Markov number

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A Markov number or Markoff number is a positive integer x, y or z that is part of a solution to the Markov Diophantine equation

$x^2 + y^2 + z^2 = 3xyz.\,$

The first few Markov numbers are

1, 2, 5, 13, 29, 34, 89, 169, 194, 233, 433, 610, 985, 1325, ... (sequence A002559 in OEIS)

appearing as coordinates of the Markov triples

(1, 1, 1), (1, 1, 2), (1, 2, 5), (1, 5, 13), (2, 5, 29), (1, 13, 34), (1, 34, 89), (2, 29, 169), (5, 13, 194), (1, 89, 233), (5, 29, 433), (89, 233, 610), etc.

The Markov numbers, arranged in a binary tree.

There are infinitely many Markov numbers and Markov triples. The symmetry of the Markov equation allows us to rearrange the order of the coordinates, so a Markov triple $(a, b, c)$ may be normalized, as above, by assuming that $a\le b\le c$ . Aside from the two smallest triples, every Markov triple $(a, b, c)$ consists of three distinct integers. The unicity conjecture states that for a given Markov number $c$ , there is exactly one normalized solution having $c$ as its largest element. (A030452 lists Markov numbers that appear in solutions where one of the other two terms is 5).

The Markov numbers can also be arranged in a binary tree. The largest number at any level is always about a third from the bottom. All the Markov numbers on the regions adjacent to 2's region are odd-indexed Pell numbers (or numbers n such that $2 n 2 - 1$ is a square, A001653), and all the Markov numbers on the regions adjacent to 1's region are odd-indexed Fibonacci numbers (A001519). Thus, there are infinitely many Markov triples of the form

$(1, F_{2n - 1}, F_{2n + 1}),\,$

where F_x is the xth Fibonacci number. Likewise, there are infinitely many Markov triples of the form

$(2, P_{2n - 1}, P_{2n + 1}),\,$

where P_x is the xth Pell number.

Odd Markov numbers are 1 more than multiples of 4, while even Markov numbers are 2 more than multiples of 32.^[1]

Knowing one Markov triple (x, y, z) one can find another Markov triple, of the form $(x, y,3 x y - z)$ . Markov numbers are not always prime; but members of a Markov triple are always coprime. It's not necessary that $x < y < z$ in order for the $(x, y,3 x y - z)$ to yield another triple. In fact, if one doesn't change the order of the members before applying the transform again, it returns the same triple one started with. Thus, starting with (1, 1, 2) and trading y and z before each iteration of the transform lists Markov triples with Fibonacci numbers. Starting with that same triplet and trading x and z before each iteration gives the triples with Pell numbers.

In 1979, Don B. Zagier proved that the nth Markov number is asymptotically given by

$m_n = \tfrac13 e^{C\sqrt{n}+o(1)} \quad\text{with } C = 2.3523418721 \ldots$ .

Moreover he pointed out that $x 2 + y 2 + z 2 = 3 x y z + 4 / 9$ , an extremely good approximation of the original Diophantine equation, is equivalent to $f (x) + f (y) = f (z)$ with f(t) = arcosh(3t/2).^[2]

The nth Lagrange number can be calculated from the nth Markov number with the formula

$L_n = \sqrt{9 - {4 \over {m_n}^2}}.\,$

Markov numbers are named after the Russian mathematician Andrey Markov. Due to the different but equally valid ways of transliterating Cyrillic, the term is written as "Markoff numbers" in some literature. But in this particular case, "Markov" might be preferable because "Markoff number" might be misunderstood as "mark-off number."