The Fibonacci sequence is $0, 1, 1, 2, 3, 5, 8, 13, 21, 34,\ldots$, where each term after the first two is the sum of the two previous terms.

Can we find the next Fibonacci number if we are given any Fibonacci number?

For example, if $n = 8$ then the answer should be $13$ because $13$ is the next Fibonacci number after $8$.

J. W. Tanner
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    Only if you exclude 1, since it would return two unique and valid answers – Stian Yttervik Oct 13 '19 at 20:37
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    The obvious answer, although not very satisfying, is "calculate the Fibonacci sequence until you reach the number you were given, then go one more". I suggested an edit to exclude that, since I don't think it's what OP was looking for, but apparently the reviewers felt differently. – Geoffrey Brent Oct 15 '19 at 03:34
  • There is an explicit formula for F(n) that makes the given info about a value of F(n-1) unnecessary. But Matt Daly's answer is spot on as to the right idea, although there is an issue about rounding vs. floor vs. ceiling vs. precision... – Dr. Michael W. Ecker Oct 15 '19 at 18:59

4 Answers4


Given a Fibonacci number $n$, let $m$ be the next Fibonacci number. The Fibonacci sequence has the property that for any three consecutive elements $r,s,t$, we have $rt=s^2\pm 1$ (proof is by induction, which you might like to try $-$ the choice of signs alternates). And we know that the previous Fibonacci number is $m-n$. So we have $$m(m-n)=n^2\color{red}{\pm} 1$$ This is a quadratic equation in $m$, with solutions $m=\frac12(n\color{blue}{\pm}\sqrt{5n^2\color{red}{\pm} 4})$. We know that $m\ge n$, so $m$ must equal $\frac12(n\color{blue}{+}\sqrt{5n^2\color{red}{\pm} 4})$. And we can choose between $\color{red}{+}4$ and $\color{red}{-}4$ because only one of $\sqrt{5n^2\color{red}{+}4}$ and $\sqrt{5n^2\color{red}{-}4}$ can be an integer (with the single exception of $n=1$).

So the answer is whichever one of $\frac12(n+\sqrt{5n^2+4})$ and $\frac12(n+\sqrt{5n^2-4})$ is an integer.

Note that the single exception $n=1$ occurs twice in the Fibonacci sequence, so there are indeed two possible answers in this case.

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The ratio of any two consecutive entries in the Fibonacci sequence rapidly approaches $\varphi=\frac{1+\sqrt5}2$. So if you multiply your number by $\frac{1+\sqrt5}2$ and round to the nearest integer, you will get the next term unless you're at the very beginning of the sequence.

