A179238 - OEIS

%I #48 Feb 14 2024 10:34:50

%S 1,2,3,5,13,18,31,80,111,191,493,684,1177,3038,4215,7253,18721,25974,

%T 44695,115364,160059,275423,710905,986328,1697233,4380794,6078027,

%U 10458821,26995669,37454490,64450159,166354808,230804967,397159775

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

%C If n == 2 (mod 3), a(n) is in A179237, prefaced with a 1: (1, 2, 13, 80, 493, ...).

%C Conjecture relating palindromic continued fractions to the tangents of angles A and 2A:

%C Tan 2A is rational when tan(A) is irrational, iff tan(A) is a convergent to an infinitely repeating palindromic continued fraction.

%C Conversely, if tan(A) is such a convergent, then tan(2A) must be rational.

%C 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.

%C Conjecture proved by the author, Jul 05 2010. - _Gary W. Adamson_, Mar 25 2014

%C 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

%H Gary W. Adamson, <a href="http://qntmpkt.blogspot.com/2010/07/proof-of-adamsons-conjecture-on.html">Proof of Adamson's Conjecture on Palindromic Continued Fractions, re: A179238</a>.

%H Marcia Edson, Scott Lewis and Omer Yayenie, <a href="http://www.emis.de/journals/INTEGERS/papers/l32/l32.pdf">The k-periodic Fibonacci sequence and an extended Binet's formula</a>, INTEGERS 11 (2011) #A32.

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,6,0,0,1). - _R. J. Mathar_, Jul 06 2010

%F a(1) = 1; a(n+1) = a(n) + a(n-1) if n == (0,1) mod 3.

%F a(n+1) = 2*a(n) + a(n-1) if n == 2 mod 3.

%F a(n) = +6*a(n-3) +a(n-6). G.f.: x*(1+2*x+3*x^2-x^3+x^4)/(1-6*x^3-x^6). - _R. J. Mathar_, Jul 06 2010

%F a(3n+2) = A179237(n+1). - _R. J. Mathar_, Feb 14 2024

%e The first few convergents given

%e [1,...2,...1,...1,...2,...1,... =

%e .1....2....3....5...13,..18,...

%e ...

%e a(5) = 13 = 2*a(4) + a(3) = 2*10 + 3 since 5 == 2 mod 3

%e a(6) = 18 = a(5) + a(4) = 13 + 5 since 6 == 0 mod 3.

%e ...

%e Examples relating to the conjecture: Since A179238 has two recursive multipliers, we can extract the sequence A179237: (1, 2, 13, 80, 493, 3038,...) = convergents of A179238 such that a(n+1) = 2*a(n) + a(n-1). The convergent of this sequence = 1/(sqrt(10) - 3) = 6.16227766, ... = a, where 1/a = 0.1622776, ... 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.

%Y Cf. A179237.

%K nonn,easy

%O 1,2

%A _Gary W. Adamson_, Jul 04 2010