For every positive integer n, let an denote the number of ways n is obtained as a sum of terms that are all 1, 3 or 4. The order of the terms also matters. Prove that a2006a2007a2008 is a perfect cube.
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For every positive integer n, let an denote the number of ways n is obtained as a sum of terms that are all 1, 3 or 4. The order of the terms also matters. Prove that a2006a2007a2008 is a perfect cube.
While I am unable to prove this, I have found the following to be true:
a2006 and a2008 are squares of consecutive Fibonacci numbers
a2007 is the product of those numbers
As such, a2006a2008 = a20072
and a2006a2007a2008 = a20073