Counting function for Fibonacci numbers Are there some results about "Fibonacci-counting function" - the function counting the number of Fibonacci numbers less than or equal to some real number x?
 A: Thanks to all!   Maybe the answer is (achille hui's version): $$\pi_F(x)=\left\lfloor\log_{\phi}\sqrt{5} \ \left(\lfloor x\rfloor+\frac12\right)\right\rfloor, \ x\geq2 $$
A: The function $\pi_F(x)$ cannot have the form found in lesobrod’s answer. The argument is too long for a comment, so I’ve made it into an ‘answer’, but of course it doesn’t actually answer the original question.
Let $\widehat\varphi=-\frac1\varphi$, the other root of $x^2-x-1=0$. We know that $$F_n=\frac{\varphi^n-\widehat\varphi^n}{\sqrt5}\;,$$ so $$\log_\varphi\left(\sqrt5F_n+\widehat\varphi^n\right)=n\;.$$ 
Fix $a\in\Bbb R$, and consider the function
$$f(x)=\log_\varphi\left(\sqrt5x+a\right)\;.$$
If $a\le 0$, then $\left\lfloor f(F_{2n})\right\rfloor<2n$, since $\sqrt5F_{2n}+a<\sqrt5F_{2n}+\widehat\varphi^{2n}=\varphi^{2n}$. Thus, we may assume that $a>0$. Since $\widehat\varphi^n\to 0$ as $n\to\infty$, 
$$\epsilon_n=\frac{a-\widehat\varphi^{2n}}{\sqrt5}>0$$
for all sufficiently large $n$.
Let $x=F_{2n}-\epsilon_n$; then $a-\sqrt5\epsilon_n=\widehat\varphi^{2n}$, and
$$2n=\log_\varphi\left(\sqrt5F_{2n}+\widehat\varphi^{2n}\right)=\log_\varphi\left(\sqrt5\left(F_{2n}-\epsilon_n\right)+a\right)=f(x)=\left\lfloor f(x)\right\rfloor\;,$$
but $\pi_F(x)\le 2n-1$, since $x<F_{2n}$.
For suitable values of $a$ such a function can agree with $\pi_F$ on $\Bbb Z^+$, but not on all of $\Bbb R^+$.
