cp's OEIS Frontend

First appearance of n in A097266.

Conjecture: a(15)=a(38)=0.

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).

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.