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:
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.
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:
$$1+\lfloor{\log_\phi(\sqrt{5}(N+\frac{1}{2}))}\rfloor$$
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 $F_n > N$; then the number you are looking for is $n-1$.
