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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.
1
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
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
KEYWORD
nonn
AUTHOR
M. F. Hasler, Oct 11 2018
STATUS
approved