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Is there a simplification for the following recursive fraction :

$$\frac{\frac{\frac{n}{W(n)}}{W\left(\frac{n}{W(n)}\right)}}{W\left(\frac{\frac{n}{W(n)}}{W\left(\frac{n}{W(n)}\right)}\right)}$$

The above formula uses a recursion 3 times. I'm looking for a simplification when we have such a finite recursion, for instance when this one appears $i$ times. I would like to remove the recursion, i.e. obtain a single fraction.

Thank you.

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    The question is unclear. What is it that you mean by "simplifying"? – 2017-02-08
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    @SimplyBeautifulArt Thank for your comment. I mean "reducing to a single fraction", without any recursion, using simple terms (except the Lambert function). Post edited ! :-) – 2017-02-08
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    Since $W(n)\approx\log(n)-\log\log(n)$ for large $n$s, I doubt there is any substantial simplification. What is the source of this problem? – 2017-02-08
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    @JackD'Aurizio This a problem of mine but, in fact, I'm wondering if there exists a result on that. – 2017-02-08

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Since $u=W(u)e^{W(u)}$, it follows that $\frac u{W(u)}=e^{W(u)}$ and we get

$$\frac{\frac{\frac{n}{W(n)}}{W\left(\frac{n}{W(n)}\right)}}{W\left(\frac{\frac{n}{W(n)}}{W\left(\frac{n}{W(n)}\right)}\right)}=e^{W\left(e^{W(e^{W(n)})}\right)}$$

Which is the best simplification I can see.

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    @Dingo13: as said in the comments, it is $\approx\frac{n}{\log(n)^3}$. – 2017-02-08
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    @JackD'Aurizio that might look nice due to the exponentials. Or you've already calculated that... – 2017-02-08
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    Thank you again. Why do you use $\approx$ for asymptotic? – 2017-02-08
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    @Dingo13 he uses it as an approximation. But I suppose it's also an asymptote. – 2017-02-08
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    @SimplyBeautifulArt Can we use asymptotic equivalence? – 2017-02-08
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    @Dingo13 yes, Jack's result fits nicely if that's what you aimed for. – 2017-02-08
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    @SimplyBeautifulArt Ok, thank you again ! – 2017-02-08
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    No problem! :-) – 2017-02-08