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 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 (list; graph; refs; listen; history; text; internal format)
 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 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 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 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(n-1). 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

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Last modified December 15 09:05 EST 2019. Contains 329995 sequences. (Running on oeis4.)