|
| |
|
|
A294177
|
|
List of positive integer triples (b,c,d) where b > c > d are coprime and 1/b^2 + 1/c^2 + 1/d^2 = 1/r^2 and r is rational, ordered by b then c.
|
|
1
|
|
|
|
3, 2, 1, 4, 3, 1, 5, 3, 2, 5, 4, 1, 6, 5, 1, 7, 4, 3, 7, 5, 2, 7, 6, 1, 8, 2, 1, 8, 5, 3, 8, 7, 1, 9, 3, 2, 9, 5, 4, 9, 7, 2, 9, 7, 6, 9, 8, 1, 9, 8, 6, 10, 7, 3, 10, 9, 1, 11, 6, 5, 11, 7, 4, 11, 8, 3, 11, 9, 2, 11, 10, 1, 11, 10, 8, 12, 7, 5, 12, 11, 1, 13, 7, 6, 13, 8, 5, 13, 9, 4, 13, 10, 3, 13, 11, 2, 13, 12, 1, 14, 5, 3, 14, 8, 3, 14, 8
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,1
|
|
|
COMMENTS
|
It seems that all integers occur.
b, c, d cannot all be odd (see (8, 2, 1) for example).
Also: coprime triples (b, c, d), all different, such that (bc)^2 + (bd)^2 + (cd)^2 is a square. As squares are 0 or 1 mod 4, we can use this expression to prove that at least one element of the triples must be even. As b, c and d are coprime, at most two elements of the triples are even. - David A. Corneth, Feb 11 2018
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
1/3^2 + 1/2^2 + 1/1^2 = 1/(6/7)^2.
|
|
|
MATHEMATICA
|
n = 16; lst = {}; Do[
Do[Do[If[GCD[b, c, d] == 1, r = Sqrt[1/(1/b^2 + 1/c^2 + 1/d^2)];
If[Element[r, Integers] || Element[r, Rationals],
lst = AppendTo[lst, {b, c, d}]]], {d, c - 1}], {c, b - 1}], {b, n}]; lst//Flatten
|
|
|
PROG
|
(PARI) upto(n) = {my(res = List(), sd, stepd); for(b = 3, n, for(c = 2, b - 1, if((b - c) % 2 == 0, sd = b % 2 + 1; stepd = 2, sd = 1; stepd = 1); forstep(d = sd, c - 1, stepd, if(issquare((b*d)^2 + (b*c)^2 + (c*d)^2) && gcd([b, c, d]) == 1, listput(res, [b, c, d]))))); concat(Vec(res))} \\ David A. Corneth, Dec 29 2018
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
