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One example is ½(3+√17). How do we get this?
Consider the binomial equation x²-ax-b=0, with a odd, b even, 0 < b < a.
Let u be the positive root, and v be the negative root.
Create the sequence S as S(n) = u^n + v^n.
We have that S(1) = a
and S(2) = a² + 2b
Notice that S(1) is odd and S(2) is odd.
Now note that a = u+v and b = -uv, giving S(n) = a*S(n-1) + b*S(n-2).
This implies that S(n) is odd for all n > 0.
Next, by the nature of how a and b were restricted, we see that:
-1 < v < 0, because v=-b/u, and b < a < u.
Now, [u^n] = [S(n) – v^n],
so [u^n] = S(n) when n is odd and
[u^n] = S(n)-1 when n is even.
Therefore, u has the desired property.
As an example, pick a=3, b=2,
u = ½(3+√17)
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