A019426 Continued fraction for tan(1/3).

0, 2, 1, 7, 1, 13, 1, 19, 1, 25, 1, 31, 1, 37, 1, 43, 1, 49, 1, 55, 1, 61, 1, 67, 1, 73, 1, 79, 1, 85, 1, 91, 1, 97, 1, 103, 1, 109, 1, 115, 1, 121, 1, 127, 1, 133, 1, 139, 1, 145, 1, 151, 1, 157, 1, 163, 1, 169, 1, 175, 1, 181, 1, 187, 1, 193, 1, 199, 1, 205, 1, 211, 1, 217, 1, 223, 1

Offset: 0

Author

David W. Wilson

Keywords

• nonn
• easy
• cofr

Comments

The simple continued fraction expansion of 3*tan(1/3) is [1; 25, 1, 3, 1, 61, 1, 7, 1, 97, 1, 11, 1, ..., 36*n + 25, 1, 4*n + 3, 1, ...], while the simple continued fraction expansion of (1/3)*tan(1/3) is [0; 8, 1, 1, 1, 43, 1, 5, 1, 79, 1, 9, 1, 115, 1, 13, 1, ..., 36*n + 7, 1, 4*n + 1, 1, ...]. See my comment in A019425. - Peter Bala, Sep 30 2023

Examples

			0.346253549510575491038543565... = 0 + 1/(2 + 1/(1 + 1/(7 + 1/(1 + ...)))). - _Harry J. Smith_, Jun 13 2009
		

Links

• Harry J. Smith, Table of n, a(n) for n = 0..20000
• Khalil Ayadi, Chiheb Ben Bechir, and Maher Saadaoui, Continued Fractions with Predictable Patterns and Transcendental Numbers, J. Int. Seq. (2025) Vol. 28, Art. No. 25.1.4.
• G. Xiao, Contfrac
• Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Crossrefs

Cf. A161012 (decimal expansion of tan(1/3)).

Cf. continued fractions for tan(1/m): A019425 (m=2), A019427 (m=4), A019428 (m=5), A019429 (m=6), A019430 (m=7), A019431 (m=8), A019432 (m=9), A019433 (m=10), A093178 (m=1).

Programs

Magma

[n le 1 select 2*n else 1+3*(1-(-1)^n)*(n-1)/2: n in [0..80]]; // Bruno Berselli, Sep 21 2012

Mathematica

ContinuedFraction[Tan[1/3], 80] (* Bruno Berselli, Sep 21 2012 *)

PARI

{ allocatemem(932245000); default(realprecision, 88000); x=contfrac(tan(1/3)); for (n=0, 20000, write("b019426.txt", n, " ", x[n+1])); } \\ Harry J. Smith, Jun 13 2009

Formula

From Bruno Berselli, Sep 21 2012: (Start)

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

a(n) = 2*a(n-2)-a(n-4) with n>4, a(0)=0, a(1)=2, a(2)=1, a(3)=7, a(4)=1.

a(n) = 1+3*(1-(-1)^n)*(n-1)/2 with n>1, a(0)=0, a(1)=2.

For k>0: a(2k) = 1, a(4k+1) = 2*a(2k+1)-1 and a(4k+3) = 2*a(2k+1)+5, with a(0)=0, a(1)=2. (End)