1,1

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 FB = 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 FB = a(n).

FB is the middle length with FC < FB < FA (remember a < b < c).

Project Euler, Problem 143 - Investigating the Torricelli point of a triangle.

a(n) = A351476(n) - A351801(n) - A351803(n).

FB = sqrt(((2*a*c)^2 - (a^2+c^2-d^2)^2)/3) / d. - Jinyuan Wang, Feb 19 2022

For the 2nd triple in A336328, i.e., (73, 88, 95), we get A336329(2) = FA + FB + FC = 440/7 + 325/7 + 264/7 = 147, hence A351477(2) = 7 and a(2) = 325.

(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*c)^2 - (a^2+c^2-d^2)^2)/3)/d), ", "); ); ); ); ); } \\ Michel Marcus, Mar 01 2022

Cf. A336328 (primitive triples), A336329 (FA + FB + FC), A336330 (smallest side), A336331 (middle side), A336332 (largest side), A336333 (perimeter), A351801 (FA numerator), this sequence (FB numerator), A351803 (FC numerator), A351477 (common denominator of FA, FB, FC), A351476 (other 'FA + FB + FC').

Sequence in context: A253694 A253701 A255804 * A050240 A105683 A160971

Adjacent sequences: A351799 A351800 A351801 * A351803 A351804 A351805

Bernard Schott, Feb 19 2022

More terms from Jinyuan Wang, Feb 19 2022

approved