A351803 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).
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
Offset: 1
Keywords
Examples
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.
Links
- Project Euler, Problem 143 - Investigating the Torricelli point of a triangle.
Crossrefs
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).
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*a*b)^2 - (a^2 + b^2 - d^2)^2)/3)/d), ", ");););););} \\ Michel Marcus, Mar 02 2022
Formula
FC = sqrt(((2*a*b)^2 - (a^2 + b^2 - d^2)^2)/3) / d. - Jinyuan Wang, Feb 19 2022
Extensions
More terms from Jinyuan Wang, Feb 19 2022
Comments