$f(x+1)=f(x)+f(\alpha\cdot x)$

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I try to find an analytic increasing function $f_\alpha$ ($0\le\alpha\le1$) from $\mathbb R$ (or $\mathbb R^+$) to $\mathbb R$ such that for all $x$ $$f_\alpha(x+1)=f_\alpha(x)+f_\alpha(\alpha\cdot x)$$ and $f_\alpha(0)=1$. Hence $f(1)=2$.

Of course, there are some easy solutions $f_0(x)=x+1$ and $f_1(x)=2^x$, but how to find $f_{\frac{1}{2}}$ or any other $f_\alpha$ ? A way to find the sequence $\beta_n^{(\alpha)}$ such that $f_\alpha(x)=\sum_n\beta_n^{(\alpha)}x^n$ would be nice too.