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A351476 If F is the Fermat point of a triangle ABC with A < B < C < 2*Pi/3, where AB, BC, CA, FA, FB and FC are all positive integers, then, this sequence gives the sum FA + FB + FC when gcd(a, b, c) = A351477(n).

%I A351476 #33 Mar 11 2022 12:37:52
%S A351476 784,1029,6845,80089,24843,109561,109561,5239,24037,47045,27735,6760,
%T A351476 477481,21904,57967,186245,365403,20280,400445,356168,159953,190463,
%U A351476 718205,836405,11809,178771,1432443,1127307,22984,34295,22477,157339,6723649,44403,974408
%N A351476 If F is the Fermat point of a triangle ABC with A < B < C < 2*Pi/3, where AB, BC, CA, FA, FB and FC are all positive integers, then, this sequence gives the sum FA + FB + FC when gcd(a, b, c) = A351477(n).
%C A351476 Inspired by Project Euler, Problem 143 (see link) where such a triangle is called a "Torricelli triangle".
%C A351476 Differs from A336329 where ABC is a primitive integer-sided triangle with A < B < C < 2*Pi/3 and only FA+FB+FC is an integer; in fact, terms of A336329 are the sum of 3 fractions of the form FA = p/t, FB = q/t, FC = r/t but (p+q+r)/t is an integer. Here, FA, FB and FC are all integers and the sums FA+FB+FC are displayed according to same order as in A336329. The corresponding common denominators t of the fractions (p/t, q/t, r/t) are in A351477.
%C A351476 If FA + FB + FC = d, then we have this "beautifully symmetric equation" between a, b, c and d (see Martin Gardner):
%C A351476 3*(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2.
%C A351476 => d = sqrt(((a^2 + b^2 + c^2)/2) + (1/2) * sqrt(6*(a^2*b^2 + b^2*c^2 + c^2*a^2) - 3*(a^4 + b^4 + c^4))).
%D A351476 Martin Gardner, Mathematical Circus, Elegant triangles, First Vintage Books Edition, 1979, p. 65.
%H A351476 Project Euler, <a href="https://projecteuler.net/problem=143">Problem 143 - Investigating the Torricelli point of a triangle</a>.
%H A351476 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/FermatPoints.html">Fermat points</a>.
%F A351476 a(n) = A336329(n) * A351477(n).
%F A351476 If FA + FB + FC = d, then
%F A351476 d^2 = (1/2) * (a^2 + b^2 + c^2) + 2 * S * sqrt(3) where S = area of triangle ABC.
%e A351476 a(1) = FA + FB + FC = 325 + 264 + 195 = 784, corresponding to first triple (399, 455, 511) whose gcd = 7.
%e A351476 a(6) = FA + FB + FC = 70720 + 34200 + 4641 = 109561, corresponding to triple (36741, 73151, 92680) whose gcd = 331.
%e A351476 a(7) = FA + FB + FC = 91200 + 12376 + 8985 = 109561, corresponding to triple (16219, 94335, 97976) whose gcd = 331.
%o A351476 (PARI) lista(nn) = {my(d); for(c=4, nn, for(b=ceil(c/sqrt(3)), c-1, for(a=1+(sqrt(4*c^2-3*b^2)-b)\2, b-1, if(gcd([a, b, c])==1 && issquare(d=6*(a^2*b^2+b^2*c^2+c^2*a^2)-3*(a^4+b^4+c^4)) && issquare(d=(a^2+b^2+c^2+sqrtint(d))/2), d = sqrtint(d); my(s = numerator(sqrtint(((2*b*c)^2 - (b^2+c^2-d^2)^2)/3)/d) + numerator(sqrtint(((2*a*b)^2 - (a^2+b^2-d^2)^2)/3)/d) + numerator(sqrtint(((2*a*c)^2 - (a^2+c^2-d^2)^2)/3)/d)); print1(s, ", ");););););} \\ _Michel Marcus_, Mar 02 2022
%Y A351476 Cf. A336328, A336329, A336330, A336331, A336332, A336333, A351477, A351801, A351802, A351803.
%K A351476 nonn
%O A351476 1,1
%A A351476 _Bernard Schott_, Feb 12 2022
%E A351476 More terms from _Jinyuan Wang_, Feb 17 2022