0,2

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

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).

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)

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

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

(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

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).

Sequence in context: A373458 A280691 A092666 * A333599 A335405 A369924

Adjacent sequences: A019423 A019424 A019425 * A019427 A019428 A019429

nonn,easy,cofr

approved