A179238 Numerators in convergents to infinitely repeating period 3 palindromic continued fraction [1,2,1,...].

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

Offset: 1

Author

Gary W. Adamson, Jul 04 2010

Keywords

• nonn
• easy

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

Examples

			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 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.
		

Links

• Gary W. Adamson, Proof of Adamson's Conjecture on Palindromic Continued Fractions, re: A179238.
• Marcia Edson, Scott Lewis and Omer Yayenie, The k-periodic 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

Crossrefs

Cf. A179237.

Formula

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

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

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

a(3n+2) = A179237(n+1). - R. J. Mathar, Feb 14 2024