A010127 Continued fraction for sqrt(23).

4, 1, 3, 1, 8, 1, 3, 1, 8, 1, 3, 1, 8, 1, 3, 1, 8, 1, 3, 1, 8, 1, 3, 1, 8, 1, 3, 1, 8, 1, 3, 1, 8, 1, 3, 1, 8, 1, 3, 1, 8, 1, 3, 1, 8, 1, 3, 1, 8, 1, 3, 1, 8, 1, 3, 1, 8, 1, 3, 1, 8, 1, 3, 1, 8, 1, 3, 1, 8, 1, 3, 1, 8, 1, 3, 1, 8, 1, 3, 1, 8, 1, 3, 1, 8, 1, 3, 1, 8

Offset

0,1

References

Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See Table 95 at p. 262.

Harold Davenport, The Higher Arithmetic, Cambridge University Press, 8th ed., 2008, p. 97.

James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 275-276.

Links

Harry J. Smith, Table of n, a(n) for n = 0..20000

G. Xiao, Contfrac.

Index entries for continued fractions for constants.

Index entries for linear recurrences with constant coefficients, signature (0,0,0,1).

Formula

From Amiram Eldar, Nov 12 2023: (Start)

Multiplicative with a(2) = 3, a(2^e) = 8 for e >= 2, and a(p^e) = 1 for an odd prime p.

Dirichlet g.f.: zeta(s) * (1 + 1/2^(s-1) + 5/4^s). (End)

From Stefano Spezia, Aug 17 2024: (Start)

G.f.: (4 + x + 3*x^2 + x^3 + 4*x^4)/(1 - x^4).

E.g.f.: (5*cos(x) + 11*cosh(x) + 2*sinh(x) - 8)/2. (End)

Example

4.795831523312719541597438064... = 4 + 1/(1 + 1/(3 + 1/(1 + 1/(8 + ...)))). - Harry J. Smith, Jun 03 2009

Mathematica

ContinuedFraction[Sqrt[23], 300] (* Vladimir Joseph Stephan Orlovsky, Mar 05 2011 *)
PadRight[{4}, 120, {8, 1, 3, 1}] (* Harvey P. Dale, Oct 23 2024 *)

Prog

(PARI) { allocatemem(932245000); default(realprecision, 17000); x=contfrac(sqrt(23)); for (n=0, 20000, write("b010127.txt", n, " ", x[n+1])); } \\ Harry J. Smith, Jun 03 2009

Crossrefs

Cf. A010479 (decimal expansion).

Sequence in context: A089612 A353776 A292269 * A263022 A326690 A353275

Adjacent sequences: A010124 A010125 A010126 * A010128 A010129 A010130

Keyword

nonn,cofr,easy,mult

Author

N. J. A. Sloane

Status

approved