OFFSET
1,1
COMMENTS
Inspired by Project Euler, Problem 143 (see link).
The triples of sides (a,b,c) with a < b < c are in increasing order of largest side.
For the corresponding primitive triples and miscellaneous properties and references, see A336328.
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.
For the terms of the data, every FA, FB, FC is a fraction but FA + FB + FC is an integer (see example).
This sequence is not increasing. For example, a(5) = 283 for triangle with largest side = 205 while a(6) = 273 for triangle with largest side = 208. Also, a(6) = a(7) = 331 show that two distinct triangles can have the same minimum possible integer distance FA + FB + FC.
REFERENCES
Martin Gardner, Mathematical Circus, Elegant triangles, First Vintage Books Edition, 1979, p. 65.
LINKS
FORMULA
For triangle (a, b, c) whose area is S, and d = FA+FB+FC, then
d = sqrt((1/2)*(a^+b^2+c^2) + 2*S*sqrt(3)), also,
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))), or
3*(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2.
EXAMPLE
For first triple (57, 65, 73), d = 112 is solution of
3*(57^4 + 65^4 + 73^4 + d^4) = (57^2 + 65^2 + 73^2 + d^2)^2, hence, 112 is a term because d = FA + FB + FC = 264/7 + 195/7 + 325/7 = 112.
PROG
(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), print1(sqrtint(d), ", "))))); \\ Jinyuan Wang, Jul 20 2020
CROSSREFS
KEYWORD
nonn
AUTHOR
Bernard Schott, Jul 18 2020
EXTENSIONS
More terms from Jinyuan Wang, Jul 20 2020
STATUS
approved