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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.

%I A375098 #22 Aug 09 2025 03:36:34
%S A375098 3,9,11,41,123,153,571,1713,2131,7953,23859,29681,110771,332313,
%T A375098 413403,1542841,4628523,5757961,21489003,64467009,80198051,299303201,
%U A375098 897909603,1117014753,4168755811,12506267433,15558008491,58063278153,174189834459,216695104121
%N 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.
%C A375098 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.
%C A375098 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.
%C A375098 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,...}.
%C A375098 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).
%H A375098 Christian N. K. Anderson, <a href="/A375098/b375098.txt">Table of n, a(n) for n = 1..1000</a>
%H A375098 Christian N. K. Anderson, <a href="/A375098/a375098.txt">Table of most cube-like Pythagorean quadruples and the corresponding and the error quotient for d=1..10000 </a>
%H A375098 Christian N. K. Anderson, <a href="/A375098/a375098_1.txt">Table of d, a, b, c and error quotient for n = 1..50.</a>
%H A375098 Christian N. K. Anderson, <a href="/A375098/a375098.nb">Mathematica code</a> for generating this sequence with different amounts of rigorous error checking.
%F A375098 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),
%F A375098 For n == 1 (mod 3), a(n) = 4*a(n-2)-a(n-1) OR a(n) = floor(a(n-1)*(2+sqrt(3))),
%F A375098 For n == 2 (mod 3), a(n) = 3*a(n-1).
%e A375098 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.
%e A375098 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.
%e A375098 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.
%e A375098 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.
%t A375098 (* An efficient program is provided in the links section. *)
%Y A375098 Cf. A096907, A096908, A096909, A096910 for lists of a, b, c, and d of the 10,000 first Pythgorean quads, sorted by ascending d.
%Y A375098 Contains all terms in A001835 and A079935 except the leading 1s.
%Y A375098 Cf. A001653: The 2D equivalent of this sequence (i.e., right triangle whose legs are closest to equal)
%K A375098 nonn
%O A375098 1,1
%A A375098 _Christian N. K. Anderson_, Mark K. Transtrum, and David D. Allred, Jul 29 2024