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In my book they calculated the $28$th Fibonacci number and said $F_{28} = 3 \times 13 \times 19 \times 281 = 317811$. This made me wonder if there was an easier way to find the $28$th Fibonacci number than by doing it by hand.

The Short One
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Puzzled417
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  • If I had to compute $F_{28}$, I would certainly do it be successive additions the obvious way. Finding a shortcut solution would take more time. –  Aug 10 '16 at 18:52
  • If I had to compute $F_{28}$, I would certainly do it be successive additions the obvious way. Finding a shortcut solution would take more time. Maybe with the doubling formula ? –  Aug 10 '16 at 19:21
  • $F_{28} = \underbrace{3}{= F_4} \times \underbrace{\overbrace{13}^{=F_7} \times \color{red}{29}}{= F_{14}} \times 281$ – achille hui Aug 10 '16 at 19:28
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    Calling this a duplicate of this one(among others). At least I didn't see the answers here adding anything to the answers from 6 years ago. I recall having posted links to those answers on other sites (or other answers from approximately the same year) – Jyrki Lahtonen Aug 10 '16 at 20:14
  • @achillehui That's true because they form a divisibility sequence.. But that doesn't lead to a fast general way to compute them. – Bill Dubuque Aug 10 '16 at 20:48

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They can be computed by matrix exponentiation, which is quick by repeated squaring. Recall $$ \left(\begin{array}{ccc} \,1 & 1 \\\ 1 & 0 \end{array}\right)^{\large n}\ =\ \left(\begin{array}{ccc} F_{\large n+1} & F_{\large n} \\\ F_{\large n} & F_{\large n-1} \end{array}\right) $$

More simply $ $ if $\: (a,b)_{\large n} := (f_{\large n-1},\,f_{\large n}),\ $ then applying the above yields

that a $\:\rm\color{#0a0}{shift}\:$ is $\ (a,b)_{\large n}\color{#0a0}\Rightarrow(b,\,a+b)_{\large n+1}\ $

and $\,\rm\color{#c00}{doubling}$ $\ (a,b)_{\large n} \color{#c00}\Rightarrow (a^2\!+b^2,\,b^2\!+2ab)_{\large 2n}.\ $ For example, let's compute your $\, F_{\large 28}$

$$(0,1)_{\large 1}\!\color{#0a0}\Rightarrow (1,1)_{\large 2}\color{#0a0}\Rightarrow (1,2)_{\large 3} \color{#c00}\Rightarrow (5,8)_{\large 6}\color{#0a0}\Rightarrow (8,13)_{\large 7}\color{#c00}\Rightarrow (233,377)_{\large 14}\color{#c00}\Rightarrow(\ldots,317811)_{\large 28}$$

is computable using only mental arithmetic (except possibly the final step).

Community
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Bill Dubuque
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The simplest direct calculation is to take the integer nearest to $\frac{\varphi^n}{\sqrt5}$, where $\varphi=\frac12(1+\sqrt5)$; see here.

Brian M. Scott
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  • Why do you think they wrote the prime factorization first? – Puzzled417 Aug 10 '16 at 18:41
  • @Puzzled417: I have no idea. Apparently they had some use for it, but it’s not connected with any nice way of calculating the Fibonacci numbers. – Brian M. Scott Aug 10 '16 at 18:42
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    @Puzzled417 perhaps they were trying to point out some property by means of example which was to be proven in a later example or exercise. For example, $F_{4n}$ is always a multiple of three for any natural number $n$. (Notice a pattern: $\color{red}{0},1,1,2,\color{red}{3}, 5, 8, 13, \color{red}{21}, 34, 55, 89, \color{red}{144},\dots$) (If this wasn't included in your textbook, it is a nice problem that I recommend trying to prove). In particular, $F_{28}$ also follows this pattern. – JMoravitz Aug 10 '16 at 18:49
  • Could you compute that by hand ? –  Aug 10 '16 at 18:52
  • @Yves: Yes. Would I? Of course not: for that I’d use the recurrence. – Brian M. Scott Aug 10 '16 at 18:55
  • @BrianM.Scott: with how many decimals do you need to perform the computation ? –  Aug 10 '16 at 19:19
  • @Yves: I’ve no idea what the exact number is, and I really don’t care; it’s clearly small enough to allow hand computation, though the computation would be quite unpleasant. I do know that starting with $6$ decimal places for $\varphi$ and $\sqrt5$ is sufficient, though I don’t know exactly how many places have to be kept in the intermediate calculations. Do you have a point with all this? – Brian M. Scott Aug 10 '16 at 19:27
  • @BrianM.Scott: yep, looking a little further than the well-known formulas. This would be instructive. –  Aug 10 '16 at 19:30
  • @Yves: I doubt that it was well-known to the OP. Feel free to offer an answer that you prefer, if you have one; I intend to ignore any further oblique carping. – Brian M. Scott Aug 10 '16 at 19:40
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There is a formula which is now called Binet's Formula, but was allready known to Euler, Daniel Bernoulli and de Moivre,

$$F_n=\frac{\phi^n}{\sqrt{5}}+(-1)^{n+1}\frac{\phi^{-n}}{\sqrt{5}},$$ in which $\phi=(1+\sqrt{5})/2$ is the golden ratio.

The result of Brian M. Scott follows by an estimation of the second term.

MrYouMath
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There are identities that relate $F_n$ and $F_{kn}$ like this one:

$$F_{4n} = 4F_nF_{n+1}(F_{n+1}^2 + 2F_n^2) - 3F_n^2(F_n+2F_{n+1}^2)$$

Here you can express $F_{28}$ in terms of $F_7=13$ and $F_8=21$.

John
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Unfortunately this requires (sometimes) a calculator to compute. You could potentially use the following: Let $\phi = \frac{1+\sqrt5}{2}$

$F_{n}=\frac{\phi^x-(1-\phi)^x}{\sqrt5}$

Ian L
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