A327342 a(n) gives the number of distinct odd prime divisors of m(n) = A002559(n) (Markoff numbers).

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, 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, 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

Offset

1,12

Comments

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).

References

Martin Aigner, Markov's Theorem and 100 Years of the Uniqueness Conjecture, Springer, 2013.

Links

Table of n, a(n) for n=1..100.

Formula

a(n) = number of distinct odd prime divisors of m(n) = A002559(n), for n >= 1.

a(n) = A005087(A002559(n)). - Michel Marcus, Sep 18 2023

Example

For the examples a(6) = 1 and a(12) = 2 see A327343.

Crossrefs

Cf. A002559, A005087, A324250, A327343.

Sequence in context: A058745 A275333 A108393 * A297828 A062245 A062246

Adjacent sequences: A327339 A327340 A327341 * A327343 A327344 A327345

Keyword

nonn,easy

Author

Wolfdieter Lang, Sep 11 2019

Status

approved