How many numbers are in the Fibonacci sequence below a given threshold? Assuming I'm asked to generate Fibonacci numbers up to N, how many numbers will I generate? I'm looking for the count of Fibonacci numbers up to N, not the Nth number.

So, as an example, if I generate Fibonacci numbers up to 25, I will generate:


1, 1, 2, 3, 5, 8, 13, 21
that's 8 numbers


How do I calculate this mathematically for an arbitrary "n"?
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.
I don't think you will be able to find a closed-form formula for the number of Fibonacci numbers less than $N$ - if you had chosen $N = 26$ instead of $N = 25$, you would have gotten the same answer, but if you had set $N = 34$, the answer would have suddenly jumped to $9$. As has been suggested before, you can use Binet's formula:

nth Fibonacci number $F_n= (\phi^n - (-\phi)^{-n})/(\sqrt 5)$ where $\phi = (1 + (\sqrt 5))/2$

Simply plug in values of $n$ until $F_n &gt; N$; then the number you are looking for is $n-1$.