

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



5, 32, 221, 1517, 7565, 10400, 71285, 257045, 338720, 488597, 1687397, 3348896, 8732021, 15800621, 22953677, 75533477, 157326845, 296631725, 376282400, 514518485, 741527357, 1078334240, 1945074605, 7391012837, 10076746685, 12768548000, 16843627085, 24001135925, 34830756896, 50658755621, 83909288237, 164358078917, 342312755621, 347220276512, 781553243021, 1636268213885, 2244540316037, 2379883179965, 3756053306912, 7713367517021
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OFFSET

1,1


COMMENTS

Subsequence of A079896.
For the Markoff form f_{m(n)}(x, y) = m(n)*F_{m(n)}(x, y) of Cassels (pp. 3139), see the comments on A305310. Some references are given in A002559, A305308 and A305310.
f_m(x, y) is an indefinite binary quadratic form because the discriminant is positive.
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.
To find reduced forms one needs f(n) := ceiling(sqrt(D(n))) which is 3*m(n) because (3*m1)^2 < 9*m^2  4 < (3*m)^2, due to 6*m(n) > 5, for n >= 1.
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


REFERENCES

J. W. S. Cassels, An Introduction to Diophantine Approximation, Cambridge University Press, 1957, Chapter II, The Markoff Chain, pp. 1844.


LINKS

Table of n, a(n) for n=1..40.


FORMULA

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


EXAMPLE

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


CROSSREFS

Cf. A002559, A079896, A305308, A305310.
Sequence in context: A297068 A024064 A164594 * A199486 A277756 A208632
Adjacent sequences: A305309 A305310 A305311 * A305313 A305314 A305315


KEYWORD

nonn,easy


AUTHOR

Wolfdieter Lang, Jun 26 2018


STATUS

approved



