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.

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

Author

M. F. Hasler, Oct 11 2018

Keywords

• nonn

Comments

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.

Links

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

Crossrefs

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

Programs

PARI

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.

Formula

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