$f(6)=144$ and $f(n+3) = f(n+2)\{f(n+1)+f(n)\}$, Then $f(7) =$? Given that $f(6)=144$ and $f(n+3) = f(n+2) \cdot\Big(f(n+1)+f(n)\Big)$  $[$For $n = 1,2,3,4]$
Then find the value of $f(7)$.
The solution is not unique but all of them are positive integers.
I can't find a way out.

Letting $a=f(1)$, $b=f(2)$, $c=f(3)$, $d=f(4)$, $e=f(5)$, $144=f(6)$, and $g=f(7)$ for notational simplicity, we have
$$\begin{align}
d&=c(b+a)\\
e&=d(c+b)\\
144&=e(d+c)\\
g&=144(e+d)\\
\end{align}$$
If I've done the algebra correctly, we can eliminate $d$ and $e$ leaving
$$144=c^2(a+b+1)(a+b)(b+c)$$
and
$$g=144c(a+b+1)(a+b)$$
If we assume that $a$, $b$, and $c$ are positive integers, the first equation leaves only a handful of possibilities, such as $(a,b,c)=(7,1,1)$.  In particular, since $a+b$ and $a+b+1$ must be consecutive divisors of $144$, $a+b$ can only be $2$, $3$, or $8$.
