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%I #13 Feb 19 2022 10:22:18
%S 57,73,43,127,97,111,49,95,296,152,323,147,285,255,247,469,403,871,
%T 561,657,559,1083,833,1057,485,507,1072,760,767,379,211,195,1208,952,
%U 1443,1023,1051,889,1240,1209,1249,1423,1005,1679,1568,1843,193,485,1512
%N Smallest 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 Inspired by Project Euler, Problem 143 (see link).
%C For the corresponding primitive triples and miscellaneous properties and references, see A336328.
%C If FA + FB + FC = d, then we have this "beautifully symmetric equation" between a, b, c and d (see Martin Gardner):
%C 3*(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2.
%C This sequence is not increasing. For example, a(2) = 73 for triangle with largest side = 95 while a(3) = 43 for triangle with largest side = 152.
%D Martin Gardner, Mathematical Circus, Elegant triangles, First Vintage Books Edition, 1979, p. 65.
%H Project Euler, <a href="https://projecteuler.net/problem=143">Problem 143 - Investigating the Torricelli point of a triangle</a>.
%F a(n) = A336328(n, 1).
%e a(1) = 57 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 Cf. A336328 (triples), A336329 (FA + FB + FC), this sequence (smallest side), A336331 (middle side), A336332 (largest side), A336333 (perimeter).
%Y Cf. A072054 (smallest sides: primitives and multiples).
%K nonn
%O 1,1
%A _Bernard Schott_, Jul 21 2020
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