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%I #16 Sep 18 2023 07:39:51
%S 0,0,1,1,1,1,1,1,1,1,1,2,2,2,1,1,2,1,2,1,2,2,2,1,1,2,1,2,2,2,1,2,3,3,
%T 2,1,2,1,1,2,2,2,2,3,1,1,3,1,1,2,3,2,2,2,3,2,1,3,3,2,3,1,3,3,2,2,2,3,
%U 1,2,2,2,2,2,2,1,3,2,1,4,1,2,2,4,2,1,3,3,4,3,1,1,2,2,2,1,3,2,3,2
%N a(n) gives the number of distinct odd prime divisors of m(n) = A002559(n) (Markoff numbers).
%C 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).
%C 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).
%D Martin Aigner, Markov's Theorem and 100 Years of the Uniqueness Conjecture, Springer, 2013.
%F a(n) = number of distinct odd prime divisors of m(n) = A002559(n), for n >= 1.
%F a(n) = A005087(A002559(n)). - _Michel Marcus_, Sep 18 2023
%e For the examples a(6) = 1 and a(12) = 2 see A327343.
%Y Cf. A002559, A005087, A324250, A327343.
%K nonn,easy
%O 1,12
%A _Wolfdieter Lang_, Sep 11 2019
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