A020807 -id:A020807 - cp's OEIS frontend

A002193 Decimal expansion of square root of 2.

1, 4, 1, 4, 2, 1, 3, 5, 6, 2, 3, 7, 3, 0, 9, 5, 0, 4, 8, 8, 0, 1, 6, 8, 8, 7, 2, 4, 2, 0, 9, 6, 9, 8, 0, 7, 8, 5, 6, 9, 6, 7, 1, 8, 7, 5, 3, 7, 6, 9, 4, 8, 0, 7, 3, 1, 7, 6, 6, 7, 9, 7, 3, 7, 9, 9, 0, 7, 3, 2, 4, 7, 8, 4, 6, 2, 1, 0, 7, 0, 3, 8, 8, 5, 0, 3, 8, 7, 5, 3, 4, 3, 2, 7, 6, 4, 1, 5, 7

Offset: 1

N. J. A. Sloane

Sometimes called Pythagoras's constant.
Its continued fraction expansion is [1; 2, 2, 2, ...] (see A040000). - Arkadiusz Wesolowski, Mar 10 2012
The discovery of irrational numbers is attributed to Hippasus of Metapontum, who may have proved that sqrt(2) is not a rational number; thus sqrt(2) is often regarded as the earliest known irrational number. - Clark Kimberling, Oct 12 2017
From Clark Kimberling, Oct 12 2017: (Start)
In the first million digits,
  0 occurs  99814 times;
  1 occurs  99925 times;
  2 occurs 100436 times;
  3 occurs 100190 times;
  4 occurs 100024 times;
  5 occurs 100155 times;
  6 occurs  99886 times;
  7 occurs 100008 times;
  8 occurs 100441 times;
  9 occurs 100121 times. (End)
Diameter of a sphere whose surface area equals 2*Pi. More generally, the square root of x is also the diameter of a sphere whose surface area equals x*Pi. - Omar E. Pol, Nov 10 2018
Sqrt(2) = 1 + area of region bounded by y = sin x, y = cos x, and x = 0. - Clark Kimberling, Jul 03 2020
Also aspect ratio of the ISO 216 standard for paper sizes. - Stefano Spezia, Feb 24 2021
The standard deviation of a roll of a 5-sided die. - Mohammed Yaseen, Feb 23 2023
From Michal Paulovic, Mar 22 2023: (Start)
The length of a unit square diagonal.
The infinite tetration (power tower) sqrt(2)^(sqrt(2)^(sqrt(2)^(...))) equals 2 from the identity (x^(1/x))^((x^(1/x))^((x^(1/x))^(...))) = x where 1/e <= x <= e. (End)

			1.41421356237309504880168872420969807856967187537694807317667...

John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 24, 182.
Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, Section 1.1.
David Flannery, The Square Root of 2, Copernicus Books Springer-Praxis Pub. 2006.
Martin Gardner, Gardner's Workout, Chapter 2 "The Square Root of 2=1.414213562373095..." pp. 9-19 A. K. Peters MA 2002.
Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §3.4 Irrational Numbers and §4.4 Powers and Roots, pp. 84, 145.
Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 64-67.
B. Rittaud, Le fabuleux destin de sqrt(2), Le Pommier, Paris 2006.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 60, page 605.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987, pp. 34-35.

Harry J. Smith, Table of n, a(n) for n = 1..20000
D. and J. Ensley, Review of "The Square Root of 2" by D. Flannery.
Steven R. Finch, Errata and Addenda to Mathematical Constants, arXiv:2001.00578 [math.HO], 2020.
M. F. Jones, 22900D approximations to the square roots of the primes less than 100, Math. Comp., 22 (1968), 234-235.
I. Khavkine, PlanetMath.org, square root of 2 is irrational.
Jason Kimberley, Index of expansions of sqrt(d) in base b.
C. E. Larson, (Avoiding) Proof by Contradiction: sqrt(2) is Not Rational, arXiv:2005.03878 [math.HO], 2020.
Robert Nemiroff and Jerry Bonnell, The Square Root of Two to 1 Million Digits.
Robert Nemiroff and Jerry Bonnell, The Square Root of Two to 5 million digits.
Robert Nemiroff and Jerry Bonnell, The first 10 million digits of the square root of 2.
Simon Plouffe, Plouffe's Inverter, The square root of 2 to 10 million digits.
Simon Plouffe, Generalized expansion of real constants.
M. Ripa and G. Morelli, Retro-analytical Reasoning IQ tests for the High Range, 2013.
Vladimir Ivanovich Smirnov, A course of higher mathematics, vol. 1 , Pergamon Press, 1964, p. 3.
Horace S. Uhler, Many-Figure Approximations To Sqrt(2), And Distribution Of Digits In Sqrt(2) And 1/Sqrt(2), Proc. Nat. Acad. Sci. U. S. A. 37, (1951). 63-67.
Eric Weisstein's World of Mathematics, Pythagoras's Constant.
Eric Weisstein's World of Mathematics, Square Root.
Index entries for algebraic numbers, degree 2.

