2.54 centimeters or 0.0254 meters. See 299792458 for the definition of the meter. See also 1609.344.

The reciprocal 1/2.54 is 0.3937007874015748031496...; see 39.37 and 99.9998.

2.5665438321713888444675291063322857517829728287023...

This is a real number that encodes the prime numbers in a way similar to the terms of a continued fraction. It is associated with Don Blazys, who claims to link it to the fine-structure constant, among other things.

If you know the sequence you want to encode, an approximation of a constant like this can be computed in this way:

100 ' Modified continued-fraction encoding for strictly-increasing 101 ' integer sequence. In this example we encode the prime numbers. 101 ' Terms are used in reverse order 989 ' add more terms here 990 bl = 29; 991 bl = ((1/bl)+1) * 23; 992 bl = ((1/bl)+1) * 19; 993 bl = ((1/bl)+1) * 17; 994 bl = ((1/bl)+1) * 13; 995 bl = ((1/bl)+1) * 11; 996 bl = ((1/bl)+1) * 7; 997 bl = ((1/bl)+1) * 5; 998 bl = ((1/bl)+1) * 3; 999 bl = ((1/bl)+1) * 2; 1000 ' We're done 1010 print bl; 1999 end

Using enough terms you can get a value with many digits of precision; the original terms of the sequence are obtained by reversing the calculation, like so:

integer part of 2.566543... is 2
1/((2.566543.../2)-1) = 3.530176...
integer part of 3.530176... is 3
1/((3.530176.../3)-1) = 5.658487...
integer part of 5.658487... is 5
1/((5.658487.../5)-1) = 7.593155...
(and so on...)

Each step of the calculation removes the integer part of the number, then takes the reciprocal to make a new value that has a nonzero integer part. By using a sufficiantly precise starting value, this iteration can "generate" arbitrarily many terms of the sequence used to generate the constant, in this case the prime numbers. If one had an infinitely precise starting value, one could generate all the primes. See also 2.313036....

As a general-purpose way to encode a sequence of integers, this method is less universal than the normal continued fractions method, because the encoding is one-to-one only when the sequence being encoded is a steadily-increasing sequence.

2.618033..

The square of Φ, the golden ratio. Among other things, it is the value of both sides of the equality:

Φ+1 = Φ²

2.66514414269022518865029724987313984827421131371465... = 2^√2

This number is known as the Gelfond-Schneider constant. Along with e^π, it was shown to be transcendental by Gelfond's theorem (sometimes called the Gelfond-Schneider theorem). This theorem shows that if x is not 0 or 1 but is algebraic, and y is both algebraic and irrational, then x^y is transcendental. It was proven independently by Gelfond and Schneider in 1934.

2.685452...

Khintchine's constant is the geometric mean of the terms of the continued fraction for most (but not all) real numbers. Rational numbers, roots of 3^rd order polynomials and certain other classes of real numbers have patterns in their continued fraction expansion — so they do not obey this principle. However, the real numbers that do obey this average outnumber those that don't, and they outnumber by such a great extent, that if you picked a real number at "random" the probability that it would not have this average is 1/infinity.

2.718281828459045... (e)

(basis of natural logarithms)

e, base of natural logarithms, Euler's number, etc. Competes with the square root of 2 for second-most-famous non-integer after π. e is the limit of

(1 + 1/X)^X

as X goes to infinity. An efficient way to calculate e goes like this:

e = 2
i = 2
f = 1
repeat until satisfied
     f = f/i
     e = e + f
     i = i + 1
end repeat

which uses the well-known infinite series: e is the sum of the reciprocals of the factorials of all the natural numbers.

The solution of x^x=27. :-)

There are so many properties of small numbers like this that it is somewhat tedious to list them all (but if you're interested look here). 3 is a prime, a triangular number, a Fibonacci number, the number of spatial dimensions of our universe, etc. etc.

There are many occurrences of a 3-times repetition in rituals, customs, stories, etc. Some examples include: the three knocks of Black Rod when calling the House of Commons to hear the address of the Sovereign at the opening of Parliament; three verses in the standard song format; the three pigs' houses (straw, wood, stone/brick) in the children's tale. The use of three (rather than two or four) is probably due to the fact that three is the minimum necessary to establish a pattern (such as a regular tempo) or to convey the impression of an ongoing sequence or succession.

