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A351801
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a(n) = A351477(n) * FA 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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325, 440, 5016, 39360, 14800, 70720, 91200, 3864, 9405, 30429, 11704, 4669, 250096, 11704, 32640, 81840, 203000, 7208, 218120, 199325, 99360, 76760, 359352, 342912, 8184, 122200, 595595, 621387, 12600, 26040, 19320, 137344, 3108105, 24955, 409640, 58400, 1520
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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 FA = 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 FA = a(n).
FA is the largest length with FC < FB < FA (remember a < b < c).
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LINKS
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FORMULA
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FA = sqrt(((2*b*c)^2 - (b^2 + c^2 - d^2)^2)/3) / d. - Jinyuan Wang, Feb 19 2022
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EXAMPLE
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For the 1st triple in A336328, i.e., (57, 65, 73), we get A336329(1) = FA + FB + FC = 325/7 + 264/7 + 195/7 = 112, hence A351477(1) = 7 and a(1) = 325.
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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*b*c)^2 - (b^2 + c^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), this sequence (FA numerator), A351802 (FB numerator), A351803 (FC numerator), A351477 (common denominator of FA, FB, FC), A351476 (other 'FA + FB + FC').
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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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