A019425 -id:A019425 - 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)

A019430 Continued fraction for tan(1/7).

Original entry on oeis.org

0, 6, 1, 19, 1, 33, 1, 47, 1, 61, 1, 75, 1, 89, 1, 103, 1, 117, 1, 131, 1, 145, 1, 159, 1, 173, 1, 187, 1, 201, 1, 215, 1, 229, 1, 243, 1, 257, 1, 271, 1, 285, 1, 299, 1, 313, 1, 327, 1, 341, 1, 355, 1, 369, 1, 383, 1, 397, 1, 411, 1, 425, 1, 439, 1, 453, 1, 467, 1, 481, 1, 495, 1, 509, 1

Offset: 0

David W. Wilson

			0.14383695943619093528003059... = 0 + 1/(6 + 1/(1 + 1/(19 + 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. A161016 (decimal expansion), A019425 through A019433.

[0, 6] cat [(-1+3*(-1)^n-7*(-1+(-1)^n)*n)/2: n in [2..80]]; // Vincenzo Librandi, Jan 03 2016

Block[{$MaxExtraPrecision=1000},ContinuedFraction[Tan[1/7],80]] (* Harvey P. Dale, Feb 01 2013 *)
Join[{0, 6}, LinearRecurrence[{0, 2, 0, -1}, {1, 19, 1, 33}, 100]] (* Vincenzo Librandi, Jan 03 2016 *)

{ allocatemem(932245000); default(realprecision, 96000); x=contfrac(tan(1/7)); for (n=0, 20000, write("b019430.txt", n, " ", x[n+1])); } \\ Harry J. Smith, Jun 14 2009

Vec(x*(x^4-x^3+7*x^2+x+6)/((x-1)^2*(x+1)^2) + O(x^100)) \\ Colin Barker, Sep 08 2013

From Colin Barker, Sep 08 2013: (Start)
a(n) = (-1+3*(-1)^n-7*(-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+7*x^2+x+6) / ((x-1)^2*(x+1)^2). (End)

A019432 Continued fraction for tan(1/9).

Original entry on oeis.org

0, 8, 1, 25, 1, 43, 1, 61, 1, 79, 1, 97, 1, 115, 1, 133, 1, 151, 1, 169, 1, 187, 1, 205, 1, 223, 1, 241, 1, 259, 1, 277, 1, 295, 1, 313, 1, 331, 1, 349, 1, 367, 1, 385, 1, 403, 1, 421, 1, 439, 1, 457, 1, 475, 1, 493, 1, 511, 1, 529, 1, 547, 1, 565, 1, 583, 1, 601, 1, 619, 1, 637, 1, 655

Offset: 0

David W. Wilson

The odd-indexed terms from and after a(3) are equal to 18n+7. - Harvey P. Dale, Sep 26 2021

			0.11157062783380058372650480... = 0 + 1/(8 + 1/(1 + 1/(25 + 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. A161018 (decimal expansion), A019425 through A019433.

Block[{$MaxExtraPrecision=1000}, ContinuedFraction[Tan[1/9],100]] (* or *) LinearRecurrence[{0,2,0,-1},{0,8,1,25,1,43},80] (* Harvey P. Dale, Sep 26 2021 *)

{ allocatemem(932245000); default(realprecision, 98000); x=contfrac(tan(1/9)); for (n=0, 20000, write("b019432.txt", n, " ", x[n+1])); } \\ Harry J. Smith, Jun 14 2009

Vec(x*(x^4-x^3+9*x^2+x+8)/((x-1)^2*(x+1)^2) + O(x^100)) \\ Colin Barker, Sep 08 2013

From Colin Barker, Sep 08 2013: (Start)
a(n) = (-1+3*(-1)^n-9*(-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+9*x^2+x+8) / ((x-1)^2*(x+1)^2). (End)

A133265 Diagonal of the A135356 triangle.

Original entry on oeis.org

2, 2, 2, 4, 2, 6, 2, 8, 2, 10, 2, 12, 2, 14, 2, 16, 2, 18, 2, 20, 2, 22, 2, 24, 2, 26, 2, 28, 2, 30, 2, 32, 2, 34, 2, 36, 2, 38, 2, 40, 2, 42, 2, 44, 2, 46, 2, 48, 2, 50, 2, 52, 2, 54, 2, 56, 2, 58, 2, 60, 2, 62, 2, 64, 2, 66, 2, 68, 2, 70, 2, 72, 2, 74, 2, 76, 2, 78, 2, 80

Offset: 0

Paul Curtz, Dec 20 2007

Regular continued fraction expansion of 2*sin(1/2)/( cos(1/2) - sin(1/2) ) = 2.40822 34423 35827 84841 ... = 2 + 1/(2 + 1/(2 + 1/(4 + 1/(2 + 1/(6 + 1/(2 + 1/(8 + 1/(2 + ... )))))))). Cf. A019425. - Peter Bala, Feb 15 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A019425.

