Project Euler 25 Solution: 1000-digit Fibonacci number
The Fibonacci sequence is defined by the recurrence relation:
Fn = Fn−1 + Fn−2, where F1 = 1 and F2 = 1.
Hence the first 12 terms will be:
F1 = 1
F2 = 1
F3 = 2
F4 = 3
F5 = 5
F6 = 8
F7 = 13
F8 = 21
F9 = 34
F10 = 55
F11 = 89
F12 = 144
The 12th term, F12, is the first term to contain three digits.
What is the index of the first term in the Fibonacci sequence to contain 1000 digits?
Solution
Lemma: Length \(L(n)\) of number \(n\): \[ \begin{array}{rl} & 10^{d-1} \leq n < 10^d\\ \Leftrightarrow & (d-1) \cdot log(10) \leq log(n) < d \cdot log(10)\\ \Leftrightarrow & (d-1) \leq log(n) / log(10) < d\\ \Leftrightarrow & d \leq log_{10}(n) + 1 < d + 1\\ \Leftrightarrow & d = \lfloor1 + log_{10}(n)\rfloor\\ \Leftrightarrow & L(n) = \lfloor 1 + log_{10}(n)\rfloor \end{array} \]
The nth fibonacci number has a well known closed formula using the golden ratio:
\[F(n) = \frac{\Phi^n - (-\Phi)^{-n}}{\sqrt{5}}\]
As \(n\) will become quite large, we can ignore the subtrahend of this formula and simplify the expression a bit, or more formally:
\[\lim\limits_{n\to\infty} F(n) =\lim\limits_{n\to\infty} \frac{\Phi^n - (-\Phi)^{-n}}{\sqrt{5}} = \frac{\Phi^n}{\sqrt{5}}\]
Okay, now let's bring together the two things:
\[ \begin{array}{rl} L(F(n)) &= \lfloor 1 + log_{10}(\Phi^n / sqrt(5))\rfloor\\ &= \lfloor 1 + n \cdot log_{10}(\Phi) - 0.5 \cdot log_{10}(5)\rfloor \end{array} \]
We now can use this for our 1000 digit fibonacci number and solve:
\[ \begin{array}{rl} & 1000 < 1 + d \cdot log_{10}(\Phi) - 0.5 \cdot log_{10}(5)\\ \Leftrightarrow& 999 < d \cdot log_{10}(\Phi) - 0.5 \cdot log_{10}(5)\\ \Leftrightarrow& d > (999 + 0.5 \cdot log_{10}(5)) / log_{10}(\Phi)\\ \Leftrightarrow& d > (999 \cdot log(10) + 0.5 \cdot log(5)) / log(\Phi)\\ \Rightarrow& d = \lceil(999 \cdot log(10) + 0.5 \cdot log(5)) / log(\Phi)\rceil \end{array} \]
Compressing all constants and formulating it for a general \(n\), the implementation can then be stated as:
function solution(n) {
return Math.ceil(4.78497 * n - 3.1127);
}
solution(1000);