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Subsequence of A079896.

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.

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*m-1)^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

For these Markoff forms see Cassels, p. 31. A link to the two original Markoff references is given in A305308.

MF(n) = f_{m(n)}(x, y) = m(n)*F_{m(n)}(x, y) = m(n)*x^2 + (3*m(n) - 2*k(n))*x*y + (l(n) - 3*k(n))*y^2, with the Markoff number m = m(n) = A002559(n) and l(n) = (k(n)^2 + 1)/m(n), for n >= 1.

Every m(n) is proved to appear as largest member of a Markoff triple MT(n) = (m_1(n), m_2(n), m(n)), with positive integers m_1(n) < m_2(n) < m(n) for n >= 3 (MT(1) = (1, 1, 1) and MT(2) = (1, 1, 2)) satisfying the Markoff equation m_1(n)^2 + m_2(n)^2 + m(n)^2 = 3*m_1(n)*m_2(n)*m(n). The famous Markoff uniqueness conjecture is that m(n) as largest member determines exactly one ordered triple MT(n). See, e.g., the Aigner reference, pp. 38-39, and Corollary 3.5, p. 48. [In numerating the sequence with n related to A002559(n) this conjecture is assumed to be true. - Wolfdieter Lang, Jul 29 2018]

The nonnegative integers k(n) are defined for the Markoff forms given by Cassels by k(n) = min{k1(n), k2(n)}, where m_1(n)*k1(n) - m_2(n) == 0 (mod m(n)), with 0 <= k1(n) < m(n), and m_2(n)*k2(n) - m_1(n) == 0 (mod m(n)), with 0 <= k2(n) < m(n). The k1 and k2 sequences are k1 = [0, 1, 2, 5, 17, 13, 34, 99, 119, 89, 179, 233, 577, 818, 610, 1777, 1597, 3363, 2673, 2923, 5609, 4181, 6089, 10946, 19601, 22095, 26536, 31881, 38447, 28657, 39916, 51709, 114243, 75025, 113922, 263522, 206855, 196418, 396263, 572063, ...], and k2 = [0, 1, 3, 8, 12, 21, 55, 70, 75, 144, 254, 377, 408, 507, 987, 1120, 2584, 2378, 3793, 4638, 3468, 6765, 8612, 17711, 13860, 15571, 16725, 19760, 23763, 46368, 56641, 83428, 80782, 121393, 180763, 162867, 292538, 317811, 249755, 353702, ...].

The discriminant of the form MF(n) = f_{m(n)}(x, y) is D(n) = 9*m(n)^2 - 4. D(n) = A305312(n), for n >= 1. Because D(n) > 0 (not a square) this is an indefinite binary quadratic form, for n >= 1. See Cassels Fig. 2 on p. 32 for the Markoff tree with these forms.

The quadratic irrational xi, determined by the solution with positive square root of f_{m(n)}(x, 1) = 0, is xi(n) = ((2*k - 3*m) + sqrt(D))/(2*m) (the argument n has been dropped). The regular continued fraction is eventually periodic, but not purely periodic. One can find equivalent Markoff forms determining purely periodic quadratic irrationals. The corresponding k sequence is given in A305311.

For the approximation of xi(n) with infinitely many rationals (in lowest terms) Perron's unimodular invariant M(xi) enters. For quadratic irrationals M(xi) < 3, and the values coincide with the discrete Lagrange spectrum < 3: M(xi(n)) = Lagrange(n) = sqrt{D(n)}/m(n), n >= 1. For n=1..4 see A002163, A010466, A200991 and A305308.

The second member m_2 of the Markoff (Markov) triple MT(n) = (m_1(n), m_2(n), m(n)) with m_1(n) <= m_2(n) <= m(n), for n >= 1, with m(n) = A002559(n) is given in A305314(n). For n>=3 the inequalities are strict. The existence of MT(n) with largest number m(n) is proved. The uniqueness is conjectured. The Markoff equation is (the argument n is dropped) m_1^2 + m_2^2 + m^2 = 3*m_1*m_2*m. See the references under A002559.

See A305313 for comments, and A002559 for references.

Only the odd-indexed Fibonacci numbers appear in the Markoff sequence. The n-th term is about 0.1673*n^2.

The indefinite binary quadratic form F(n,x,y) = x^2 - 3*m(n)*x*y + y^2 = [1, -3*m(n), 1] representing -m(n)^2 with m(n) = A002559(n), determines Markoff triples MT(n) = (x(n) = A305313(n), y(n) = A305314(n), m(n)) with x(n) < y(n) < m(n), for n >= 3. For n = 1 and 2: x(n) = y(n) = 1. The Frobenius-Markoff conjecture is that this solution is unique. This form F(n,x,y) has discriminant D(n) = (3*m(n))^2 - 4 = a(n)*(a(n) + 4) = A305312(n) > 0.

