cp's OEIS Frontend

The Markov tree is a complete, infinite binary tree. Vertices are labeled by triples. The root vertex is (1, 5, 2). The left child of (a, b, c) is (a, 3*a*b - c, b); its right child is (b, 3*b*c - a, c). The sequence is a triangle read by rows consisting of the middle element of each triple, which is always the largest element of the triple. Row r contains 2^r elements.

The tree contains contains exactly one representative of each class of permutation equivalent nonsingular solutions to Markov's equation, a^2 + b^2 + c^2 = 3 * a * b * c. Nonsingular solutions are those in which a, b, and c are three distinct numbers. The two singular triples (1, 1, 1) and (1, 2, 1) are omitted in this sequence.

A consequence of Markov's equation is that the recurrence for the tree may be reformulated as follows: the left child of (a, b, c) is (a, (a^2 + b^2) / c, b); its right child is (b, (b^2 + c^2) / a, c).

An open problem is to prove the uniqueness conjecture, which asserts that the largest element of a triple determines the other two.

Frobenius proposed assigning a rational number index in (0,1) to each vertex of the tree, and hence to each term in this sequence. This is the Farey index, obtained by assigning the triple (0/1, 1/2, 1/1) to the root vertex and using the following rules to assign triples to the rest of the tree: the vertex labeled (u/v, w/x, y/z) with w = u + u and x = v + z has left child (u/v, (u+w)/(v+x), w/x) and right child (w/x, (w+y)/(x+z), y/z). The Farey index is the center element of the triple. Each rational number in (0, 1) appears as the Farey index of exactly one vertex of the tree. The index of a(n) is A007305(n+2) / A007306(n+2).

A sequence of leftward steps in the tree produces odd-indexed Fibonacci numbers, A001519, which have Farey indices of the form 1 / n. A sequence of rightward steps in the tree produces odd-indexed Pell numbers, A001653, which have Farey indices of the form (n - 1) / n. A sequence of leftward steps followed by a single rightward step produces A350922, corresponding to Farey indices of the form 2 / (2 * n + 1). Alternating steps right, left, right, left, right, ... produces A064098, which corresponds to Farey indices of the form F(n) / F(n + 1), where F(n) is the n-th Fibonacci number.

The Shiraishi theorem demonstrated that there were an infinite number of cubic number triples whose sum equaled a cubic number (A226903). Through an analysis of the cubic number triples found by Russell and Gwyther, another way to prove that there are an infinite number of cubic number triples who sum equals a cubic number appeared. For any triple, one can add a zero to the end of the three numbers. The new three numbers will also equal a cubic number. For example, 3^3+4^3+5^3=6^3 can be transformed into 30^3+40^3+50^3=60^3. The number of zeros that are consistently applied to each of the numbers who cubic numbers will always create new cubic numbers. For example, 30^3+40^3+50^3=60^3 can become 300^3+400^3+500^3=600^3 and 3000^3+4000^3+5000^3=6000^3, and so on. Through experiments, the formula holds true for Pythagorean triples and Pythagorean quadruples as well. To apply the method to Pythagorean triples, 3^2+4^2=5^2 can be transformed into 30^2+40^2=50^2, 300^2+400^2=500^2, and so on. For Pythagorean quadruples, 3^2+4^2+12^2=13^2 can be transformed into 30^2+40^2+120^2=130^2 and then to 300^2+400^2+1200^2=1300^2. The property holds even beyond the second and third powers. For example, 3530^4=300^4+1200^4+2720^4+3150^4 just as 353^4=30^4+120^4+272^4+315^4. Additionally, 1440^5=270^5+840^5+1100^5+1330^5 just as 144^5=27^5+84^5+110^5+133^5. Once one set is found, it appears there can be an infinite number of similar sets for any power through this method.

The list of Russell and Gwyther also reveals that the cube of 38 can be represented as the sum of the cubes of nine unique positive integers. This is because 38^3=3^3+4^3+5^3+7^3+14^3+17^3+18^3+24^3+30^3.