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I can't see any obvious way this could be calculated. It seems to converge to a value of approximately 0.6278...

$\dfrac{1 + \dfrac{3}{4}}{2 + \dfrac{5}{6}} \approx 0.6176 $

$\dfrac{1 + \dfrac{3 + \dfrac{7}{8}}{4 + \dfrac{9}{10}}}{2 + \dfrac{5 + \dfrac{11}{12}}{6 + \dfrac{13}{14}}} \approx 0.6175 $

Going all the way up to 62 gives a result of 0.627841944566, so it seems to converge.

Is it possible to find a value for this? Will it have a closed form solution?

  • 0
    This might help https://en.wikipedia.org/wiki/Continued_fraction – 2017-01-12
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    Not to detract from your question, but I find the following different expression more natural: $$1+\frac{2+\frac{4+\frac{8+\cdots}{9+\cdots}}{5+\frac{10+\cdots}{11+\cdots}}}{3+\frac{6+\frac{12+\cdots}{13+\cdots}}{7+\frac{14+\cdots}{15+\cdots}}}$$ That's a proper binary tree, that is. And here the powers of two are apparent. – 2017-01-12
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    @anonymous I considered if it would be an example of a continued fraction when I was tagging the post, but it doesn't seem to me like it is since each fraction has a numerator. That being said I'm honestly not sure. – 2017-01-12
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    @Rahul $$0 + \dfrac{1 + \dfrac{3\cdots}{4\cdots}}{2 + \dfrac{5\cdots}{6\cdots}}$$ – 2017-01-12
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    More compactly (thanks @Axoren), your expression is $f(0)$, where $f$ is the (unique?) function that satisfies $f(n) = n + \frac{f(2n+1)}{f(2n+2)}$ and $n \le f(n) \le n+1$. – 2017-01-12
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    @Rahul I just came to the same equation just now. I think tackling this as a recurrence relation is the right way to go about it. – 2017-01-12
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    @Axoren I haven't got any experience dealing with solving recurrence relations, and W|A isn't very helpful so it seems like it is quite a complex one. Should I edit the question to add this information to it or make a new post? – 2017-01-12
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    Looking at the function f(n) with a bit of Python code to estimate it makes it look almost perfectly linear. Whether this is actually the case I don't know. [See here](https://i.imgur.com/em1UK1h.png) – 2017-01-12
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    Can you try plotting $f(n)-n$ instead? – 2017-01-12

1 Answers 1

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Define $$f_m(n) = \begin{cases} n+\cfrac{f_m(2n+1)}{f_m(2n+2)} & \text{if $n

One can show via reverse induction over $n=m,\ldots,1,0$ that $f_m(n) \in [n, n+1]$. So in the limit, defining $f(n)=\lim_{m\to\infty}f_m(n)$, we have $f(n) \in [n, n+1]$.

Using interval arithmetic we can then obtain rigorous bounds on $f(n)$. Define the interval-valued function $$[f]_m(n) = \begin{cases} n+\cfrac{[f]_m(2n+1)}{[f]_m(2n+2)} & \text{if $n

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    P.S. My preferred number $1+\frac{2+\frac{4+\frac{8+\cdots}{9+\cdots}}{5+\frac{10+\cdots}{11+\cdots}}}{3+\frac{6+\frac{12+\cdots}{13+\cdots}}{7+\frac{14+\cdots}{15+\cdots}}}$ lies in $[\underbrace{1.73022677823852}\!12699, \underbrace{1.73022677823852}\!22390]$. – 2017-01-12
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    +1 This interval arithmetic approach is awesome and provides a good basis for testing exact results. That last statement just needs to be proved and you have bounds around the solution which are arbitrarily tight. – 2017-01-12