A375098 Diagonals of a Euclidian solid such that there exists a Pythagorean quadruple d^2=a^2+b^2+c^2 that is more cube-like than any prior value of d.

3, 9, 11, 41, 123, 153, 571, 1713, 2131, 7953, 23859, 29681, 110771, 332313, 413403, 1542841, 4628523, 5757961, 21489003, 64467009, 80198051, 299303201, 897909603, 1117014753, 4168755811, 12506267433, 15558008491, 58063278153, 174189834459, 216695104121

Offset

1,1

Comments

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).

Links

Christian N. K. Anderson, Table of n, a(n) for n = 1..1000

Christian N. K. Anderson, Table of most cube-like Pythagorean quadruples and the corresponding and the error quotient for d=1..10000

Christian N. K. Anderson, Table of d, a, b, c and error quotient for n = 1..50.

Christian N. K. Anderson, Mathematica code for generating this sequence with different amounts of rigorous error checking.

Formula

For n == 0 (mod 3), a(n) = 4*a(n-2)-a(n-3) OR a(n)=floor(a(n-1)*(2+sqrt(3))/3),

For n == 1 (mod 3), a(n) = 4*a(n-2)-a(n-1) OR a(n)=floor(a(n-1)*(2+sqrt(3)),

For n == 2 (mod 3), a(n) = 3*a(n-1).

Example

3 is in the sequence because 3^2=1^2+2^2+2^2 is the smallest Pythagorean quad, with an error of one part in 4.344.
6 is NOT in the sequence because {6,2,4,4} is the most cube-like Pythagorean quad, but only ties the previous record without breaking it.
7 is NOT in the sequence because the most cube-like quad {7,2,3,6} has an error of one part in 2.2, worse than that for d=3.
9 is in the sequence NOT because of {9,3,6,6} which ties the previous record, but because {9,4,4,7} improves on the previous record with an error of one part in 4.958.

Mathematica

An efficient program is provided in the links section.

Crossrefs

Cf. A096907, A096908, A096909, A096910 for lists of a, b, c, and d of the 10,000 first Pythgorean quads, sorted by ascending d.

Contains all terms in A001835 and A079935 except the leading 1s.

Cf. A001653: The 2D equivalent of this sequence (i.e., right triangle whose legs are closest to equal)

Sequence in context: A135365 A088877 A069871 * A061957 A136984 A168400

Adjacent sequences: A375095 A375096 A375097 * A375099 A375100 A375101

Keyword

nonn

Author

Christian N. K. Anderson, Mark K. Transtrum, and David D. Allred, Jul 29 2024

Status

approved