3.14159264581... = 3 + ¹/₈ + ¹/₆₁ + ¹/₅₀₂₀

The closest approximation to π using three terms of the greedy Egyptian fraction algorithm. It is somewhat more accurate than this one but requires a larger denominator term.

3.141592653589... = π (pi)

(π (pi), the best-known irrational constant)

The most well-known non-integral mathematical constant, the subject of several books, etc. Pi shows up in many places you don't expect it to (for instance, see 1.644934....)

There are lots of ways calculate π. One of the most efficient is Chudnovsky's series sum, which looks like this:

A = 545140134; B = 13591409; C = 640320;
D = C³; // 262537412640768000
E = C/12; // 53360
F = C/64; // 10005
G = C²/96; // 4270934400
total = 0;
n = 0;
repeat until satisfied:
total = total + -1ⁿ ((6n)!(A n + B))/(n!³(3n)!Dⁿ)
n = n + 1;
end repeat
π = E √C / total
π = 1 / (√F / G) total

The last two lines are equivalent, and come from the two different versions of the Chudnovsky calculation I have seen. Every time you repeat the loop adds more than 14 digits to the accuracy of the approximation.

Take a large integer N. Round up to the nearest multiple of N-1. Then round that number up to the nearest multiple of N-2. Continue until you get to the nearest multiple of 1, and call this final value F. For sufficiently large initial values of N, the ratio N²/F comes arbitrarily close to π. (This is from mathpages.com via MathWorld¹⁰¹; more esoteric formulas at the footnote.)

Transcendental numbers

One of the famous problems from ancient times is called squaring the circle. The original problem was to use a straightedge and compass to determine the size of a square whose area is the same as that of a given circle. This is essentially equivalent to finding a precise formula for π that involves integers, the operations + - × /, and a few simple others like square root.

This problem eventually evolved into a more refined version, that of finding an integer polynomial to which π is a root. Numbers that can be expressed this way are algebraic, those that are not are transcendental. For various reasons it became important to be able to determine if certain numbers are algebraic or transcendental.

The transcendental nature of π was first shown by Lindemann in 1882. See also 1.632526919438... and 2.66514414269....

Digits of Pi

The following formula allows one to compute the Nth digit (in base 16) of π without computing all the preceding digits:

π = SUM [ (4/8n+1 - 2/8n+4 - 1/8n+5 - 1/8n+6) × 16^-n ]

The following poem is a mnemonic for remembering the first 42 digits of π. The number of letters in each word corresponds to a digit, with 10-letter words for '0' digits. It was written in 1995 by Michael Keith based on "The Raven" by Edgar Allen Poe:

3.1 Poe, E. - 415 Near A Raven 926 Midnights so dreary, 535 tired and weary, 8979 Silently pondering volumes extolling 32384 all by-now obsolete lore. 62643 During my rather long nap 383 the weirdest tap! 27950 An ominous vibrating sound disturbing 288 my chamber's antedoor. 419716 "This", I whispered quietly, "I ignore".

If you like this kind of thing, Mr Keith finished "The Raven" and went on to write several other modified versions of well-known bits of literature, then made an entire story around it, which he calls Cadaeic Cadenza. It has about 3800 words corresponding to the first 3834 digits of π. He later created the book Not A Wake which does the same thing in 10000 words, and many other interesting examples of constrained writing, all available at his website cadaeic.net.

3.14159267204... = (e (φ + 6))/5 - 1

One of many close approximations to π. This one achieves the closest possible approximation using 5 single digits or constants and 4 operations. Here are some others: with three constants: √4e-1 = 3.14215...; with four constants: ln((7² - e)/2) = 3.1415998...; with six constants (sort of): (4-φ^√e)×(2/√7+1) = 3.141592653650... and √3+√2+1/(ln(7)-6³) = 3.1415926535515... . (If this sort of thing amuses you, you might want to check out my inverse equation solver; the command line for this problem is: ries 3.1415926535897932 -Ox -x -NpSCL .)

