Of late, regarding the Harmonious Pythagorean Tetrahedra, I thought it prudent to consider both apex singularity solutions inclusively. The resulting object is a concave hexahedron where the base right triangle exists in the medial plane.
By manipulating the associated geometric construction, the requisite Pythagorean areas are distributed among the lateral faces of the upper and lower “tetrahedra”. In other words, I divided the area of each lateral face in the single-solution case (tetrahedron) and applied half to each in the two-solution case such that their sum achieves Pythagorean compliance for the developed hexahedra.
This exploration led to a fun tangent of interest as detailed below. Note that simply changing the common divisor of the equation below will produce different integer sequences and thus a different mapping of Pythagorean Triples. In this particular example I have chosen a common divisor of 2.
Established is a right triangle where a is the short cathetus, b the long and c the hypotenuse.
As an extension of 0 = c2 – b2 – a2 and when applied to a Pythagorean Triple, the square root of c squared divided by two, less the same for b and again for a, yields the negative square root of an even integer.
√(c2/2) – √(b2/2) – √(a2/2) = -√+ℤ
For example, where: a = 3 , b = 4 , c = 5
-√+ℤ = √(52/2) – √(42/2) – √(32/2)
-√+ℤ = √(25/2) – √(16/2) – √(9/2)
-√+ℤ = √12.5 – √8 – √4.5
-√+ℤ = 3.535533905932738 – 2.82842712474619 – 2.121320343559643
-√+ℤ = -1.414213562373095
ℤ = 2
Solved for both primitives and multiples, these integers then sorted in ascending order produce the following sequence that when mapped provide an alternative “tree” of Pythagorean Triples.
2,8,8,18,18,18,32,32,32,50,50,50,72,72,72,72,72,72,98,98,98,128,128,128,128,162,162,162,162,162,200,200,200,200,200,200,242,242,242,288,
288,288,288,288,288,288,288,288,338,338,338,392,392,392,392,392,392,450,450,450,450,450,450,450,450,450,512,512,512,512,512,578,578,
578,648,648,648,648,648,648,648,648,648,648,722,722,722,800,800,800,800,800,800,800,800,800,882,882,882,882,882,882,882,882,882,968,
968,968,968,968,1058,1058, …
Mapping result:
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In the graph below, an incomplete population is depicted. Each integer occurrence is represented by the a/b ratio of a Pythagorean Triple. Curves indicate the ascending sequential paths of integer single occurrence with lower integer values corresponding to larger a/b ratios.
Steve,
This is a nice follow-up to your earlier posts about Harmonious Pythagorean Tetrahedra, and the mapping patterns that you show are very interesting!
I was wondering if the alternative “tree” of Pythagorean Triples might already appear on OEIS. The sequence https://oeis.org/A078644 (1, 2, 3, 3, 3, 6, 3, 4, 5, 6, 3, 9, 3, 6, 9, 5, 3, 10, 3, 9, 9…) does match up until the 21st term, but unless there is a typo, the 22nd term (multiple of 968) is 6 instead of your 5.
Many thanks for the interesting development of your Harmonious Pythagorean Tetrahedra Conjecture.
William
Nice to hear from you William and I hope you are well.
The 2,8,8,18,18,18,32,32,32,50,50,50,72,72,72,72,72,72,98,… integer sequence matches most closely to https://oeis.org/A093907 and I have already contacted the author Dr. Guillermo Restrepo of the Max Planck Institute for Mathematics in the Sciences. He graciously responded and ask for additional information which I have sent. Unfortunately, he has not had time to evaluate. The 2004 OEIS submission from himself and Leonardo Pachon was titled “Number of elements in the n-th period of the periodic table as predicted by the Aufbau principle”. The significant difference of the two sequences are the multiple occurrences of like values vs. those of paired occurrence. The Aufbau principle is fascinating and I anxiously await Dr. Restrepo’s response.
Having said that, the sequence you point out, I missed completely. A preliminary check does indicate that 5 is correct in the 22nd term however I will effort a more thorough investigation. As the https://oeis.org/A078644 sequence references Pythagorean primitives, I am hopeful of the typographic error.
I find often get a little too close and cannot see the forest for the trees. Many thanks for checking behind me. I will let you know as conclusions appear.
Steve