There is no such sequence. Suppose that such a sequence exists. First, see that $$a_{n+2}a_n=a_{n+1}+a_{n-1}\text{ and }a_{n+2}a_{n+4}=a_{n+1}+a_{n+3},$$and dividing the second equation by the first we get $$\frac{a_{n+4}}{a_n}=\frac{a_{n+3}+a_{n+1}}{a_{n-1}+a_{n+1}}\implies\frac{a_{n+4}-a_{n}}{a_{n}}=\frac{a_{n+3}-a_{n-1}}{a_{n-1}+a_{n+1}} (i),$$which implies that $a_{n+4}\geq a_n\iff a_{n+3}\geq a_{n-1}$, with equality being also if and only if. Then, if $a_5\leq a_1$, the sequence would be limited by $m=\min(a_1,a_2,a_3,a_4)$, a contradiction. Thus, we must have $a_5>a_1$ and all terms of the sequence with indexes differing by a multiple of $4$ are distinct and are increasing. Thus, from $(i)$ we get $$\frac{a_{n+4}-a_n}{a_{n+3}-a_{n-1}}=\frac{a_n}{a_{n-1}+a_{n+1}}=a_{n+2}^{-1}.$$By multiplying these equations, we get $$a_{n+4}-a_n=(a_5-a_1)\prod_{4\leq i\leq n+2}a_i^{-1}.$$But since the sequence is unbounded and $(a_{4k+i})_{k\geq0}$ is increasing, there must be some index $i$ for which, whenever $n>\frac{N}{4}$ for some $N$, we have $$a_{4n+i}>2m^{-3}$$which implies that for $n>N$ we have $$a_na_{n+1}a_{n+2}a_{n+3}>2,$$since $a_k>m$, for each $k\geq 5$. See that this implies that $$\prod_{4\leq i\leq n+2}a_i=\Omega(2^{cn}),$$for some $c>0$, which implies that $$a_{n+4}-a_n=O(2^{dn}),$$for some $d<0$. But this implies that the growth of the sequence is bounded above by a geometric progression with ratio smaller than $1$, which obviously converges, a contradiction. Thus, there is no unbounded $\textit{phine}$ sequence. $\blacksquare$