3.1415269531... = 3 + ¹/₁₃ + ¹/₁₇ + ¹/₁₇₃

(an Egyptian value for π)

The numbers 13, 17 and 173 have been found together in ancient Egyptian records without explanation, but it is conjectured that they express an Egyptian fraction approximation to π. If so, it is a much better approximation than those attributed to Archimedes and Ptolemy.

3.14159292035... = 355/113

The approximation to π given by the ratio 355 / 113. This approximation is admired mainly because it is "efficient": the fraction has 6 digits but the approximation gives you about 7¹/₂ digits of π. It's also kind of easy to remember, if you remember the denominator first (because then you get "113, 355" which is a very simple pattern).

Continued Fractions

Along with 22/7, 355/113 is a member of the continued-fraction series for π. Continued fraction approximations can be used to get "optimal" rational approximations for any number. By "optimal" we mean the closest approximation that is possible with a given-size numerator and denominator. The continued fraction approximations for π are: ³/₁, ²²/₇, ³³³/₁₀₆, ³⁵⁵/₁₁₃, ¹⁰³⁹⁹³/₃₃₁₀₂, ¹⁰⁴³⁴⁸/₃₃₂₁₅, ²⁰⁸³⁴¹/₆₆₃₁₇, ³¹²⁶⁸⁹/₉₉₅₃₂, ⁸³³⁷¹⁹/₂₆₅₃₈₁, ^1146408/₃₆₄₉₁₃, ^4272943/_1360120, ^5419351/_1725033, ...

To generate a continued-fraction series, you just keep truncating the fraction and taking the reciprocal:

    3.141592653589... = 3 + 0.141592653589... = 3 + 1/7.062513305931...
    7.062513305931... = 7 + 1/15.996594406685...
  15.996594406685... = 15 + 1/1.003417231013...
    1.003417231013... = 1 + 1/292.634591014395...
292.634591014395... = 292 + 1/1.575818089...

This gives you the terms of the continued fraction: 3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, ... (Sloane's A1203). You use these integers to make the actual continued fraction for π:

π = 3+1/(7+1/(15+1/(1+1/(292+1/...))))

The fact that 355/113 is an "efficient" approximation (as described above) is because the next term in the series (292) is relatively high. On average, continued-fraction series have much smaller terms most of the time (see Khintchine's constant). The beginning of the π series is an exception.

If you stop a continued fraction series at any point and reduce to a simple fraction, you get an approximation:

3+1/(7+1/(15+1/(1))) = 3+1/(7+1/(16)) = 3+1/(113/16) = 3+16/113 = 355/113

If you want to generate a whole series of numerators and denominators, use an algorithm like this:

(given: X = number to be converted) N = 1 D = 0 N1 = 0 D1 = 1 repeat until satisfied: let XI = integer part of X let X = fractional part of X let X = 1 / X let N2 = N1 let N1 = N let N = N1 * XI + N2 let D2 = D1 let D1 = D let D = D1 * XI + D2 print "New approximation: ", N, "/", D end repeat

The approximation 355/113 for π was known to the Chinese mathematician Zu Chongzhi around 480 AD.

3.141666... = 3+⁸/₆₀+³⁰/_60²

(Ptolemaic value for π)

The approximation to pi used by Ptolemy.

3.141851... = 3 ¹⁴¹/₉₉₄ = (3 ¹/₇ + 3 ¹⁰/₇₁) / 2

(Archimedes' value for π)

Some say that Archimedes used a value for pi that was "between 3 ¹/₇ and 3 ¹⁰/₇₁", rather than simply 3 ¹/₇; this would be an improvement since 3 ¹⁰/₇₁ is closer to the true value of pi. 3 ¹⁴¹/₉₉₄ is the average of the two approximations.

3.14187085960562107111501317

An approximation to pi given by the unitless expression:

(Planck energy times the combined fuel economy of a Toyota Prius)
divided by
(the pressure in the centre of the earth times the width of the English Channel at its narrowest point)

which is given in this xkcd cartoon. (All of the units cancel out, as you will see if you follow the links to each of these constants, and the approximation is correct to within the limits of measurement and/or experiment. I have given only 27 digits of precision, but more may be possible in the near future — watch this space! This formula is a great commentary on all the theories described at 137.035.)

