n=2:A=(0111),λmax=12(1+√5),x2−x−1=0
n=3:A=(101011111),λmax=1+√2,x2−2x−1=0
n=4:A=(1101101101111111),λmax=12(3+√13),x2−3x−1=0
n=5:A=(1110111011101110111111111),λmax=2+√5,x2−4x−1=0
n=6:A=(111101111011110111101111011111111111),λmax=12(5+√29),x2−5x−1=0
⋮
n=n,x2−(n−1)x−1=0⟹λmax=12(n−1+√n2−2n+5)