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 A306120 Lengths of largest face diagonal in 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 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 (list; graph; refs; listen; history; text; internal format)
 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

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Last modified June 14 04:54 EDT 2021. Contains 345018 sequences. (Running on oeis4.)