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Alternate Form

For any integer n > 0, the expression FnϕFn+1 can be repeatedly expanded as below:

Fn ϕ − Fn+1 = Fnϕ − Fn− 1 − Fn
            = Fn(ϕ − 1)− Fn −1
              F
            = --n− Fn −1                          (since ϕ − 1 = 1∕ϕ)
              (ϕ   )
            =  − 1-  (F   ϕ − F )   (now, repeatedly expand as above)
                 ϕ     n−1     n
              (  1 )2
            =  − --   (Fn −2ϕ − Fn−1)
              (  ϕ )
                 1- 3
            =  − ϕ    (Fn −3ϕ − Fn−2)
              ...
              ...
              (  1 )n
            =  − --   (F0ϕ − F1)
                (ϕ   )n
            = −  − 1-                                             (4)
                   ϕ

So, the three gap-lengths in corollary 1 can also be expressed as:

                --1-
|Ft+2ϕ − Ft+3| = ϕt+2
                  1
|Ft+1ϕ − Ft+2| = ϕt+1
                 1
  |Ftϕ − Ft+1| =--t                                          (5)
                ϕ

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