A351801 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).
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
Offset: 1
Keywords
Examples
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.
Links
- Project Euler, Problem 143 - Investigating the Torricelli point of a triangle.
- Wikipedia, Fermat point.
Crossrefs
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').
Programs
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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
Formula
FA = sqrt(((2*b*c)^2 - (b^2 + c^2 - d^2)^2)/3) / d. - Jinyuan Wang, Feb 19 2022
Extensions
More terms from Jinyuan Wang, Feb 19 2022
Comments