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

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

1,2

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)

Links

G. C. Greubel, Table of n, a(n) for n = 1..1000

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

Example

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)

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 *)

Prog

(PARI) contfrac(sqrt(2)*tan(1/sqrt(2))) \\ G. C. Greubel, Oct 12 2018
(Magma) continuedFraction(Sqrt(2)*Tan(1/Sqrt(2))); // G. C. Greubel, Oct 12 2018

Crossrefs

Cf. A123168.

Sequence in context: A323072 A055032 A039930 * A368922 A327357 A334810

Adjacent sequences: A123248 A123249 A123250 * A123252 A123253 A123254

Keyword

nonn,cofr

Author

Benoit Cloitre, Oct 08 2006

Status

approved