Because -3*m(n) < 0 this form F(n,x,y) is not reduced (see e.g., the Buell reference, or the W. Lang link in A225953 for the definition).

The principal reduced form for F(n,x,y) is prF(n,X,Y) = X^2 + a(n)*X*Y - a(n)*Y^2 = [1, a(n), -a(n)]. (See, e.g., Lemma 2 of the W. Lang link in A225953 where b = a(n), f(D(n)) = ceiling(sqrt(D(n))) = 3*m(n), and D(n) and f(D(n)) have the same parity.) The relation between these forms is F(n,Y,Y-X) = prF(n,X,Y) with Y > 0, Y-X > 0, and X <= 0 (for n >= 3, X < 0).

Only the odd-indexed Pell numbers (A001653) appear in the Markoff sequence. The n-th term is about 0.5602*n^2.

These sequence members appear as exponents of 2 in the number of representative parallel primitive forms for binary quadratic forms of discriminant Disc(n) = 9*m(n)^2 - 4 and representation of -m(n)^2. The reduced (primitive) principal form of this discriminant is F_p(n; X, Y) = X^2 + b(n)*X*Y - b(n)*Y^2, written also as F_p(n) = [1, b(n), -b(n)], with b(n) = 3*m(n) - 2 = A324250(n). This form representing -m(n)^2 is important for the determination of Markoff triples MT(n).

For more details see A327343(n) = 2^a(n). The Frobenius-Markoff uniqueness conjecture on ordered triples with largest member m(n) is certainly true for m(n) if a(n) = 0 (so-called singular cases) or 1. See the Aigner reference, p. 59, Corollary 3.20, for n >= 3 (the a(n) = 1 cases).

For the definition of parallel forms for an indefinite binary quadratic form with discriminant Disc and representation of an integer k see, e.g., the Buell, Scholz-Schoeneberg references or the W. Lang link, section 3, with a scanning prescription.

For the Markoff case Disc(n) = 9*m(n) - 4 = b(n)*(b(n)+2), with m = A002559 and b = A324250.

The Markoff form MF(n;x,y) = x^2 - 3*m(n)*x*y + y^2, also written as MF(n) = [1, -3*m(n), 1], representing -m(n)^2, has as first reduced form the principal form F_p(n;X,Y) = X^2 + b(n)*X*Y - b(n)*Y^2, or F_p(n) = [1, b(n), -b(n)], where the connection is X = x-y, Y = x, or x = Y, y = Y - X. Hence X <= 0 for x <= y.

Only proper solutions (gcd(X, Y) = 1) are of interest. Also only primitive representative parallel forms FPa(n;i), for i = 1, 2, ..., #FPa(n), are considered.

In the present case it is possible to give directly the prescription for the primitive representative parallel forms (rpapfs). This is done for the even m(n) == 2 (mod 32) case and the odd case m(n) == 1 (mod 4) separately.

These rpapfs are written as FPa(n;i) = [-m(n)^2, B(n,i), -C(n,i)]. Their number a(n) = #FPa(n) can be found from congruences with an application of the Chinese remainder theorem and the lifting theorem (see Apostol, Theorem 5.26, pp. 118-119, and Theorem 5.30, pp. 121-122 (only part (a) is effective here)). The existence of two solutions for each odd prime modulus is important as input for the lifting to higher prime powers. For each of the singular cases m(1) = 1 and m(2) = 2, without odd prime divisors, there is only one rpapf.

The Frobenius-Markoff uniqueness conjecture is certainly true for m(n) if a(n) = 1 or a(n) = 2. In the latter case the two rpapfs have to be equivalent to the principal form F_p(n), because the known solution implied by the ordered triple MT(n) = (x(n), y(n), m(n)) has an unordered partner solution which after ordering becomes (x(n), y'(n), m(n')) with y'(n) = m(n) and m(n') = 3*x(n)*m(n) - y(n) >= m(n).

See A327344 for details on the congruences which determine the rpapfs.

Frobenius' conjecture on Markoff triples is that the maximal member z of the triple of positive integers (x,y,z), satisfying x^2 + y^2 + z^2 - 3*x*y*z = 0, with x <= y <= z, determines x and y uniquely. Also, each entry from A002559 (Markoff numbers) is conjectured to appear as a maximal member z. If an entry A002559(n) should not appear as z then one puts z(n) = 0 and row n will be 0, 0.

If this Frobenius conjecture is true then the row length of this array is always 2, and only positive numbers appear.