A286016 Signed continued fraction expansion with all signs negative of tanh(1).

1, 5, 2, 2, 2, 2, 9, 2, 2, 2, 2, 2, 2, 2, 2, 13, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 17, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 21, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 25, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2

Offset

1,2

Comments

For any given sequence of signs (e_1, e_2, ..., e_n, ...) one may define the signed continued fraction expansion of a real number x by using floor or ceiling in the step i according to e_i = +1 or e_i = -1. In the present case for the sequence (-1, -1, -1, -1, ...) consisting of only negative signs the ceiling is taken at each step, and the formulas with x_0 = x are a_n = ceiling(x_n) and x_{n+1} = 1/(a_n - x_n). See chapter 1 and 2 of the book by Perron.

Links

Table of n, a(n) for n=1..86.

Adolf Hurwitz, Über die Kettenbrüche, deren Teilnenner arithmetische Reihen bilden, Vierteljahrsschrift der Naturforschenden Gesellschaft in Zürich, Jahrg. 41 (1896), Jubelband 2, 34-64. Reprinted in Hurwitz's Mathematische Werke.

Oskar Perron, Die Lehre von den Kettenbrüchen, Teubner, Leipzig, 1930.

Formula

Using an obvious condensed notation we get for the sequence 1, 5, 2^(4), 9, 2^(8), 13, 2^(12), 17, 2^(16), 21, 2^(20), ... where 2^(m) means m copies of 2.

Example

a(2) = 5, a(3) = a(4) = a(5) = a(6) = 2, a(7) = 9, etc. These numbers are obtained from the partial quotients xj as follows:
x2 =  (1 +  e^2)/( 2 + 0e^2) ~4.17 so that a(2)=ceiling(x2)=5;
x3 =  (2 + 0e^2)/( 9 - e^2)  ~1.21 so that a(3)=ceiling(x3)=2;
x4 =  (9 -  e^2)/(16 - 2e^2) ~1.31 so that a(4)=ceiling(x4)=2;
x5 = (16 - 2e^2)/(23 - 3e^2) ~1.46 so that a(5)=ceiling(x5)=2;
x6 = (23 - 3e^2)/(30 - 4e^2) ~1.87 so that a(6)=ceiling(x6)=2;
x7 = (30 - 4e^2)/(37 - 5e^2) ~8.11 so that a(7)=ceiling(x7)=9.
The pairs of integers appearing in the xj's are obtained as the principal or as every other of the non-principal approximating fractions of e^2 in the sense of the A. Hurwitz reference.

Maple

x:=(exp(1)-exp(-1))/(exp(1)+exp(-1)):b:=ceil(x): x1:=1/(b-x):L:=[b]:
for k from 0 to 40 do:
b1:=ceil(x1): x1:=1/(b1-x1): L:=[op(L), b1]: od: print(L);

Crossrefs

Cf. A004273 (continued fraction of tanh(1)), A280135, A280136.

Sequence in context: A081119 A303579 A306966 * A119320 A363252 A070962

Adjacent sequences: A286013 A286014 A286015 * A286017 A286018 A286019

Keyword

nonn,cofr

Author

Kutlwano Loeto, Apr 30 2017

Extensions

More terms from Jinyuan Wang, Jul 02 2022

Status

approved