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A305312 Discriminant a(n) of the indefinite binary quadratic Markoff form m(n)*F_{m(n)}(x, y) with m(n) = A002559(n), for n >= 1. 6

%I #15 Jul 30 2018 15:06:44

%S 5,32,221,1517,7565,10400,71285,257045,338720,488597,1687397,3348896,

%T 8732021,15800621,22953677,75533477,157326845,296631725,376282400,

%U 514518485,741527357,1078334240,1945074605,7391012837,10076746685,12768548000,16843627085,24001135925,34830756896,50658755621,83909288237,164358078917,342312755621,347220276512,781553243021,1636268213885,2244540316037,2379883179965,3756053306912,7713367517021

%N Discriminant a(n) of the indefinite binary quadratic Markoff form m(n)*F_{m(n)}(x, y) with m(n) = A002559(n), for n >= 1.

%C Subsequence of A079896.

%C For the Markoff form f_{m(n)}(x, y) = m(n)*F_{m(n)}(x, y) of Cassels (pp. 31-39), see the comments on A305310. Some references are given in A002559, A305308 and A305310.

%C f_m(x, y) is an indefinite binary quadratic form because the discriminant is positive.

%C a(n) is also the discriminant D(n) = a(n) of the indefinite binary quadratic form determining the Markoff triple MT(n) = (x(n), y(n), m(n)) if the largest member is m(n) = A002559(n) and x(n) <= y(n) <= m(n). This is the form x^2 - 3*m*x*y + y^2 = -m^2 (with dropped argument n), or in reduced version X^2 + b*X*Y - b*Y^2 = -m^2, with b = b(n) = 3*m(n) - 2, where X = X(n) = y(n) - x(n) and Y = Y(n) = y(n). The uniqueness of such Markoff triples MT(n) with given largest members m(n) is a conjecture.

%C To find reduced forms one needs f(n) := ceiling(sqrt(D(n))) which is 3*m(n) because (3*m-1)^2 < 9*m^2 - 4 < (3*m)^2, due to 6*m(n) > 5, for n >= 1.

%C If the forms for a Markoff triple with largest member m are numerated with n giving m as m(n) = A002559(n)as in the present entry then the uniqueness conjecture is assumed to be true. Otherwise certain m(n) will lead to several different forms. - _Wolfdieter Lang_, Jul 30 2018

%D J. W. S. Cassels, An Introduction to Diophantine Approximation, Cambridge University Press, 1957, Chapter II, The Markoff Chain, pp. 18-44.

%F a(n) = 9*m(n)^2 - 4 = 9*A002559(n)^2 - 4, n >= 1.

%e a(5) = 7565 because 9*29^2 - 4 = 7565.

%Y Cf. A002559, A079896, A305308, A305310.

%K nonn,easy

%O 1,1

%A _Wolfdieter Lang_, Jun 26 2018

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