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A351803
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a(n) = A351477(n) * FC where F is the Fermat point of a primitive integer-sided triangle ABC with A < B < C < 2*Pi/3 and FA + FB + FC = A336329(n).
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5
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195, 264, 765, 13464, 3515, 4641, 5985, 360, 6120, 5096, 7616, 435, 111360, 1785, 7752, 47957, 80475, 6307, 74613, 50715, 16640, 52800, 123845, 181608, 520, 24000, 265200, 94600, 885, 3264, 1357, 6120, 1721400, 3128, 162393, 2409, 384, 122507, 27720, 22575, 12383
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OFFSET
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1,1
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COMMENTS
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Inspired by Project Euler, Problem 143 (see link) where such a triangle is called a "Torricelli triangle".
For the corresponding primitive triples, miscellaneous properties and references, see A336328.
Equivalently, a(n) is the numerator of the fraction FC = a(n) / A351477(n).
Also, 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, when FA + FB + FC = d = A351476(n), we have FC = a(n).
FC is the smallest length with FC < FB < FA (remember a < b < c).
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LINKS
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FORMULA
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FC = sqrt(((2*a*b)^2 - (a^2 + b^2 - d^2)^2)/3) / d. - Jinyuan Wang, Feb 19 2022
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EXAMPLE
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For the 3rd triple in A336328, i.e., (43, 147, 152), we get A336329(3) = FA + FB + FC = 5016/37 + 1064/37 + 765/37 = 185, hence A351477(3) = 37 and a(3) = 765.
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PROG
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(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); print1(numerator(sqrtint(((2*a*b)^2 - (a^2 + b^2 - d^2)^2)/3)/d), ", "); ); ); ); ); } \\ Michel Marcus, Mar 02 2022
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CROSSREFS
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Cf. A336328 (primitive triples), A336329 (FA + FB + FC), A336330 (smallest side), A336331 (middle side), A336332 (largest side), A336333 (perimeter), A351801 (FA numerator), A351802 (FB numerator), this sequence (FC numerator), A351477 (common denominator of FA, FB, FC), A351476 (FA + FB + FC other).
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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