assume Ai is one of term in the sequence.
|x x x Ai| + | x x Ai x| + | X Ai X X| + | Ai X X X| + ....... (X is 1 or -1)
Ai will always appear 4 times in the sequence.
we can rewrite as
a b c (d =Ai) e f g . ...
a b c d b c d e c d e f d e f g
Ai=1/-1, => a(bc) + (b(c)e) + (c(e)f) + (ef)g = 0
case1
if abc + bce = 0, a = -e
if cef + efg = 0, c = -g
case2
if abc + cef = 0, ab = -ef
if bce + efg = 0, bc = -gf
case 3
if abc + efg = 0, abc=-efg
if bce + cef = 0, b=-f
from case 1 and 3, a = -e, b=-f, c=-g, but it can't apply to case 2.
if case 1 is true and case 3 is true, then abc + cef = -1*(bce + efg) as the unique solution.
we can rewrite the sequence again
X,Y,Z,Ai,-X,-Y,-Z,-Ai....
and xyzAi + yzAi-x +Zai-x-y, + Ai-x-y-z = 0
