A351476 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, this sequence gives the sum FA + FB + FC when gcd(a, b, c) = A351477(n).

784, 1029, 6845, 80089, 24843, 109561, 109561, 5239, 24037, 47045, 27735, 6760, 477481, 21904, 57967, 186245, 365403, 20280, 400445, 356168, 159953, 190463, 718205, 836405, 11809, 178771, 1432443, 1127307, 22984, 34295, 22477, 157339, 6723649, 44403, 974408

Offset: 1

Author

Bernard Schott, Feb 12 2022

Keywords

• nonn

Comments

Inspired by Project Euler, Problem 143 (see link) where such a triangle is called a "Torricelli triangle".

Differs from A336329 where ABC is a primitive integer-sided triangle with A < B < C < 2*Pi/3 and only FA+FB+FC is an integer; in fact, terms of A336329 are the sum of 3 fractions of the form FA = p/t, FB = q/t, FC = r/t but (p+q+r)/t is an integer. Here, FA, FB and FC are all integers and the sums FA+FB+FC are displayed according to same order as in A336329. The corresponding common denominators t of the fractions (p/t, q/t, r/t) are in A351477.

If FA + FB + FC = d, then we have this "beautifully symmetric equation" between a, b, c and d (see Martin Gardner):

3*(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2.

=> d = sqrt(((a^2 + b^2 + c^2)/2) + (1/2) * sqrt(6*(a^2*b^2 + b^2*c^2 + c^2*a^2) - 3*(a^4 + b^4 + c^4))).

Examples

			a(1) = FA + FB + FC = 325 + 264 + 195 = 784, corresponding to first triple (399, 455, 511) whose gcd = 7.
a(6) = FA + FB + FC = 70720 + 34200 + 4641 = 109561, corresponding to triple (36741, 73151, 92680) whose gcd = 331.
a(7) = FA + FB + FC = 91200 + 12376 + 8985 = 109561, corresponding to triple (16219, 94335, 97976) whose gcd = 331.
		

References

• Martin Gardner, Mathematical Circus, Elegant triangles, First Vintage Books Edition, 1979, p. 65.

Links

• Project Euler, Problem 143 - Investigating the Torricelli point of a triangle.
• Eric Weisstein's World of Mathematics, Fermat points.

Crossrefs

Cf. A336328, A336329, A336330, A336331, A336332, A336333, A351477, A351801, A351802, A351803.

Programs

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); my(s = numerator(sqrtint(((2*b*c)^2 - (b^2+c^2-d^2)^2)/3)/d) + numerator(sqrtint(((2*a*b)^2 - (a^2+b^2-d^2)^2)/3)/d) + numerator(sqrtint(((2*a*c)^2 - (a^2+c^2-d^2)^2)/3)/d)); print1(s, ", ");););););} \\ Michel Marcus, Mar 02 2022

Formula

a(n) = A336329(n) * A351477(n).

If FA + FB + FC = d, then

d^2 = (1/2) * (a^2 + b^2 + c^2) + 2 * S * sqrt(3) where S = area of triangle ABC.

Extensions

More terms from Jinyuan Wang, Feb 17 2022