Cf. A020807, A010503, A001790, A005187.

Cf. A004539 (binary version).

-- After Michael B. Porter's PARI program.
a002193 n = a002193_list !! (n-1)
a002193_list = w 2 0 where
w x r = dig : w (100 * (x - (20 * r + dig) * dig)) (10 * r + dig)
where dig = head (dropWhile (\d -> (20 * r + d) * d < x) [0..]) - 1
-- Reinhard Zumkeller, Nov 22 2013

Digits:=100; evalf(sqrt(2)); # Wesley Ivan Hurt, Dec 04 2013

RealDigits[N[2^(1/2), 128]] (* Vladimir Joseph Stephan Orlovsky, Dec 25 2008 *)

fpprec: 100$ ev(bfloat(sqrt(2))); /* Martin Ettl, Oct 17 2012 */

default(realprecision, 20080); x=sqrt(2); for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b002193.txt", n, " ", d)); \\ Harry J. Smith, Apr 21 2009

r=0; x=2; /* Digit-by-digit method */
for(digits=1,100,{d=0;while((20*r+d)*d <= x,d++);
d--; /* while loop overshoots correct digit */
print(d);x=100*(x-(20*r+d)*d);r=10*r+d}) \\ Michael B. Porter, Oct 20 2009

\\ Works in v2.15.0; n = 100 decimal places
my(n=100); digits(floor(10^n*quadgen(8))) \\ Michal Paulovic, Mar 22 2023

Sqrt(2) = 14 * Sum_{n >= 0} (A001790(n)/2^A005187(floor(n/2)) * 10^(-2n-1)) where A001790(n) are numerators in expansion of 1/sqrt(1-x) and the denominators in expansion of 1/sqrt(1-x) are 2^A005187(n). 14 = 2*7, see A010503 (expansion of 1/sqrt(2)). - Gerald McGarvey, Jan 01 2005
Limit_{n -> +oo} (1/n)*(Sum_{k = 1..n} frac(sqrt(1+zeta(k+1)))) = 1/(1+sqrt(2)). - Yalcin Aktar, Jul 14 2005
sqrt(2) = 2 + n*A167199(n-1)/A167199(n) as n -> infinity (conjecture). - Mats Granvik, Oct 30 2009
sqrt(2) = limit as n goes to infinity of A179807(n+1)/A179807(n) - 1. - Mats Granvik, Feb 15 2011
sqrt(2) = Product_{l=0..k-1} 2*cos((2*l+1)*Pi/(4*k)) = (Product_{l=0..k-1} R(2*l+1,rho(4*k)) - 1), identical for k >= 1, with the row polynomials R(n, x) from A127672 and rho(4*k) := 2*cos(Pi/(4*k)) is the length ratio (smallest diagonal)/side in a regular (4*k)-gon. From the product formula given in a Oct 21 2013 formula contribution to A056594, with n -> 2*k, using cos(Pi-alpha) = - cos(alpha) to obtain 2 for the square of the present product. - Wolfdieter Lang, Oct 22 2013
If x = sqrt(2), 1/log(x - 1) + 1/log(x + 1) = 0. - Kritsada Moomuang, Jul 10 2020
From Amiram Eldar, Jul 25 2020: (Start)
Equals Product_{k>=0} (1 + (-1)^k/(2*k + 1)).
Equals Sum_{k>=0} binomial(2*k,k)/8^k. (End)
Equals i^(1/2) + i^(-1/2). - Gary W. Adamson, Jul 11 2022
Equals (sqrt(2) + (sqrt(2) + (sqrt(2) + ...)^(1/3))^(1/3))^(1/3). - Michal Paulovic, Mar 22 2023
Equals 1 + Sum_{k>=1} (-1)^(k-1)/(2^(2*k)*(2*k - 1))*binomial(2*k,k) [Newton]. - Stefano Spezia, Oct 15 2024
From Antonio Graciá Llorente, Dec 19 2024: (Start)
Equals Sum_{k>=0} 2*k*binomial(2*k,k)/8^k.
Equals Product_{k>=2}  k/sqrt(k^2 + 1).
Equals Product_{k>=0} (6*k + 3)/((6*k + 3) - (-1)^k).
Equals Product_{k>=1} (2*k + 1)/((2*k + 1) + (-1)^k).
Equals Product_{k>=0} ((4*k + 3)*(4*k + 1 + (-1)^k))/((4*k + 1)*(4*k + 3 + (-1)^k)). (End)
Equals hypergeom([1/2, 1/2], [1/2], 1/2). - Stefano Spezia, Jan 05 2025

A041085 Denominators of continued fraction convergents to sqrt(50).

