cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-8 of 8 results.

A019425 Continued fraction for tan(1/2).

Original entry on oeis.org

0, 1, 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, 1, 144, 1, 148, 1, 152, 1, 156, 1
Offset: 0

Views

Author

David W. Wilson

Keywords

Comments

From Peter Bala, Nov 17 2019: (Start)
The simple continued fraction expansion for tan(1/2) may be derived by setting z = 1/2 in Lambert's continued fraction tan(z) = z/(1 - z^2/(3 - z^2/(5 - ... ))) and, after using an equivalence transformation, making repeated use of the identity 1/(n - 1/m) = 1/((n - 1) + 1/(1 + 1/(m - 1))).
The same approach produces the simple continued fraction expansions for the numbers tan(1/n), n*tan(1/n) and 1/n*tan(1/n) for n = 1,2,3,.... [added Oct 03 2023: and, even more generally, the simple continued fraction expansions for the numbers d*tan(1/n) and 1/d*tan(1/n), where d divides n. See A019429 for an example]. (End)

Examples

			0.546302489843790513255179465... = 0 + 1/(1 + 1/(1 + 1/(4 + 1/(1 + ...)))). - _Harry J. Smith_, Jun 13 2009

Links

Crossrefs

Cf. A161011 (decimal expansion). Cf. A019426 through A019433.

Programs

  • Magma
    [0,1] cat [n-1/2-(n-3/2)*(-1)^n+Binomial(1,n)- 2*Binomial(0,n): n in [2..80]]; // Vincenzo Librandi, Jan 03 2016
  • Maple
    a := n -> if n < 2 then n else ifelse(irem(n, 2) = 0, 1, 2*n - 2) fi:
seq(a(n), n = 0..80);  # Peter Luschny, Oct 03 2023
  • Mathematica
    Join[{0, 1}, LinearRecurrence[{0, 2, 0, -1}, {1, 4, 1, 8}, 100]] (* Vincenzo Librandi, Jan 03 2016 *)
  • PARI
    { allocatemem(932245000); default(realprecision, 85000); x=contfrac(tan(1/2)); for (n=0, 20000, write("b019425.txt", n, " ", x[n+1])); } \\ Harry J. Smith, Jun 13 2009

Formula

a(n) = n - 1/2 - (n-3/2)*(-1)^n + binomial(1,n) - 2*binomial(0,n). - Paul Barry, Oct 25 2007
From Philippe Deléham, Feb 10 2009: (Start)
a(n) = 2*a(n-2) - a(n-4), n>=6.
G.f.: (x + x^2 + 2*x^3 - x^4 + x^5)/(1-x^2)^2. (End)
From Peter Bala, Nov 17 2019; (Start)
Related simple continued fraction expansions:
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, ...]. (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

Views

Author

David W. Wilson

Keywords

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

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)

A019429 Continued fraction for tan(1/6).

Original entry on oeis.org

0, 5, 1, 16, 1, 28, 1, 40, 1, 52, 1, 64, 1, 76, 1, 88, 1, 100, 1, 112, 1, 124, 1, 136, 1, 148, 1, 160, 1, 172, 1, 184, 1, 196, 1, 208, 1, 220, 1, 232, 1, 244, 1, 256, 1, 268, 1, 280, 1, 292, 1, 304, 1, 316, 1, 328, 1, 340, 1, 352, 1, 364, 1, 376, 1, 388, 1, 400, 1, 412, 1, 424, 1, 436, 1
Offset: 0

Views

Author

David W. Wilson

Keywords

Examples

			0.16822721830224246125721608... = 0 + 1/(5 + 1/(1 + 1/(16 + 1/(1 + ...)))). - _Harry J. Smith_, Jun 13 2009

Links

Crossrefs

Cf. A161015 (decimal expansion). Cf. A019426 through A019433.

