cp's OEIS Frontend

a(n) ~ (1+G)*A225206(n)/Pi, where G is Catalan's constant (A006752).

The numbers x, y and z form a sigma-quadratic triple. See Dimitrov link.

Other than a(2) whose walk is comprised of 8 diagonal segments, all known terms are produced by 3- or 4-segment walks, including some with examples of both. It is conjectured that this holds true for all n >= 3.

For n = 2, the walk segments are the hypotenuses of noncongruent primitive Pythagorean triangles.

The offset is 2, because even though the graph could be defined in dimension 1 (the vertices would be the points of Z, with each point connected to its two neighbors), it would not contain any closed paths.

Removing the constraint that the diagonal segments must all have the same length gives A386251. All such walks in 2 dimensions, by diagonal segment length rather than total walk length, gives A386671.

This sequence is case k=2, A008784 is case k=1, A004613 is case k=4 of divisors of 1 + k*x^2.

From Peter M. Chema, May 08 2017 (Start): Also gives the body diagonals of all primitive Pythagorean quadruples that define square prisms, with sides [b, b, and c] and diagonal d, such that 2*b^2 + c^2 = d^2. E.g., sides [2, 2, 1], diagonal 3 = a(2); [4, 4, 7], 9 = a(3); [6, 6, 7], 11 = a(4); [12, 12, 1], 17 = a(5); [6, 6, 17] 19 = a(6); [10, 10, 23], 27 = a(7); [20, 20, 17], 33 = a(8); [24, 24, 23], 41 = a(9)... (a subsequence of A096910) (End)

Editorial note: The above comment would be better expressed by talking about right tetrahedra (also called trirectangular tetrahedra), that is, tetrahedra with vertices (b 0 0), (0 c 0), (0 0 d) (here b=c). These are the correct generalizations of Pythagorean triangles. N. J. A. Sloane, May 08 2017

From Frank M Jackson, May 23 2017: (Start)

Starting at a(2)=3, this gives the shortest side of a primitive Heronian triangle whose perimeter is 4 times its shortest side. Aka a primitive integer Roberts triangle (see Buchholz link).

Also odd and primitive terms generated by x^2 + 2y^2 with x>0 and y>0.

Also integers with all prime divisors congruent to 1 or 3 (mod 8). (End)

It is provable that all such walks must be even in total length. It is also provable that 3-segment closed walks are impossible for n < 6, and conjectured that a(n) for all n >= 6 are produced by 3-segment walks.

For n = 2, the walk segments are the hypotenuses of noncongruent primitive Pythagorean triangles.

The offset is 2, because even though the graph could be defined in dimension 1 (the vertices would be the points of Z, with each point connected to its two neighbors), it would not contain any closed paths.

Adding a constraint that the diagonal segments must all have the same length gives A385525.

A Pythagorean quadruple is a quadruple (a,b,c,d) of positive integers such that a^2 + b^2 + c^2 = d^2 with a <= b <= c. Its inradius is (a+b+c-d)/2, which is a positive integer.

For every positive integer n, there is at least one Pythagorean quadruple with inradius n.

To determine how "cube-like" a Pythagorean quad is, we use the quotient C/(C-a*b*c) where C is the volume of an ideal cube for a given diagonal d, C=(d/sqrt(3))^3. A ratio of 100 indicates that the best {a,b,c} combination creates a solid 1 part per 100 smaller than the ideal cube volume.

Contains all the terms in A001835 and A079935 except the leading 1s. For such a term b(m) contained in a(n) from those sequences, 2*b(m) will tie the current record, while 3*b(m) will tie but will also break the current record exactly half the time and thus appear as a(n+1). This means round((2+sqrt(3))*b(m)) will also be in the sequence as either a(n+1) or a(n+2) if a better quad for 3*b(n) was found.

The constant (2+sqrt(3)) follows from the recurrence relationship b(n)=4*b(n-1)-b(n-2). The constant can be represented as the continued fraction 3,1,2,1,2,1,2,1,2,.... The equivalents for the 2D case (A001653) are 3+2*sqrt(2) and {5,1,4,1,4,1,4,1,4,...}.

We conjecture this method provides complete solutions. This was confirmed directly by brute-force testing up to 1e7 (optimized using the sum-of-two-squares theorem for a given d^2-a^2 = b^2 + c^2; see link below).