A389354 - OEIS

%I #10 Oct 10 2025 23:13:01

%S 0,3,98,969,108878,312018,3694603,10078344,100468830,1086820980,

%T 4258790305,10549105808,32350651245,144668863570,343512402093,

%U 1044916071794,1207337145420,10416809232063,109731742000660,166942232236153

%N a(n) = (A388289(n)^2 - 1)/(3*2^5), for n >= 1.

%C This sequence appears in the discriminants of the reduced principal Markoff forms F_p(X, Y) with even Markoff numbers.

%C For the discriminants see A305312(n) = Disc(n) = 9*m(n)^2 - 4 = b(n)*(b(n) + 4), with b(n) = 3*m(n) - 2 = A324250(n), and A308687(n) = D(n), for all Markoff numbers m(n) = A002559(n).

%C The reduced principal indefinite binary quadratic Markoff forms are F_p(n; X, Y) = X^2 + b(n)*X*Y - b(n)*Y^2, for n >= 1, representing -m(n)^2. The Frobenius-Markoff conjecture, that there is only one Markoff triple with largest number m(n), is trivially true for n = 1 and n = 2 (the so-called singular cases), and for n >= 3 it claims that there is only one proper solution F_p(n; X, Y) = -m(n)^2, with X < 0, Y >= 1, and Y - X < m(n).

%C For even Markoff numbers m(A388290(n)) = 2*A388289(n) one has DiscEven(n) = Disc(A388290(n)) = 4*D(A388290(n)).

%C The sequence member a(n) appears in the formula for DiscEven(n) = 32*(1 + 2^2*3^3*a(n)) = 32*(1 + 108*a(n)), for n >= 1.

%F a(n) = Me(n)*(1 + 8*Me(n))/3, where Me(n) = A309376(A388290(n)), for n >= 1.

%e a(1) = (1^2 - 1)/(2^5*3) = 0.

%e a(2) = (17^2 - 1)/(2^5*3) = 288/(32*3) = 3.

%e a(2) = 1*(1 + 8*1)/3 = 3.

%Y Cf. A002559, A308687, A324250, A305312, A309376, A388289, A388290.

%K nonn,easy

%O 1,2

%A _Wolfdieter Lang_, Oct 02 2025