Original entry on oeis.org

1, 14, 197, 2772, 39005, 548842, 7722793, 108667944, 1529074009, 21515704070, 302748930989, 4260000737916, 59942759261813, 843458630403298, 11868363584907985, 167000548819115088, 2349876047052519217, 33065265207554384126, 465263588952813896981

Offset: 0

N. J. A. Sloane

For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 14's along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
a(n) equals the number of words of length n on alphabet {0,1,...,14} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
From Michael A. Allen, Apr 30 2023: (Start)
Also called the 14-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
a(n) is the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 14 kinds of squares available. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..800
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (14,1).

Cf. A000129, A020807, A040042, A041084.

Row n=14 of A073133, A172236 and A352361 and column k=14 of A157103.

[n le 2 select (14)^(n-1) else 14*Self(n-1) +Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 17 2012

with(combinat): seq(fibonacci(3*n+3,2)/5, n=0..17); # Zerinvary Lajos, Apr 20 2008

LinearRecurrence[{14, 1}, {1, 14}, 30] (* Vincenzo Librandi, Nov 17 2012 *)
Table[Fibonacci[3n + 3, 2]/5, {n, 0, 20}] (* Vladimir Reshetnikov, Sep 16 2016 *)
Convergents[Sqrt[50],20]//Denominator (* Harvey P. Dale, Aug 16 2025 *)

A041085=BinaryRecurrenceSequence(14,1,1,14)
[A041085(n) for n in range(31)] # G. C. Greubel, Sep 29 2024

a(n) = round((7+5*sqrt(2))*a(n-1)). - Vladeta Jovovic, Jun 15 2003
From Paul Barry, Feb 06 2004: (Start)
a(n) = A000129(3*n+3)/5.
a(n) = (1/20)*((10+7*sqrt(2))*(1+sqrt(2))^(3*n) + (10-7*sqrt(2))*(1-sqrt(2))^(3*n)).
a(n-1) = Sum_{i=0..n} Sum_{j=0..n-i} (n!/(i!*j!*(n-i-j)!))*A000129(2*n-i)/5. (End)
a(n) = Fibonacci(n+1, 14), the n-th Fibonacci polynomial evaluated at x=14. - T. D. Noe, Jan 19 2006
From Philippe Deléham, Nov 03 2008: (Start)
a(n) = 14*a(n-1) + a(n-2); a(0)=1, a(1)=14.
G.f.: 1/(1-14*x-x^2). (End)
a(n) = ((7+5*sqrt(2))^(n+1) - (7-5*sqrt(2))^(n+1))/(10*sqrt(2)). - Gerry Martens, Jul 11 2015

Additional term from Colin Barker, Nov 12 2013

A188582 Decimal expansion of sqrt(2) - 1.

Original entry on oeis.org

4, 1, 4, 2, 1, 3, 5, 6, 2, 3, 7, 3, 0, 9, 5, 0, 4, 8, 8, 0, 1, 6, 8, 8, 7, 2, 4, 2, 0, 9, 6, 9, 8, 0, 7, 8, 5, 6, 9, 6, 7, 1, 8, 7, 5, 3, 7, 6, 9, 4, 8, 0, 7, 3, 1, 7, 6, 6, 7, 9, 7, 3, 7, 9, 9, 0, 7, 3, 2, 4, 7, 8, 4, 6, 2, 1, 0, 7, 0, 3, 8, 8, 5, 0, 3, 8, 7, 5, 3, 4, 3, 2, 7, 6, 4, 1, 5, 7, 2, 7, 3, 5, 0, 1

