%I #8 Sep 20 2019 12:37:29
%S 0,0,39,273,1365,333,12870,46410,10878,88218,304668,107559,1576614,
%T 2852889,4144413,13637988,28406235,53558505,12085458,92899170,
%U 133886883,34633998,351194025,1334488428,1819412595,410100933,3041210445,4333538430,1118696184,9146719764,15150288153,29675764248
%N 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).
%C 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.
%C 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).
%C The d(n) sequence begins with {9, 9, 321, 2193, 10929, 2673, 102969, 371289, 87033, 705753,...}
%F 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.
%e n = 6: m(6)/2 = 17, M(6) = (17  1)/16 = 1, a(6) = 37*1*9 = 333. d(6) = 2673.
%e n = 7: m(7) = 89, M(7) = 22, a(7) = 13*22*45 = 12870. d(7) = 102969.
%e 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].
%e 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].
%e 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).
%Y Cf. A002559, A309376, A327342, A327343.
%K nonn,easy
%O 1,3
%A _Wolfdieter Lang_, Sep 11 2019
