How many numbers are in the Fibonacci sequence below a given threshold?

Solution 1:

As mentioned in the comments, Binet's formula,

$$F_n=\frac1{\sqrt{5}}(\phi^n-(-\phi)^{-n})$$

where $\phi=\frac12(1+\sqrt 5)$ is the golden ratio, is a closed-form expression for the Fibonacci numbers. See this related question for a few proofs.

As for counting how many Fibonacci numbers are there that are less than or equal to a given number $n$, one can derive an estimate from Binet's formula. The second term in the formula can be ignored for large enough $n$, so

$$F_n\approx\frac{\phi^n}{\sqrt{5}}$$

Solving for $n$ here gives

$$n=\frac{\log\,F_n+\frac12\log\,5}{\log\,\phi}$$

Taking the floor of that gives a reasonable estimate; that is, the expression

$$\left\lfloor\frac{\log\,n+\frac12\log\,5}{\log\,\phi}\right\rfloor$$

can be used to estimate the number of Fibonacci numbers $\le n$. This is inaccurate for small $n$, but does better for large $n$.

It turns out that by adding a fudge term of $\frac12$ to $n$, the false positives of the previous formula disappear. (Well, at least in the range I tested.) Thus,

$$\left\lfloor\frac{\log\left(n+\frac12\right)+\frac12\log\,5}{\log\,\phi}\right\rfloor$$

gives better results.

Solution 2:

Assuming we take $F_0=0$ and $F_1=1$, for all $n\geq 0$, the value of the $F_n$ is the nearest integer to $\frac{\phi^n}{\sqrt{5}}$, where $\phi = \frac{1+\sqrt{5}}{2}$.

So, the number of Fibonnaci numbers less than $N$ is the number of $n\geq 0$ such that:

$$ \frac{\phi^n}{\sqrt{5}} \leq N+\frac{1}{2}$$

Solving this, we see that $n \leq \log_\phi(\sqrt{5}(N+\frac{1}{2}))$. That means that the number of $n$ is:

$$1+\lfloor{\log_\phi(\sqrt{5}(N+\frac{1}{2}))}\rfloor$$