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.

A123251 Continued fraction for sqrt(2)*tan(1/sqrt(2)).

Original entry on oeis.org

1, 4, 1, 3, 1, 12, 1, 7, 1, 20, 1, 11, 1, 28, 1, 15, 1, 36, 1, 19, 1, 44, 1, 23, 1, 52, 1, 27, 1, 60, 1, 31, 1, 68, 1, 35, 1, 76, 1, 39, 1, 84, 1, 43, 1, 92, 1, 47, 1, 100, 1, 51, 1, 108, 1, 55, 1, 116, 1, 59, 1, 124, 1, 63, 1, 132, 1, 67, 1, 140, 1, 71, 1, 148, 1, 75, 1, 156, 1, 79, 1
Offset: 1

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Author

Benoit Cloitre, Oct 08 2006

Keywords

Comments

This continued fraction allows us to see that tan(1/sqrt(2)), sin(1/sqrt(2)), cos(1/sqrt(2)) are irrational. More generally, for any fixed positive integer m, the continued fraction for sqrt(m)*tan(1/sqrt(m)) is given by a(12*n-11) = a(12*n-9) = a(12*n-7) = a(12*n-5) = a(12*n-3) = a(12*n-1) = 1; a(12*n-10) = 12*m*n - 9*m - 2; a(12*n-8) = 12*n-9; a(12*n-6) = 12*m*n - 5*m - 2; a(12*n-4) = 12*n-5; a(12*n-2) = 12*m*n - m - 2; a(12*n) = 12*n-1.
From Peter Bala, Oct 02 2023: (Start)
Further to the above, the simple continued fraction expansion for sqrt(2)*tan(sqrt(2)/2) may be derived by setting z = sqrt(2)/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))) together with further equivalence transformations.
The same approach can be used to find the simple continued fraction expansions for the numbers sqrt(N)*tan(sqrt(N)/(k*N)) and (sqrt(N)/N)*tan(sqrt(N)/(k*N)) for k >= 1. An example is given below. (End)

Examples

			From _Peter Bala_ Oct 03 2023: (Start)
For k > 1, the simple continued fraction expansion of sqrt(2)*tan(sqrt(2)/(2*k)) is [0; k - 1, 1, 2*3*k - 2, 1, 5*k - 2, 1, 2*7*k - 2, 1, 9*k - 2, 1, 2*11*k - 2, 1, 13*k - 2,  1, 2*15*k - 2, 1, ...], and the simple continued fraction expansion of (sqrt(2)/2)*tan(sqrt(2)/(2*k)) is [0; 2*k - 1, 1, 3*k - 2, 1, 2*5*k - 2, 1, 7*k - 2, 1, 2*9*k - 2, 1, 11*k - 2, 1, 2*13*k - 2, 1, 15*k - 2, 1, ...]. (End)
		

Crossrefs

Cf. A123168.

Programs

  • Magma
    continuedFraction(Sqrt(2)*Tan(1/Sqrt(2))); // G. C. Greubel, Oct 12 2018
  • Maple
    cfrac(sqrt(2)*tan(1/sqrt(2)),81,'quotients'); # Muniru A Asiru, Oct 13 2018
  • Mathematica
    ContinuedFraction[Sqrt[2]*Tan[1/Sqrt[2]], 100] (* G. C. Greubel, Oct 12 2018 *)
  • PARI
    contfrac(sqrt(2)*tan(1/sqrt(2))) \\ G. C. Greubel, Oct 12 2018
    

Formula

For n >= 1 we have a(12*n-11) = a(12*n-9) = a(12*n-7) = a(12*n-5) = a(12*n-3) = a(12*n-1) = 1; a(12*n-10) = 24*n-20; a(12*n-8) = 12*n-9; a(12*n-6) = 24*n-12; a(12*n-4) = 12*n-5; a(12*n-2) = 24*n-4; a(12*n) = 12*n-1.
Empirical g.f.: x*(x^7 - x^6 + 4*x^5 - x^4 + 3*x^3 + x^2 + 4*x + 1) / ((x-1)^2*(x+1)^2*(x^2+1)^2). - Colin Barker, Jun 28 2013
a(2*n-1) = 1, a(4*n) = 4*n-1 and a(4*n-2) = 8*n-4 for n >= 1. - Peter Bala, Oct 02 2023