%I #18 Feb 19 2022 04:54:50
%S 7,7,37,283,91,331,331,13,43,97,43,13,691,37,91,193,349,13,283,211,97,
%T 91,379,409,7,97,691,613,13,19,13,91,2593,19,349,43,1,337,97,169,37,
%U 19,31,409,3217,67,571,169,241,43,67,157,4171,3601,889,1591,811,1,139
%N a(n) is the common denominator of FA, FB and FC, where F is the Fermat point of the integer-sided triangle ABC with A < B < C < 2*Pi/3 such that FA + FB + FC = A336329(n).
%C Inspired by Project Euler, Problem 143 (see link).
%C For the corresponding primitive triples, miscellaneous properties and references, see A336328.
%H Leisure Maths Entertainment Forum, <a href="http://kuing.orzweb.net/viewthread.php?tid=6994">The primitive integer triangles with nonzero rational distances between three vertices and 1st isogonic center</a>, Chinese blog.
%H Project Euler, <a href="https://projecteuler.net/problem=143">Problem 143 - Investigating the Torricelli point of a triangle</a>.
%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/FermatPoints.html">Fermat points</a>.
%F a(n) = A351476(n)/A336329(n).
%F a(n) is the common denominator of fractions FA, FB, FC when FA = sqrt(((2*b*c)^2 - (b^2+c^2-d^2)^2)/3) / d, FB = sqrt(((2*a*c)^2 - (a^2+c^2-d^2)^2)/3) / d, FC = sqrt(((2*a*b)^2 - (a^2+b^2-d^2)^2)/3) / d, with a = (A336328(n,1), b = (A336328(n,2), c = (A336328(n,3)) and d = A336329(n) (formulas FA, FB, FC from _Jinyuan Wang_, Feb 17 2022).
%e For 1st triple (57, 65, 73) in A336328, we get A336329(1) = FA + FB + FC = 325/7 + 264/7 + 195/7 = 112, hence a(1) = 7.
%e For 3rd triple (43, 147, 152) in A336328, we get A336329(3) = FA + FB + FC = 5016/37 + 1064/37 + 765/37 = 185, hence a(3) = 37.
%Y Cf. A336328 (primitive triples), A336329 (FA + FB + FC), A336330 (smallest side), A336331 (middle side), A336332 (largest side), A336333 (perimeter).
%Y Cf. A351476.
%K nonn
%O 1,1
%A _Bernard Schott_, Feb 12 2022
%E More terms from _Jinyuan Wang_, Feb 17 2022