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Answer by Oscar Lanzi for 111...11 base p = 111...11 base q

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Going a bit outside the box here.

If we define a "Fibonacci-base" representation as

$$({\ldots}a_2a_1a_0)_F=\sum_{k=1}^\infty a_kF_k$$

where $a_k\in\{0,1\}$ and $F_k$ are the Fibonacci numbers, then applying the Shanks transformation to ${\ldots}111_F$ gives $-1$. Applying the Shanks transformation to the ordinary binary representation ${\ldots}111_2$ also gives $-1$.


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