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A336331 Middle side of primitive integer-sided triangles with A < B < C < 2*Pi/3 and such that FA + FB + FC is an integer where F is the Fermat point of the triangle.

%I A336331 #10 Feb 19 2022 10:22:26
%S A336331 65,88,147,168,185,221,285,312,315,343,392,377,464,343,408,589,725,
%T A336331 901,931,1085,1120,1240,1147,1128,1208,1400,1105,1099,1005,1464,1541,
%U A336331 1544,1635,1403,1463,1387,1744,1891,1617,1505,1769,1885,2072,2345,1713,2165,2667,2784,2855
%N A336331 Middle side of primitive integer-sided triangles with A < B < C < 2*Pi/3 and such that FA + FB + FC is an integer where F is the Fermat point of the triangle.
%C A336331 Inspired by Project Euler, Problem 143 (see link).
%C A336331 For the corresponding primitive triples and miscellaneous properties and references, see A336328.
%C A336331 If FA + FB + FC = d, then we have this "beautifully symmetric equation" between a, b, c and d (see Martin Gardner):
%C A336331    3*(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2.
%C A336331 This sequence is not increasing. For example, a(11) = 392 for triangle with largest side = 407 while a(12) = 377 for triangle with largest side = 437.
%D A336331 Martin Gardner, Mathematical Circus, Elegant triangles, First Vintage Books Edition, 1979, p. 65.
%H A336331 Project Euler, <a href="https://projecteuler.net/problem=143">Problem 143 - Investigating the Torricelli point of a triangle</a>.
%F A336331 a(n) = A336328(n, 2).
%e A336331 a(1) = 65 because the first triple is (57, 65, 73) with corresponding d = FA + FB + FC = 264/7 + 195/7 + 325/7 = 112 and the symmetric relation satisfies: 3*(57^4 + 65^4 + 73^4 + 112^4) = (57^2 + 65^2 + 73^2 + 112^2)^2 = 642470409.
%Y A336331 Cf. A336328 (triples), A336329 (FA + FB + FC), A336330 (smallest side), this sequence (middle side), A336332 (largest side), A336333 (perimeter).
%Y A336331 Cf. A072053 (middle sides: primitives and multiples).
%K A336331 nonn
%O A336331 1,1
%A A336331 _Bernard Schott_, Jul 20 2020