If a recursive function $F$ is defined as $F(x)=F(2x/3) + 1$, where $F(1) = 1$, how can one define the function non-recursively?
For example, $$F(1.5)=F(3/2)=F\left(\frac{2\cdot(3/2)}3\right)+1=F(1)+1=2$$How can I find the non-recursive version of this function?
Let $a=2/3$. Thus $F(x)=F(ax)+1$. As was mentioned in the comments, if one defines $$G(x)=F\left(a^x\right),$$ then $$G(x)=F(a^{x+1})+1=G(x+1)+1.$$ Then, upon differentiation, we have that $$G'(x)=G'(x+1),\qquad G(0)=1.$$ This differential equation has a lot of solutions. Perhaps the simplest is of the form $G(x)=\alpha x+\beta$. Suppose this is the case. Then $$\alpha x+\beta=1+\beta+\alpha(x+1)\Rightarrow \alpha=-1\\ G(0)=1\Rightarrow \beta=1.$$ Thus we may write $G(x)=1-x$ as a solution. Indeed, this satisfies the differential equation. Anyway, if $y=\log_ax$, then $G(y)=F(a^y)=F(x)$. Therefore we can write $$F(x)=1-\log_{2/3}x.$$