Offset: 0

Robert G. Wilson v, Apr 04 2011

"In his Book 'The Theory of Poker,' David Sklansky coined the phrase 'Fundamental Theorem of Poker,' a tongue-in-cheek reference to the Fundamental Theorem of Algebra and Fundamental Theorem of Calculus from introductory texts on those two subjects. The constant [sqrt(2) - 1] appears so often in poker analysis that we will in the same vein go so far as to call it 'the golden mean of poker,' and we call it 'r' for short. We will see this value in a number of important results throughout this book." [Chen and Ankenman]
If a triangle has sides whose lengths form a harmonic progression in the ratio 1/(1 - d) : 1 : 1/(1 + d) then the triangle inequality condition requires that d be in the range 1 - sqrt(2) < d < sqrt(2) - 1. - Frank M Jackson, Oct 01 2013
This constant is the 6th smallest radius r < 1 for which a compact packing of the plane exists, with disks of radius 1 and r. - Jean-François Alcover, Sep 02 2014, after Steven Finch
This constant is also the largest argument of the arctangent function in the Viète-like formula for Pi given by Pi/2^(k+1) = arctan(sqrt(2 - a_(k-1))/a_k), where the index k >= 2 and the nested radicals are defined by recurrence using the relations a_k = sqrt(2 + a_(k-1)), a_1 = sqrt(2). When k = 2 the argument of the arctangent function sqrt(2 - a_1)/a_2 = sqrt(2 - sqrt(2))/sqrt(2 + sqrt(2)) = sqrt(2) - 1 is largest. Consequently, at k = 2 the Viète-like formula for Pi can be written as Pi/8 = arctan(sqrt(2 - sqrt(2))/sqrt(2 + sqrt(2))) = arctan(sqrt(2) - 1) (after Abrarov-Quine, see the article). - Sanjar Abrarov, Jan 07 2017
If r and R are respectively the inradius and the circumradius of a triangle, then the ratio r/R <= 1/2 (Euler inequality), and this maximum value 1/2 is obtained when the triangle is equilateral. Now, for a right triangle, the ratio r/R <= this constant = sqrt(2) - 1, and this maximum value sqrt(2) - 1 is obtained when the right triangle is isosceles. This is the answer to the question 1 of the Olympiade Mathématique Belge Maxi in 2008. - Bernard Schott, Sep 07 2022

			0.414213562373095048801688724209698078569671875376948073...

B. C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag, p. 140, Entry 25.
Bill Chen and Jerrod Ankenman, The Mathematics of Poker, Chpt 14 - You Don't Have To Guess: No-Limit Bet Sizing, p. 153, ConJelCo, LLC, Pittsburgh PA 2006.
Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, pp. 396 and 486.

G. C. Greubel, Table of n, a(n) for n = 0..10000
S. M. Abrarov and B. M. Quine, A Viète-like formula for pi based on infinite sum of the arctangent functions with nested radicals, figshare, 4509014, (2017).
Steven R. Finch, Errata and Addenda to Mathematical Constants, p. 62.
Olympiade Mathématique Belge, OMB 2008, Finale Maxi, Question 1.
Index entries for algebraic numbers, degree 2
Index to sequences related to Olympiads.

Cf. A002193, A014176, A020807, A120731, A182168 (sin(Pi/8)), A144981 (cos(Pi/8)).

Sqrt(2) - 1; // G. C. Greubel, Jan 31 2018

Mathematica
```
RealDigits[ Sqrt[2] - 1, 10, 111][[1]]
```

sqrt(2) - 1 \\ G. C. Greubel, Jan 31 2018

Equals exp(asinh(cos(Pi))) = exp(asinh(-1)). - Geoffrey Caveney, Apr 23 2014
Equals tan(Pi/8) = A182168 / A144981 = 1 / A014176. - Bernard Schott, Apr 12 2022
From Antonio Graciá Llorente, Mar 15 2024: (Start)
Equals Product_{k >= 0} ((8*k - 1)*(8*k + 9))/((8*k - 5)*(8*k + 13)).
Equals Product_{k >= 1} A047554(k)/A047447(k). (End)
From  Peter Bala, Mar 24 2024: (Start)
An infinite family of continued fraction expansions for this constant can be obtained from Berndt, Entry 25, by setting n = 1/2 and x = 8*k + 2 for k >= 0.
For example, taking k = 0 and k = 1 yields
Equals 1/(2 + (1*3)/(4 + (5*7)/(4 + (9*11)/(4 + (13*15)/(4 + ... + (4*n + 1)*(4*n + 3)/(4 + ...)))))) and
Equals (21/5) * 1/(10 + (1*3)/(20 + (5*7)/(20 + (9*11)/(20 + (13*15)/(20 + ... + (4*n + 1)*(4*n + 3)/(20 + ...)))))). (End)
Tan(arctan(c) + arctan(c^3)) = 1/2. - Gary W. Adamson, Apr 04 2024

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A002193 Decimal expansion of square root of 2.

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A041085 Denominators of continued fraction convergents to sqrt(50).

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A188582 Decimal expansion of sqrt(2) - 1.

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