

A179238


Numerators in convergents to infinitely repeating period 3 palindromic continued fraction [1,2,1,...].


1



1, 2, 3, 5, 13, 18, 31, 80, 111, 191, 493, 684, 1177, 3038, 4215, 7253, 18721, 25974, 44695, 115364, 160059, 275423, 710905, 986328, 1697233, 4380794, 6078027, 10458821, 26995669, 37454490, 64450159, 166354808, 230804967, 397159775
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OFFSET

1,2


COMMENTS

If n == 2 mod 3, a(n) is in A179237, prefaced with a 1: (1, 2, 13, 80, 493, ...).
Conjecture relating palindromic continued fractions to the tangents of angles A and 2A:
Tan 2A is rational when tan(A) is irrational, iff tan(A) is a convergent to an infinitely repeating palindromic continued fraction.
Conversely, if tan(A) is such a convergent, then tan(2A) must be rational.
The conjecture applies to any palindromic form such as [a,a,a,...], [a,b,a,a,b,a,...], [a,b,b,a,a,b,b,a,...], [a,b,c,c,b,a,...] and so on.
Conjecture proved by the author, Jul 05 2010.  Gary W. Adamson, Mar 25 2014
Tan(2) corrected to tan(A), which is the half angle of a right triangle with sides (a, b) and c = sqrt(a^2 + b^2), not an integer. Example: Given a right triangle with sides a = 3, b = 5, c = sqrt(34), we calculate the two acute angles 2A and divide each by two, giving us two angles A; then perform tan(A) on each. These have an algebraic equivalency of two constants: 3/(5 + sqrt(34)) = 0.276983964... and 5/(3 + sqrt(34)) = 0.566190378... The continued fraction representations of these constants are respectively barover[3,1,1,1,1,3] and barover[1,1,3,3,1,1], both palindromic. The conjecture and proof don't apply to Pythagorean triangles, in which case the tangents of the half angles would be rational.  Gary W. Adamson, Apr 07 2014


LINKS

Table of n, a(n) for n=1..34.
Gary W. Adamson, Proof of Adamson's Conjecture on Palindromic Continued Fractions, re: A179238.
Marcia Edson, Scott Lewis and Omer Yayenie, The kperiodic Fibonacci sequence and an extended Binet's formula, INTEGERS 11 (2011) #A32.
Index entries for linear recurrences with constant coefficients, signature (0,0,6,0,0,1).  R. J. Mathar, Jul 06 2010


FORMULA

a(1) = 1; a(n+1) = a(n) + a(n1) if n == (0,1) mod 3.
a(n+1) = 2*a(n) + a(n1) if n == 2 mod 3.
a(n) = +6*a(n3) +a(n6). G.f.: x*(1+2*x+3*x^2x^3+x^4)/(16*x^3x^6).  R. J. Mathar, Jul 06 2010


EXAMPLE

The first few convergents given
[1,...2,...1,...1,...2,...1,... =
.1....2....3....5...13,..18,...
...
a(5) = 13 = 2*a(4) + a(3) = 2*10 + 3 since 5 == 2 mod 3
a(6) = 18 = a(5) + a(4) = 13 + 5 since 6 == 0 mod 3.
...
Examples relating to the conjecture: Since A179238 has two recursive multipliers, we can extract the series A179237: (1, 2, 13, 80, 493, 3038,...) = convergents of A179238 such that a(n+1) = 2*a(n) + a(n1). The convergent of this series = 1/(sqrt(10)  3) = 6.16227766,... = a, where 1/a = .1622776,... the convergent to the palindromic continued fraction [6,6,6,...]. Then letting 1/a = tan(A), then tan(2A) = 1/3, rational. Similarly, taking a = 6.16227766,... = tan(A), then tan(2A) = 1/3, rational, with a and 1/a irrational.


CROSSREFS

Cf. A179237.
Sequence in context: A281598 A042261 A112596 * A041385 A108282 A042047
Adjacent sequences: A179235 A179236 A179237 * A179239 A179240 A179241


KEYWORD

nonn


AUTHOR

Gary W. Adamson, Jul 04 2010


EXTENSIONS

More terms from R. J. Mathar, Jul 06 2010


STATUS

approved



