My question is simple, but a mystery to me. Skip to the last paragraph if you're not interested in the story of my exploration.
EDIT: I seem to have misinterpreted a key detail regarding how the sequence relates to the continued fraction expansion of $\sqrt{5}$ which I've fixed in this post. See the comments for more info.
The background: I had the brilliant idea to consider the Fibonacci-like sequence which starts with $$\frac{1 \pm \sqrt{5}}{2}$$ as initial values, and see what happens. Nothing too interesting, it just devolves into $$T_n = \frac{F_n + F_{n-2}\sqrt{5}}{2}$$ fairly quickly ($\forall n \geq 4$); this has nothing to do with the presence of $\sqrt{5}$ as opposed to any other value, though.
However, the 2 in the denominator gave me something to work with: every third Fibonacci number is certainly even, which means I can reduce every third element of this sequence I've constructed. The first couple of these, reduced, are: $$\begin{array}{c|ccccc}n & 3 & 6 & 9 & 12 & 15\\ \hline T_n & 0\sqrt{5}+1 & 1\sqrt{5}+4 & 4\sqrt{5}+17 & 17\sqrt{5}+72 & 72\sqrt{5}+305 & \end{array}$$
It was at this point that I realized all I was doing was taking the $3n$th Fibonacci number and dividing it by two, which was kind of boring. I figured I'd look up the sequence on the OEIS, though, and it seems to be the sequence of denominators you get after truncating the continued fraction expansion of $\sqrt{5}$ after some number of denominators and then simplifying.
The question: why?
Recursion for the $3n\,$th Fibonacci number: let $G_n=F_{3n}$. Then we have $$\eqalign{ G_{n+1}&=F_{3n+2}+F_{3n+1}\cr F_{3n+2}&=F_{3n+1}+G_n\cr 2F_{3n+1}&=2G_n+2F_{3n-1}\cr -G_n&=-F_{3n-1}-F_{3n-2}\cr F_{3n-1}&=F_{3n-2}+G_{n-1}\ .\cr}$$ Now add all these equations: lots of things cancel and we get $$G_{n+1}=4G_n+G_{n-1}\ .$$ For the convergents $p_n/q_n$ of a continued fraction we have $$q_{n+1}=a_nq_n+q_{n-1}\ .$$ In the case of $\sqrt5$ all of the $a_n$ for $n\ge1$ are equal to $4$, so we have $$q_{n+1}=4q_n+q_{n-1}\ .$$ This is the same as the recurrence for $G_n$, so as long as the initial values match, the sequences will be identical.