A014176 Decimal expansion of the silver mean, 1+sqrt(2).

2, 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

Author

N. J. A. Sloane

Keywords

• nonn
• cons
• easy

Comments

From Hieronymus Fischer, Jan 02 2009: (Start)

Set c:=1+sqrt(2). Then the fractional part of c^n equals 1/c^n, if n odd. For even n, the fractional part of c^n is equal to 1-(1/c^n).

c:=1+sqrt(2) satisfies c-c^(-1)=floor(c)=2, hence c^n + (-c)^(-n) = round(c^n) for n>0, which follows from the general formula of A001622.

1/c = sqrt(2)-1.

See A001622 for a general formula concerning the fractional parts of powers of numbers x>1, which satisfy x-x^(-1)=floor(x).

Other examples of constants x satisfying the relation x-x^(-1)=floor(x) include A001622 (the golden ratio: where floor(x)=1) and A098316 (the "bronze" ratio: where floor(x)=3). (End)

In terms of continued fractions the constant c can be described by c=[2;2,2,2,...]. - Hieronymus Fischer, Oct 20 2010

Side length of smallest square containing five circles of diameter 1. - Charles R Greathouse IV, Apr 05 2011

Largest radius of four circles tangent to a circle of radius 1. - Charles R Greathouse IV, Jan 14 2013

An analog of Fermat theorem: for prime p, round(c^p) == 2 (mod p). - Vladimir Shevelev, Mar 02 2013

n*(1+sqrt(2)) is the perimeter of a 45-45-90 triangle with hypotenuse n. - Wesley Ivan Hurt, Apr 09 2016

This algebraic integer of degree 2, with minimal polynomial x^2 - 2*x - 1, is also the length ratio diagonal/side of the second largest diagonal in the regular octagon (not counting the side). The other two diagonal/side ratios are A179260 and A121601. - Wolfdieter Lang, Oct 28 2020

c^n = A001333(n) + A000129(n) * sqrt(2). - Gary W. Adamson, Apr 26 2023

c^n = c * A000129(n) + A000129(n-1), where c = 1 + sqrt(2). - Gary W. Adamson, Aug 30 2023

Examples

			2.414213562373095...
		

References

• B. C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag, p. 140, Entry 25.

Links

• G. C. Greubel, Table of n, a(n) for n = 1..10000
• Nicholas R. Beaton, Mireille Bousquet-Mélou, Jan de Gier, Hugo Duminil-Copin, and Anthony J. Guttmann, The critical fugacity for surface adsorption of self-avoiding walks on the honeycomb lattice is 1+sqrt(2), arXiv:1109.0358 [math-ph], 2011-2013.
• Eric Weisstein's World of Mathematics, Silver Ratio
• Wikipedia, Exact trigonometric constants
• Wikipedia, Metallic mean
• Wikipedia, Silver ratio
• Index entries for algebraic numbers, degree 2

Crossrefs

Apart from initial digit the same as A002193.

Cf. A000032, A006497, A080039, A179260, A121601.

See A098316 for [3;3,3,...]; A098317 for [4;4,4,...]; A098318 for [5;5,5,...]. - Hieronymus Fischer, Oct 20 2010

Cf. A000129, A049310.

Programs

Maple

Digits:=100: evalf(1+sqrt(2)); # Wesley Ivan Hurt, Apr 09 2016

Mathematica

RealDigits[1 + Sqrt@ 2, 10, 111] (* Or *)
RealDigits[Exp@ ArcSinh@ 1, 10, 111][[1]] (* Robert G. Wilson v, Aug 17 2011 *)
Circs[n_] := With[{r = Sin[Pi/n]/(1 - Sin[Pi/n])}, Graphics[Append[
  Table[Circle[(r + 1) {Sin[2 Pi k/n], Cos[2 Pi k/n]}, r], {k, n}],   {Blue, Circle[{0, 0}, 1]}]]] Circs[4] (* Charles R Greathouse IV, Jan 14 2013 *)

PARI

1+sqrt(2) \\ Charles R Greathouse IV, Jan 14 2013

Formula

Conjecture: 1+sqrt(2) = lim_{n->oo} A179807(n+1)/A179807(n).

Equals cot(Pi/8) = tan(Pi*3/8). - Bruno Berselli, Dec 13 2012, and M. F. Hasler, Jul 08 2016

Silver mean = 2 + Sum_{n>=0} (-1)^n/(P(n-1)*P(n)), where P(n) is the n-th Pell number (A000129). - Vladimir Shevelev, Feb 22 2013

Equals exp(arcsinh(1)) which is exp(A091648). - Stanislav Sykora, Nov 01 2013

Limit_{n->oo} exp(asinh(cos(Pi/n))) = sqrt(2) + 1. - Geoffrey Caveney, Apr 23 2014

exp(asinh(cos(Pi/2 - log(sqrt(2)+1)*i))) = exp(asinh(sin(log(sqrt(2)+1)*i))) = i. - Geoffrey Caveney, Apr 23 2014

Equals Product_{k>=1} A047621(k) / A047522(k) = (3/1) * (5/7) * (11/9) * (13/15) * (19/17) * (21/23) * ... . - Dimitris Valianatos, Mar 27 2019

From Wolfdieter Lang, Nov 10 2023:(Start)

Equals lim_{n->oo} A000129(n+1)/A000129(n) (see A000129, Pell).

Equals lim_{n->oo} S(n+1, 2*sqrt(2))/S(n, 2*sqrt(2)), with the Chebyshev S(n,x) polynomial (see A049310). (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 + 6 for k >= 0.

For example, taking k = 0 and k = 1 yields

sqrt(2) + 1 = 15/(6 + (1*3)/(12 + (5*7)/(12 + (9*11)/(12 + (13*15)/(12 + ... + (4*n + 1)*(4*n + 3)/(12 + ... )))))) and

sqrt(2) + 1 = (715/21) * 1/(14 + (1*3)/(28 + (5*7)/(28 + (9*11)/(28 + (13*15)/(28 + ... + (4*n + 1)*(4*n + 3)/(28 + ... )))))). (End)