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%I #12 Jan 19 2019 04:15:43
%S 14,21,26,29,30,35,38,41,42,45,46,49,50,53,54,59,61,62,65,66,69,70,74,
%T 75,77,78,81,83,86,89,90,91,93,94,98,101,105,106,107,109,110,113,114,
%U 115,117,118,121,122,125,126,129,131,133,134,137,138,139,141,142,145
%N Numbers that are the sum of 3 distinct and primitive nonzero squares.
%C This sequence gives the increasingly ordered numbers m which satisfy A224772(m) > 0.
%C This sequence is a proper subsequence of A004432. The first imprimitive members of A004432 are 56, 84, 104, 116, 120, 140, 152, 164, 168, 180, 184, 196, 200, ...
%H T. D. Noe, <a href="/A224771/b224771.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) is the n-th largest number m which satisfies: m = a^2 + b^2 + c^2, with integers a, b, and c, 0 < a < b < c, and gcd(a,b,c) = 1. Such a solution is denoted by the triple (a, b, c).
%e The first triples (a, b, c) are:
%e n=1, 14: (1, 2, 3),
%e n=2, 21: (1, 2, 4),
%e n=3, 26: (1, 3, 4),
%e n=4, 29: (2, 3, 4),
%e n=5, 30: (1, 2, 5),
%e n=6, 35: (1, 3, 5),
%e n=7, 38 (2, 3, 5),
%e n=8, 41: (1, 2, 6),
%e n=9, 42: (1, 4, 5),
%e n=10, 45: (2, 4, 5),
%e ...
%e The first member with two different triples is a(18) = 62 with the triples (1, 5, 6), (2, 3, 7).
%e The first member with three different triples is a(36) = 101 with the triples (1, 6, 8), (2, 4, 9) and (4, 6, 7).
%t nn = 150; t = Table[0, {nn^2}]; Do[If[GCD[a, b, c] == 1, n = a^2 + b^2 + c^2; If[n <= nn^2, t[[n]]++]], {a, nn}, {b, a + 1, nn}, {c, b + 1, nn}]; Flatten[Position[t, _?(# > 0 &)]] (* _T. D. Noe_, Apr 20 2013 *)
%Y Cf. A224772 (multiplicities), A224773 (one half of the even members), A004432, A025442.
%K nonn
%O 1,1
%A _Wolfdieter Lang_, Apr 19 2013
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