Need an explanation of an answer regarding $\log F_n = \Theta(n)$, where $n$ is the $n$-th Fibonacci number

Question

In this answer to the question "Log Fibonacci equals Theta", I don't understand why the solution says that the base case for $n=5$, $F_5=8$. Isn't $F_5=5$, instead?

Regards

Answer 1

Yes, $F_5=5$. The fact that it is $F_6=8$ does not change the argument there. You can either decrease $A$ a bit or demand a higher starting point for the bounds. With the given $A,B$, the base case becomes $F_{11}=89$, with $11 \log (3/2)\approx 4.46, \log(89) \approx 4.49, 11\log(2) \approx 4.80$ and everything goes through.

Answer 2

I'm the author of the linked-to answer, so I can confirm that it was a failure on my part to doublecheck the convention I had assumed, that $F_0=F_1=1$, with what the OP there specified, namely $F_1=F_2=1$. I've fixed my answer there so that it now uses $n=12$, with $F_{12}=144$ as the base case. As Ross Millikan's answer here shows, $n=11$ can also be used as a base case, but verifying the inequalities is a bit easier for $n=12$.