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This question was inspired by the inactive thread How to find this value of $A$? but the focus there was on the divergence of the imaginary part.

It seems that for a given nonzero real $x$, $$\operatorname{Re}\frac{W_{k^2}(x)}{W_k(x)}=k+\frac14\operatorname{sgn}x+o(1)$$ where $k$ is a positive integer and $W_k(x)$ denotes the $k$th branch of the Lambert $W$ function.

When the branch index is negative, it also seems that $$\operatorname{Re}\frac{W_{-k^2}(x)}{W_{-k}(x)}=k+\frac12-\frac14\operatorname{sgn}x+o(1).$$

Are these asymptotics true?

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The asymptotics of $W_k(x)$ for fixed real $x$ and $|k|\to\infty$ are given by $$W_k(x)=2k\pi i-\log(2k\pi i)+\log(x)+\dfrac{\log(2k\pi i)-\log(x)}{2k\pi i}+O(\log(|k|)^2/k^2)$$ (see for instance arXiv:2012.11698v2), so your result should follow from this.

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