[(n+3+(n-1)*(-1)^(n+1))/2: n in [0..80]]; // Vincenzo Librandi, Aug 30 2011

A133265 := n -> (n+2+(n-2)*(-1)^n)/2: # Peter Luschny, Aug 30 2011

Table[(n + 3 + (n - 1) (-1)^(n + 1))/2, {n, 0, 79}] (* or *)
Table[Mod[(2 n + 5), (n (3 + (-1)^n) - (-1)^n + 7)/2], {n, 0, 79}] (* or *)
CoefficientList[Series[2 (1 + x - x^2)/((1 - x)^2*(1 + x)^2), {x, 0, 79}], x] (* Michael De Vlieger, Nov 18 2016 *)

Vec(2*(1 + x - x^2) / ((1 - x)^2 * (1 + x)^2) + O(x^100)) \\ Colin Barker, Nov 17 2016

2*(A057979 without 1, 0, first two terms).
a(n) = (n+3+(n-1)*(-1)^(n+1))/2. - Vincenzo Librandi, Aug 30 2011
a(n) = (2*n + 5) mod (n*(3 + (-1)^n) - (-1)^n + 7)/2. - Lechoslaw Ratajczak, Nov 17 2016
From Colin Barker, Nov 17 2016: (Start)
a(n) = 2*a(n-2) - a(n-4) for n>3.
G.f.: 2*(1 + x - x^2) / ((1 - x)^2 * (1 + x)^2).
(End)

A161011 Decimal expansion of tan(1/2).

Original entry on oeis.org

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

Offset: 0

Harry J. Smith, Jun 13 2009

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

			0.546302489843790513255179465780285383297551720179791246164091385932907...

Harry J. Smith, Table of n, a(n) for n = 0..20000
MathOverflow, What is the effect of adding 1/2 to a continued fraction?
Wikipedia, Lindemann-Weierstrass theorem
Index entries for transcendental numbers

Cf. A019425 (continued fraction). Cf. A049471, A161011 through A161019.

RealDigits[N[Tan[1/2],6! ]][[1]] (* Vladimir Joseph Stephan Orlovsky, Jun 13 2009 *)

default(realprecision, 20080); x=10*tan(1/2); for (n=0, 20000, d=floor(x); x=(x-d)*10; write("b161011.txt", n, " ", d));

From Peter Bala, Nov 17 2019: (Start)
Related simple continued fraction expansions:
tan(1/2) = [0; 1, 1, 4, 1, 8, 1, 12, 1, 16, 1, 20, 1, ...]. See A019425.
2*tan(1/2) = [1, 10, 1, 3, 1, 26, 1, 7, 1, 42, 1, 11, 1, 58, 1, 15, 1, 74, 1, 19, 1, 90, ...]
(1/2)*tan(1/2) = [0; 3, 1, 1, 1, 18, 1, 5, 1, 34, 1, 9, 1, 50, 1, 13, 1, 66, 1, 17, 1, 82, ...].
tan(1/2)/(1 - tan(1/2)) = [1, 4, 1, 8, 1, 12, 1, 16, 1, 20, 1, 24, ...]
2*tan(1/2)/(1 - tan(1/2)) = [2, 2, 2, 4, 2, 6, 2, 8, 2, 10, 2, 12, ...]
4*tan(1/2)/(1 - tan(1/2)) = [4, 1, 4, 2, 4, 3, 4, 4, 4, 5, 4, 6, 4, 7, ...]. (End)

A019428 Continued fraction for tan(1/5).

Original entry on oeis.org

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

David W. Wilson

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

			0.20271003550867248332135827... = 0 + 1/(4 + 1/(1 + 1/(13 + 1/(1 + ...)))). - _Harry J. Smith_, Jun 13 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. A161014 (decimal expansion), A019425 through A019433.

[0,4] cat [(-1+3*(-1)^n-5*(-1+(-1)^n)*n)/2: n in [2..80]]; // Vincenzo Librandi, Jan 03 2016

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

{ 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

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

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)

A019431 Continued fraction for tan(1/8).

Original entry on oeis.org

0, 7, 1, 22, 1, 38, 1, 54, 1, 70, 1, 86, 1, 102, 1, 118, 1, 134, 1, 150, 1, 166, 1, 182, 1, 198, 1, 214, 1, 230, 1, 246, 1, 262, 1, 278, 1, 294, 1, 310, 1, 326, 1, 342, 1, 358, 1, 374, 1, 390, 1, 406, 1, 422, 1, 438, 1, 454, 1, 470, 1, 486, 1, 502, 1, 518, 1, 534, 1, 550, 1, 566, 1, 582

Offset: 0

David W. Wilson

			0.12565513657513096779267821... = 0 + 1/(7 + 1/(1 + 1/(22 + 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. A161017 (decimal expansion), A019425 through A019433.