Programs

  • Mathematica
    Block[{$MaxExtraPrecision=1000},ContinuedFraction[Tan[1/6],100]] (* Harvey P. Dale, May 14 2014 *)
  • PARI
    { allocatemem(932245000); default(realprecision, 95000); x=contfrac(tan(1/6)); for (n=0, 20000, write("b019429.txt", n, " ", x[n+1])); } \\ Harry J. Smith, Jun 13 2009

Formula

Conjecture: a(n) = (-1+3*(-1)^n-6*(-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+6*x^2+x+5) / ((x-1)^2*(x+1)^2). - Colin Barker, May 28 2013
From Peter Bala, Nov 18 2019: (Start)
a(2*n) = 1 and a(2*n+1) = 12*n + 4, both for n >= 1.
The above conjectures are correct. The simple continued fraction expansion for tan(1/6) may be derived by setting z = 1/6 in Lambert's continued fraction tan(z) = z/(1 - z^2/(3 - z^2/(5 - ... ))) and then, after using an equivalence transformation, making repeated use of the identity 1/(n - 1/m) = 1/((n - 1) + 1/(1 + 1/(m - 1))).
A similar approach produces the related simple continued fraction expansions
2*tan(1/6) = [0, 2, 1, 34, 1, 13, 1, 82, 1, 25, 1, 130, 1, 37, 1, 178, 1, 49, ...], with denominators c(2*n) = 1, c(4*n+1) = 12*n + 1, both for n >= 1, and c(4*n+3) = 48*n + 34 for n >= 0;
3*tan(1/6) = [0; 1, 1, 52, 1, 8, 1, 124, 1, 16, 1, 196, 1, 24, 1, 268, 1, 32, ...];
6*tan(1/6) = [1; 106, 1, 3, 1, 250, 1, 7, 1, 394, 1, 11, 1, 538, 1, 15, 1, 682,..];
(1/2)*tan(1/6) = [0, 11, 1, 7, 1, 58, 1, 19, 1, 106, 1, 31, 1, 154, 1, 43, 1, ...];
(1/3)*tan(1/6) = [0, 17, 1, 4, 1, 88, 1, 12, 1, 160, 1, 20, 1, 232, 1, 28, 1, ...];
(1/6)*tan(1/6) = [0, 35, 1, 1, 1, 178, 1, 5, 1, 322, 1, 9, 1, 466, 1, 13, 1, ...];
(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

Views

Author

David W. Wilson

Keywords

Examples

			0.14383695943619093528003059... = 0 + 1/(6 + 1/(1 + 1/(19 + 1/(1 + ...)))). - _Harry J. Smith_, Jun 14 2009

Links

Crossrefs

Cf. A161016 (decimal expansion), A019425 through A019433.

Programs

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

Formula

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

Views

Author

David W. Wilson

Keywords

Comments

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

Examples

			0.11157062783380058372650480... = 0 + 1/(8 + 1/(1 + 1/(25 + 1/(1 + ...)))). - _Harry J. Smith_, Jun 14 2009

Links

Crossrefs

Cf. A161018 (decimal expansion), A019425 through A019433.

Programs

  • Mathematica
    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 *)
  • PARI
    { 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
  • PARI
    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

Formula

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)

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

Views

Author

David W. Wilson

Keywords

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

Examples

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

Links

Crossrefs

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

Programs

  • Magma
    [0,4] cat [(-1+3*(-1)^n-5*(-1+(-1)^n)*n)/2: n in [2..80]]; // Vincenzo Librandi, Jan 03 2016
  • Mathematica
     Join[{0, 4}, LinearRecurrence[{0, 2, 0, -1}, {1, 13, 1, 23}, 100]] (* Vincenzo Librandi, Jan 03 2016 *)
  • 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

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)

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

Views

Author

David W. Wilson

Keywords

Examples

			0.12565513657513096779267821... = 0 + 1/(7 + 1/(1 + 1/(22 + 1/(1 + ...)))). - _Harry J. Smith_, Jun 14 2009

Links

Crossrefs

Cf. A161017 (decimal expansion), A019425 through A019433.

Programs

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

Formula

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)

A161019 Decimal expansion of tan(1/10).

Original entry on oeis.org

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

Views

Author

Harry J. Smith, Jun 14 2009

Keywords

Comments

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

Examples

			0.100334672085450545058080045781111536819004804576442040022208065798032...

Links

Crossrefs

Cf. A019433 (continued fraction).

Programs

  • Mathematica
    RealDigits[Tan[1/10], 10, 120][[1]] (* Amiram Eldar, Jun 27 2023 *)
  • PARI
    default(realprecision, 20080); x=10*tan(1/10); for (n=0, 20000, d=floor(x); x=(x-d)*10; write("b161019.txt", n, " ", d));
Showing 1-8 of 8 results.

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