A019428 Continued fraction for tan(1/5).

0, 4, 1, 13, 1, 23, 1, 33, 1, 43, 1, 53, 1, 63, 1, 73, 1, 83, 1, 93, 1, 103, 1, 113, 1, 123, 1, 133, 1, 143, 1, 153, 1, 163, 1, 173, 1, 183, 1, 193, 1, 203, 1, 213, 1, 223, 1, 233, 1, 243, 1, 253, 1, 263, 1, 273, 1, 283, 1, 293, 1, 303, 1, 313, 1, 323, 1, 333, 1, 343, 1, 353, 1, 363, 1

Offset

0,2

Comments

The simple continued fraction expansion of 5*tan(1/5) begins [1; 73, 1, 3, 1, 173, 1, 7, 1, 273, 1, 11, 1, 373, 1, 15, 1, 473, 1, 19, 1, 573, ...], while the simple continued fraction expansion of (1/5)*tan(1/5) begins [0; 24, 1, 1, 1, 123, 1, 5, 1, 223, 1, 9, 1, 323, 1, 13, 1, 423, 1, 17, 1, 523, ...]. See my comment in A019425. - Peter Bala, Sep 30 2023

Links

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

G. Xiao, Contfrac

Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Formula

From Colin Barker, Sep 08 2013: (Start)

a(n) = (-1 + 3*(-1)^n - 5*(-1 + (-1)^n)*n)/2 for n > 1.

a(n) = 2*a(n-2) - a(n-4) for n > 5.

G.f.: x*(x^4-x^3+5*x^2+x+4) / ((x-1)^2*(x+1)^2). (End)

Example

0.20271003550867248332135827... = 0 + 1/(4 + 1/(1 + 1/(13 + 1/(1 + ...)))). - Harry J. Smith, Jun 13 2009

Mathematica

 Join[{0, 4}, LinearRecurrence[{0, 2, 0, -1}, {1, 13, 1, 23}, 100]] (* Vincenzo Librandi, Jan 03 2016 *)

Prog

(PARI) { allocatemem(932245000); default(realprecision, 93000); x=contfrac(tan(1/5)); for (n=0, 20000, write("b019428.txt", n, " ", x[n+1])); } \\ Harry J. Smith, Jun 13 2009
(PARI) Vec(x*(x^4-x^3+5*x^2+x+4)/((x-1)^2*(x+1)^2) + O(x^100)) \\ Colin Barker, Sep 08 2013
(Magma) [0, 4] cat [(-1+3*(-1)^n-5*(-1+(-1)^n)*n)/2: n in [2..80]]; // Vincenzo Librandi, Jan 03 2016

Crossrefs

Cf. A161014 (decimal expansion), A019425 through A019433.

Sequence in context: A002564 A287640 A322078 * A303547 A184753 A324186

Adjacent sequences: A019425 A019426 A019427 * A019429 A019430 A019431

Keyword

nonn,cofr,easy

Author

David W. Wilson

Status

approved