15
$\begingroup$

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"?

  • 0
    With regards to the closed-form expression, such a formula you can see here: http://en.wikipedia.org/wiki/Fibonacci_number#Closed-form_expression2011-09-26

3 Answers 3

16

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.

  • 0
    Tha$n$ks a lot, this explains everything!2011-09-26
3

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:

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

-1

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 $n-1$.

  • 0
    I corrected the answer according to Binet's formula: http://mathworld.wolfram.com/BinetsFibonacciNumberFormula.html2012-02-25