[0,7] cat [(-1+3*(-1)^n-8*(-1+(-1)^n)*n)/2: n in [2..80]]; // Vincenzo Librandi, Jan 03 2016

Join[{0, 7}, LinearRecurrence[{0, 2, 0, -1}, {1, 22, 1, 38}, 100]] (* Vincenzo Librandi, Jan 03 2016 *)
Block[{$MaxExtraPrecision=1000}, ContinuedFraction[Tan[1/8],100]] (* Harvey P. Dale, Jul 14 2025 *)

{ allocatemem(932245000); default(realprecision, 97000); x=contfrac(tan(1/8)); for (n=0, 20000, write("b019431.txt", n, " ", x[n+1])); } \\ Harry J. Smith, Jun 14 2009

Vec(x*(x^4-x^3+8*x^2+x+7)/((x-1)^2*(x+1)^2) + O(x^100)) \\ Colin Barker, Sep 08 2013

From Colin Barker, Sep 08 2013: (Start)
a(n) = (-1+3*(-1)^n-8*(-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+8*x^2+x+7) / ((x-1)^2*(x+1)^2). (End)

A265722 Number of ON (black) cells in the n-th iteration of the "Rule 1" elementary cellular automaton starting with a single ON (black) cell.

Original entry on oeis.org

1, 0, 1, 4, 1, 8, 1, 12, 1, 16, 1, 20, 1, 24, 1, 28, 1, 32, 1, 36, 1, 40, 1, 44, 1, 48, 1, 52, 1, 56, 1, 60, 1, 64, 1, 68, 1, 72, 1, 76, 1, 80, 1, 84, 1, 88, 1, 92, 1, 96, 1, 100, 1, 104, 1, 108, 1, 112, 1, 116, 1, 120, 1, 124, 1, 128, 1, 132, 1, 136, 1, 140

Offset: 0

Robert Price, Dec 14 2015

			From _Michael De Vlieger_, Dec 14 2015: (Start)
First 12 rows, replacing zeros with "." for better visibility of ON cells, followed by the total number of 1's per row at right:
                      1                        =   1
                    . . .                      =   0
                  . . 1 . .                    =   1
                1 1 . . . 1 1                  =   4
              . . . . 1 . . . .                =   1
            1 1 1 1 . . . 1 1 1 1              =   8
          . . . . . . 1 . . . . . .            =   1
        1 1 1 1 1 1 . . . 1 1 1 1 1 1          =  12
      . . . . . . . . 1 . . . . . . . .        =   1
    1 1 1 1 1 1 1 1 . . . 1 1 1 1 1 1 1 1      =  16
  . . . . . . . . . . 1 . . . . . . . . . .    =   1
1 1 1 1 1 1 1 1 1 1 . . . 1 1 1 1 1 1 1 1 1 1  =  20
(End)

S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 55.

Colin Barker, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Elementary Cellular Automaton
Index entries for sequences related to cellular automata
Index to Elementary Cellular Automata

Cf. A265718, A265720, A265721.

rule = 1; rows = 30; Table[Total[Table[Take[CellularAutomaton[rule, {{1},0},rows-1,{All,All}][[k]], {rows-k+1, rows+k-1}], {k,1,rows}][[k]]], {k,1,rows}]

Conjectures from Colin Barker, Dec 14 2015 and Apr 16 2019: (Start)
a(n) = 1/2*(-2*(-1)^n*n+2*n+3*(-1)^n-1).
a(n) = 2*a(n-2) - a(n-4) for n>3.
G.f.: (1-x^2+4*x^3) / ((1-x)^2*(1+x)^2).
(End)
a(n) = A019425(n), n>1. - R. J. Mathar, Jan 10 2016

A158496 Expansion of (1-4x+x^2)/(1+x^2)^2.

Original entry on oeis.org

1, -4, -1, 8, 1, -12, -1, 16, 1, -20, -1, 24, 1, -28, -1, 32, 1, -36, -1, 40, 1, -44, -1, 48, 1, -52, -1, 56, 1, -60, -1, 64, 1, -68, -1, 72, 1, -76, -1, 80, 1, -84, -1, 88, 1, -92, -1, 96, 1, -100, -1, 104, 1, -108, -1, 112, 1, -116, -1, 120, 1, -124, -1, 128, 1, -132, -1

Offset: 0

Paul Barry, Mar 20 2009

Hankel transform of A158495.

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

Cf. A019425.

CoefficientList[Series[(1-4x+x^2)/(1+x^2)^2,{x,0,70}],x] (* or *) LinearRecurrence[{0,-2,0,-1},{1,-4,-1,8},70] (* Harvey P. Dale, Mar 06 2012 *)

Vec((1-4*x+x^2)/(1+x^2)^2 + O(x^100)) \\ Altug Alkan, Jan 10 2016

a(n) = (n+3/2-(n+1/2)*(-1)^n)*(-1)^C(n+1,2).

a(0)=1, a(1)=-4, a(2)=-1, a(3)=8, a(n) = -2*a(n-2)-a(n-4). - Harvey P. Dale, Mar 06 2012

cp's OEIS Frontend

A019433 Continued fraction for tan(1/10).

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A019426 Continued fraction for tan(1/3).

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A019430 Continued fraction for tan(1/7).

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A019432 Continued fraction for tan(1/9).

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A133265 Diagonal of the A135356 triangle.

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A161011 Decimal expansion of tan(1/2).

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A019428 Continued fraction for tan(1/5).

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