A019432 -id:A019432 - cp's OEIS frontend

A019433 Continued fraction for tan(1/10).

0, 9, 1, 28, 1, 48, 1, 68, 1, 88, 1, 108, 1, 128, 1, 148, 1, 168, 1, 188, 1, 208, 1, 228, 1, 248, 1, 268, 1, 288, 1, 308, 1, 328, 1, 348, 1, 368, 1, 388, 1, 408, 1, 428, 1, 448, 1, 468, 1, 488, 1, 508, 1, 528, 1, 548, 1, 568, 1, 588, 1, 608, 1, 628, 1, 648, 1, 668, 1, 688, 1, 708, 1, 728

Offset: 0

David W. Wilson

From Peter Bala, Oct 04 2023: (Start)
Related simple continued fractions expansions (see my comments in A019425):
tan(1/(10*k)) = [0; 10*k - 1, 1, 30*k - 2, 1, 50*k - 2, 1, 70*k - 2, 1, 90*k - 2, 1, ...] for k >= 1.
If d is a divisor of 10 with d*d' = 10 then the simple continued fraction expansion of d*tan(1/10) begins [0; d' - 1, 1, 30*d - 2, 1, 5*d' - 2, 1, 70*d - 2, 1, 9*d' - 2, 1, 110*d - 2, 1, 13*d' - 2, ...], while the simple continued fraction expansion of (1/d)*tan(1/10) begins [ 0; 10*d - 1, 1, 3*d'- 2, 1, 50*d - 2, 1, 7*d' - 2, 1, 90*d - 2, 1, 11*d' - 2, 1, 130*d - 2, ...]. (End)

			0.10033467208545054505808004... = 0 + 1/(9 + 1/(1 + 1/(28 + 1/(1 + ...)))). - _Harry J. Smith_, Jun 14 2009

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).

Cf. A161019 (decimal expansion), A019425 through A019432.

LinearRecurrence[{0,2,0,-1},{0,9,1,28,1,48},80] (* or *) Join[{0,9},Riffle[NestList[20+#&,28,40],1,{1,-1,2}]] (* Harvey P. Dale, Jul 23 2023 *)

{ allocatemem(932245000); default(realprecision, 99000); x=contfrac(tan(1/10)); for (n=0, 20000, write("b019433.txt", n, " ", x[n+1])); } \\ Harry J. Smith, Jun 14 2009

Vec(x*(x^4-x^3+10*x^2+x+9)/((x-1)^2*(x+1)^2) + O(x^100)) \\ Colin Barker, Sep 08 2013

From Colin Barker, Sep 08 2013: (Start)
a(n) = -1/2+(3*(-1)^n)/2+5*n-5*(-1)^n*n for n>1.
a(n) = 2*a(n-2)-a(n-4) for n>5.
G.f.: x*(x^4-x^3+10*x^2+x+9) / ((x-1)^2*(x+1)^2). (End)

A019426 Continued fraction for tan(1/3).

Original entry on oeis.org

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

David W. Wilson

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

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

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).

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).

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

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

{ 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

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)

A161018 Decimal expansion of tan(1/9).

Original entry on oeis.org

1, 1, 1, 5, 7, 0, 6, 2, 7, 8, 3, 3, 8, 0, 0, 5, 8, 3, 7, 2, 6, 5, 0, 4, 8, 0, 0, 2, 4, 9, 8, 2, 9, 4, 6, 4, 5, 0, 2, 5, 4, 9, 2, 3, 4, 1, 2, 7, 7, 4, 0, 0, 7, 3, 1, 1, 1, 7, 6, 8, 4, 7, 2, 5, 0, 7, 6, 3, 7, 2, 3, 9, 4, 8, 2, 2, 3, 5, 8, 5, 2, 3, 5, 7, 9, 4, 1, 4, 5, 7, 4, 1, 1, 0, 6, 6, 1, 3, 2, 8, 4, 0, 7, 8, 6

Offset: 0

Harry J. Smith, Jun 14 2009

By the Lindemann-Weierstrass theorem, this constant is transcendental. - Charles R Greathouse IV, May 13 2019

			0.111570627833800583726504800249829464502549234127740073111768472507637...

Harry J. Smith, Table of n, a(n) for n = 0..20000
Index entries for transcendental numbers.

Cf. A019432 (continued fraction).

RealDigits[Tan[1/9], 10, 120][[1]] (* Amiram Eldar, Jun 27 2023 *)

default(realprecision, 20080); x=10*tan(1/9); for (n=0, 20000, d=floor(x); x=(x-d)*10; write("b161018.txt", n, " ", d));

cp's OEIS Frontend

A019433 Continued fraction for tan(1/10).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

PARI

Formula

A019426 Continued fraction for tan(1/3).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A161018 Decimal expansion of tan(1/9).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI