A327344 9 + 8*a(n) appears in a congruence which determines representative parallel primitive binary quadratic forms for discriminant 9*m(n)^2 - 4 and representation -m(n)^2, where m(n) = A002559(n) (Markoff numbers).

0, 0, 39, 273, 1365, 333, 12870, 46410, 10878, 88218, 304668, 107559, 1576614, 2852889, 4144413, 13637988, 28406235, 53558505, 12085458, 92899170, 133886883, 34633998, 351194025, 1334488428, 1819412595, 410100933, 3041210445, 4333538430, 1118696184, 9146719764, 15150288153, 29675764248

Offset: 1

Author

Wolfdieter Lang, Sep 11 2019

Keywords

• nonn
• easy

Comments

See A327343 for the relevance of representative parallel primitive binary quadratic forms (rpapfs) for discriminant Disc(n) = 9*m(n)^2 - 4 and representation -m(n)^2 for the determination of ordered Markoff triples.

These rpapfs FPa(n) = [-m(n)^2, B(n), - C(n)] are determined by the solutions of the congruence z^2 - d(n) == 0 (mod (m(n)/2)^2), where d(n) = 9 + 8*a(n) if m(n) = A002559(n) is even, and the integer z(n) = 1 + 2*J(n) = B(n)/4 is from the set {1, 3, ..., 2*(m(n)/2)^2 - 1}. The members C(n) = 7 + (z^2 - d(n))/(m(n)/2)^2. In the odd m(n) case the congruence is B(n)^2 - d(n) == 0 (mod m(n)^2) , where d(n) = 9 + 8*a(n) and B(n) = 2*j(n) + 1 from the set {1, 3, ..., 2*m(n)^2 -1}. The member C(n) = 1 + (1/4)*(B(n)^2 - d(n))/m(n)^2. The different solutions are then FPa(n;i), for i = 1, 2, ..., #FPa(n), with #FPa(n) = A327343(n) = 2^A327342(n).

The d(n) sequence begins with {9, 9, 321, 2193, 10929, 2673, 102969, 371289, 87033, 705753,...}

Examples

			n = 6: m(6)/2 = 17, M(6) = (17 - 1)/16 = 1, a(6) = 37*1*9 = 333. d(6) = 2673.
n = 7: m(7) = 89, M(7) = 22, a(7) = 13*22*45 = 12870. d(7) = 102969.
The two (#FPa(6) = 2^1) solutions z(6) = B(6)/4 are z(n;1) = 19 and z(n;2) = 1559. They lead to FPa(6;1) = [-34^2, 76, +1] and FPa(6;2) = [-34^2, 2236, -1079].
The two (#FPa(7) = 2^1) solutions B(7) are B(7;1) = 199 and B(7;2) = 15643 (the upper bound was 2*m(7)^2 - 1 = 15841), leading to FPa(7;1) = [-89^2, 199, +1] and FPa(7;2) = [-89^2, 15643, -7721].
In both cases the second solution leads to the ordered Markoff triples MT(6) = (1, 13, 34) and MT(7) = (1, 34, 89). The other solution leads to the unordered triples (1, 34, 13) and (1, 89, 34).
		

Crossrefs

Cf. A002559, A309376, A327342, A327343.

Formula

a(n) = (d(n) - 9)/8 = 37*M(n)*(1 + 8*M(n)) with M(n) = A309376(n) = (m(n)/2 -1)/16 if m(n) is even, and a(n) = (d(n) - 9)/8 = 13*M(n)*(1 + 2*M(n)) with M(n) = A309376(n) = (m(n)-1)/4 if m(n) is odd.