  • float ans = ((1+sqrt(5))/2.0)*n; return ceil(ans); This is not giving correct result. – ppSpp Oct 12 '19 at 15:43
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    @sr123 He did say "round to the nearest integer". That's not what `ceil` does, so of course you don't get the correct result. From checking the first few values, actually *rounding* will get you correct results unless you're starting from `0` or from `1`. – HTNW Oct 12 '19 at 22:58
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    yes float ans = ((1+sqrt(5))/2.0)*n; return round(ans); is working !! – ppSpp Oct 13 '19 at 04:31
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    Trying to do this in floating point is a bad idea; Fibonacci numbers quickly grow beyond the ability of floating point to accurately represent, so your computation will suffer from rounding error. – user2357112 Oct 13 '19 at 06:52
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    The error when using $x\varphi$ for the next term is smaller than $\pm\frac1{2x}$, apart from the very beginning of the sequence, thus justifying the rounding – Henry Oct 13 '19 at 11:10
  • To clarify Henry's comment: in abstract real number arithmetic, the "round to the nearest integer" part of this answer is safe due to error bounds, except at the very start of the sequence. This doesn't mean floating point rounding error is also safe. If you try to compute this in floating point, floating point precision limitations will cause problems as soon as you start hitting Fibonacci numbers that take more than bits of precision than your floating point format has. This is especially bad with single-precision `float`. – user2357112 Oct 13 '19 at 12:15
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    This is a math site, not a computer site, and we work with abstract real number arithmetic here, not floating point. However, anyone actually trying to compute a Fibonacci number by multiplying another Fibonacci number by $(1+\sqrt{5})/2$ is probably going to use a computer, and representation issues will be important. – user2357112 Oct 13 '19 at 12:26
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    It seems that floating-point precision first causes this to break down at the 79th Fibonacci number; at least in Python (64-bit floats), `round((1 + sqrt(5))/2 * 8944394323791464)` is 14472334024676222, while the 79th term is 14472334024676221. Using an arbitrary-precision float type, such as `gmpy2.mpfr` with precision set large enough, can go farther (e.g. precision=1000 works until the 1444th term; precision=2000 works for 2884 terms.) – Nick Matteo Oct 13 '19 at 20:44
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    @NickMatteo, yep, 64-bit IEEE (double) floats have an effective mantissa of 53 bits, so the maximum exact integer that fits is `9007199254740992`, a bit more than F(78), but less than F(79). (of course the size of the mantissa would also affect the accuracy of `sqrt(5)` itself) – ilkkachu Oct 13 '19 at 21:32
  • Similarly we can find previous fibonacci number if given a fibonacci number prev_fib = n / ( ( 1 + sqrt( 5 ) ) /2.0 ) and round it. – ppSpp Oct 15 '19 at 20:03
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    @user2357112 While I agree that Henry's argument lacks complete rigor, I disagree with your statement "This is not a computer site". Mathematics is the art of computing. Henry's argument can be improved by noting that $F_n=\big(\varphi^{n+1}-(-\varphi)^{-(n+1)}\big)/\sqrt{5}$ and that $\varphi^{-n}/\sqrt5$ whenever $n\geq1$. There is nothing "dirty" in rounding to the nearest integer., – Tom-Tom Oct 16 '19 at 09:13
  • @CloudJR I don't know who you are, who has stand to say for "most of the mathematicians". I don't think you can make a statement like this without a bit of reframing. But I'm OK if you're just expressing *your* opinion. Again, there was nothing in the question implying using only integers, you have added this implicitely in you mind and then reproached that others haven't ? It's somewhat lacking rigor for someone who speaks in the name of "most mathematicians". – Tom-Tom Mar 03 '20 at 20:45

$n\in\mathbb{N}$ is a Fibonacci number iff $5n^2-4$ or $5n^2+4$ is a square. In the former case $n=F_{2k+1}$ while in the latter case $n=F_{2k}$. Assuming $n\geq 2$, in the former case $F_{2k+2}=\lfloor \varphi n \rfloor$ and in the latter case $F_{2k+1}=\lceil \varphi n\rceil $, with $\varphi=\frac{1+\sqrt{5}}{2}$.

Example: if $n=8$ we have that $5\cdot 8^2+4=18^2$, hence $n$ is a Fibonacci number with even index and the next Fibonacci number is $\lceil 8\varphi \rceil =13$.

Jack D'Aurizio
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Along similar lines to Matthew Daly's answer:

Binet's formula gives an exact value for the n'th Fibonacci number where numbering starts with $F_0=0$ and $F_1=1$:

$F_n = \frac{(1+\sqrt{5})^n-(1-\sqrt{5})^n}{2^n \sqrt{5}}$

$\sqrt{5} F_n=(\frac{1+\sqrt{5}}{2})^n - (\frac{1-\sqrt{5}}{2})^n$

$=\phi^n - (\frac{-1}{\phi})^n$, where $\phi = \frac{1+\sqrt{5}}{2}$ (using the identity that $\phi-1=\frac{1}{\phi}$, which is easy to prove).

From there it's easy to show that for $n>2$, $|$log$_\phi(F_n\sqrt{5})-n|<0.5$. (Hint: as $n$ becomes large, the first of these two terms becomes very large and the second goes to zero.)

If $F_n=1$ obviously the question is unanswerable, and if $F_n=0$ it's trivial. If $F_n>1$ then $n>2$ and so we can calculate $n$ by rounding log$_\phi(F_n\sqrt{5})$ to the nearest integer.

Now we have $n$, simply apply Binet's formula in the forwards direction and we're done.

Geoffrey Brent
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    Good answer, the log inequality requiring $1 + \frac{1}{(\phi + 1)^n} < \sqrt{\phi}$ ($n$ odd) and $\frac{1}{\sqrt{\phi}} < 1 - \frac{1}{(\phi + 1)^n}$ ($n$ even), which is easily checked directly for $n \geq 1$ and $n \geq 2$ respectively. – Ross Ure Anderson Oct 16 '19 at 18:46