3.142857...

22/7 (or 3+¹/₇), an approximation of π used in the time of Archimedes. The difference between 22/7 and π is given by the integral

∫_{0 to 1} [ x⁴ (1 - x)⁴ / (1 + x²) dx ]

3.160493827160... = 256/81 = 4⁴/3⁴ = 3+¹/₉+¹/₂₇+¹/₈₁

4⁴/3⁴ is an approximation to π used by the Egyptians, and it is also sort of close to √10, but it has a much more interesting property:

64/27 × 4/3 = 64/27^4/3 = 256/81.

There are an infinite number of pairs of rational numbers a and b such that a^b=a^.b. To show this, start with:

x^y = x^.y

then define L to be the log base x of y:

L = log_xy

then the first equation becomes:

x^{x^L} = x x^L

which (for x>1) reduces to:

x^L = L+1

Solving for x, and substituting back for y, we get:

x = (L+1)^(1/L)
y = x^L = ((L+1)^(1/L))^L = L+1

This is the general solution for x^y = x^.y. (The same equations also happen to solve x_④y=x^y.)

To get both x and y to be rational numbers, define L to be 1/n for some integer n>1. Then you get:

x = (1/n+1)ⁿ = (n+1)ⁿ/nⁿ = ((n+1)/n)ⁿ
y = (n+1)/n
x = yⁿ

For n=1 you get x=y=2, which isn't a very interesting solution, but for n=2 you get this solution and for n=3 you get x=64/27 and y=4/3, with the product x^.y=x^y=256/81=3.160493827160...

As n gets larger, y approaches 1 and x approaches e because

e = lim (1 + 1/n)ⁿ

See also 1.632526919438... and the solution of x^y=y^x.

3.162277660168... = √10

This has sometimes been used as an approximation to π

3.375 = ²⁷/₈ = (9/4)^(3/2) = (9/4) × (3/2)

The values x=9/4 and y=3/2 give the simplest solution in rational fractions (non-integers) to the equations x^y=xy and x_④y=x^y, where _④ is the lower-valued form of the hyper4 operator.

These equations can be solved easily. We will start with the x_④y=x^y equation. The lower hyper4 function x_④y is equivalent to two exponents:

x_④y = x^{x^y-1}

so we're really solving:

x^{x^y-1} = x^y

If x>1 we have:

x^y-1 = y

and thus:

x = y^1/(y-1)

(If you change y to l you can see the similarity to the solution of x^y=y^x. The same values of l and x that solve the x^y=y^x equation also solve x_④l=x^l.)

We can also express x^y in terms of y:

x^y = (y^1/(y-1))^y = y^y/(y-1)

and now notice that this is equal to the product xy, because:

xy = yx = y^.y^1/(y-1) = y^{(y-1+1)/(y-1)} = y^y/(y-1)

Therefore, the solutions to x_④y=x^y also simultaneously solve x^y=xy.

If 1/(y-1) is an integer then x and y will both be rational numbers; this happens when y is of the form (a+1)/a, where a is any nonzero integer. Then x is:

x = ((a+1)/a)^a = (a+1)^a/a^a

The smallest non-trivial solution is when a=2, we get x=9/4 and y=3/2, and the value of x_④y = x^y = xy is 27/8=3.375.

3.5

This number is implicitly mentioned in the Bible in a few places; see 1260.

First page . . . Forward to page 3 . . . 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 5×10¹¹ 10¹⁸ 5.4×10²⁷ 10⁴⁰ 5.21...×10⁷⁸ 1.29...×10⁸⁶⁵ 10⁴⁰⁰⁰⁰ 10^9152051 10^10³⁶ 10^{10^10¹⁰⁰} — — footnotes Also, check out my large numbers and integer sequences pages.

s.30

	1	3	7	17	41	99	239		a	c	e
	1 ,	2 ,	5 ,	12 ,	29 ,	70 ,	169 ,	. . .	b ,	d ,	f