A268396 -id:A268396 - cp's OEIS frontend

A245616 Pythagorean Threesomes: triples of natural numbers defining the six legs of three Pythagorean triangles.

44, 117, 240, 240, 252, 275, 88, 234, 480, 85, 132, 720, 160, 231, 792, 132, 351, 720, 480, 504, 550, 176, 468, 960, 170, 264, 1440, 220, 585, 1200, 720, 756, 825, 320, 462, 1584, 264, 702, 1440, 308, 819, 1680, 255, 396, 2160, 960, 1008, 1100, 352, 936, 1920, 480, 693, 2376, 396, 1053, 2160, 429, 880, 2340, 340, 528, 2880

Offset: 1

Andreas Boe, Nov 05 2014

nonn

The sequence is sorted by increasing sums of triples and secondly by increasing order of first term.
The three numbers in a Pythagorean Threesome define the lengths of three sides of a tetrahedron with all integer length edges and one right angle vertex.
The sequence was calculated for the science fiction novel "The Fifth Jack" by Andreas Boe, Amazon books, 2014.
I do not have that book, but this sequence is closely related to (and may be an erroneous version of) A268396. - Arkadiusz Wesolowski, Feb 03 2016

			(44,117,240)  sqrt(44^2+117^2)=125  sqrt(117^2+240^2)=267 sqrt(240^2+44^2)=244

Andreas Boe, Table of n, a(n) for n = 1..285

Same numbers sorted gives A195816.

Cf. A268396.

x,y,sqrt(x^2+y^2) y,z,sqrt(y^2+z^2) z,x,sqrt(z^2+x^2)

Edited by N. J. A. Sloane, Feb 11 2016

A306120 Lengths of largest face diagonal in primitive Euler bricks or Pythagorean cuboids: possible values of max(d, e, f) for solutions to a^2 + b^2 = d^2, a^2 + c^2 = e^2, b^2 + c^2 = f^2 in coprime positive integers a, b, c, d, e, f.

Original entry on oeis.org

267, 373, 732, 825, 843, 1595, 1884, 2500, 2775, 3725, 3883, 6380, 6409, 8140, 8579, 9188, 9272, 9512, 11764, 12125, 13123, 14547, 14681, 14701, 19572, 20503, 20652, 24695, 25121, 25724, 29307, 30032, 30695, 31080, 32595, 34484, 37104, 37895, 38201, 38965

Offset: 1

M. F. Hasler, Oct 11 2018

nonn

These are the values obtained as sqrt(A031173(n)^2 + A031174(n)^2), sorted by size (n = 3 yields 843, n = 4 yields 732) and duplicates removed: The first duplicate is 71402500^2 = A031173(1428)^2 + A031174(1428)^2 = A031173(1626)^2 + A031174(1626)^2, there is no other among the first 3500 terms.
This considers only the face diagonals, not the space diagonals.
See the main entry A031173 for links, cross-references, and further comments.

M. F. Hasler, Table of n, a(n) for n = 1..2500

Cf. A031173, A031174, A031175, A195816, A196943, A268396.

A306120=Set(vector(1000,n,sqrtint(A031173(n)^2+A031174(n)^2)))[1..-100] \\ Discard the last 100 values, which may have holes. This is empirical: better find the smallest sqrtint(A031173(n)^2+A031174(n)^2) with n > 1000 not in the set, and discard all elements larger than that.

A306120 = { sqrtint(A031173(n)^2+A031174(n)^2); n >= 1 }.

cp's OEIS Frontend

A245616 Pythagorean Threesomes: triples of natural numbers defining the six legs of three Pythagorean triangles.

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