A351802 a(n) = A351477(n) * FB 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).
264, 325, 1064, 27265, 6528, 34200, 12376, 1015, 8512, 11520, 8415, 1656, 116025, 8415, 17575, 56448, 81928, 6765, 107712, 106128, 43953, 60903, 235008, 311885, 3105, 32571, 571648, 411320, 9499, 4991, 1800, 13875, 1894144, 16320, 402375, 42735, 805, 218925
Offset: 1
Keywords
Examples
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.
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), this sequence (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*a*c)^2 - (a^2+c^2-d^2)^2)/3)/d), ", ");););););} \\ Michel Marcus, Mar 01 2022
Formula
FB = sqrt(((2*a*c)^2 - (a^2+c^2-d^2)^2)/3) / d. - Jinyuan Wang, Feb 19 2022
Extensions
More terms from Jinyuan Wang, Feb 19 2022
Comments