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

1,1

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

Formula

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

Prog

(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.

Crossrefs

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

Sequence in context: A144662 A278307 A060402 * A232106 A049014 A186056

Adjacent sequences: A306117 A306118 A306119 * A306121 A306122 A306123

Keyword

nonn

Author

M. F. Hasler, Oct 11 2018

Status

approved