... sn 1 = sn−2+03n 1 = sn−2+ 6 3n 1 ,thensn +1 = sn−2+ 1 3n 1 +03n+03n +1 = sn−2+03n 1 + 1 3n+ 6 3n +1 = sn−2+ 6 3n 1 + 1 3n+ 6 3n +1 .Since (1, 0, 0) ≈ (0, 1, 6) ... −5wehavexn =1, xn=6and3n−2(x 1 − x 1 )+ +3(xn−2− xn−2)+xn 1 −xn 1 =2,which implies sn 1 − sn 1 ≡ 2 (mod 3), hence xn 1 − xn 1 =5(xn 1 =6, xn 1 =1) orxn 1 − xn 1 = 1( xn 1 =0,xn 1 = ... sn 1 − sn 1 = sn 1 − sn 1 = 1 3n 1 . a contradiction to Consequence 2.Hence,#sn +1 =#sn 1 +#sn 1 .From (13 ) yield sn 1 −sn 1 = 1 3n 1 , by Claim 2.2.a), xn 1 = 1. ...