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Extended answer

The approximation described below is original, explicit (in an appropriate some sense), and very accurate.

So, you want to approximate the solution $W(x)$ of $x=W(x)e^{W(x)}$, for large $x$ (the order of $x$ plays no does not play a very significant role in what follows).

Example. For $x=2000$, even $r$ as low as $80$ gives quite accurate results:\tilde W^5 (x,r) 2000,80) \approx 5.83673149492073\tilde W^6 (x,r) 2000,80) \approx 5.836731494908671,while the exact solution (according to Wims Function Calculator) is W(2000) = 5.836731494908178747....Based on many numerical results, and this approximation seems quite interesting. Here is one further example.The Omega constant $\Omega$ is the value of $W(1)$: \Omega = W(1) \approx 0.5671432904097838729999686622.With $r$ as low as $30$, we already notedget the following impressive approximations:\tilde W^1 (1,30) \approx 0.5710729200334063, \tilde W^2 (1,30) \approx 0.5674569334624368,$$ \tilde W^3 (1,30) \approx 0.5671683899602143, \tilde W^4 (1,30) \approx 0.5671452994467842,$$ \tilde W^5 (1,30) \approx 0.5671434512213455,\tilde W^6 (1,30) \approx 0.5671433032818183,\tilde W^7 (1,30) \approx 0.5671432914401158,\tilde W^8 (1,30) \approx 0.567143290492256,\tilde W^9 (1,30) \approx 0.5671432904163853,\tilde W^{10} (1,30) \approx 0.5671432904103123,\tilde W^{11} (1,30) \approx 0.5671432904098261,\tilde W^{12} (1,30) \approx 0.5671432904097873.So, $\tilde W^{12} (1,30) - W(1) \approx 3 \times 10^{-15}$.It is interesting to compare this sophisticated approximation with the standard one obtained from the converging sequence $\Omega_n \to \Omega$ defined by $W(x) \Omega_{n+1} = 5.836731494908178747...e^{-\Omega_n}$ (with initial value $\Omega_0$). For example, with $\Omega_0 = 0.5$, we only get \Omega_1 \approx 0.6065306597126334,\Omega_2 \approx 0.545239211892605,\Omega_3 \approx 0.5797030948780683,\Omega_4 \approx 0.5600646279389019,\Omega_5 \approx 0.5711721489772151,\Omega_6 \approx 0.5648629469803235,\Omega_7 \approx 0.5684380475700662,\Omega_8 \approx 0.5664094527469208,\Omega_9 \approx 0.5675596342622424,\Omega_{10} \approx 0.5669072129354714,\Omega_{11} \approx 0.5672771959707785,\Omega_{12} \approx 0.5670673518537281.

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The approximation described below is original, explicit (in an appropriate sense), and extremely very accurate. It is closely related to this question (second paragraph).

So, you want to approximate the solution $W(x)$ of $x=W(x)e^{W(x)}$, for large $x$ (the order of $x$ plays no significant role in what follows). First, define $$ \varphi (x,r) = 1 + \sum\limits_{k = 1}^{\left\lceil r \right\rceil } {\frac{{x^k [r - (k - 1)]^k }}{{k!}}} . $$ Now, consider the following series of approximations, where $r$ is assumed sufficiently large. The first one is $$ \tilde W^1 (x,r) = \frac{1}{r}\ln \varphi (x,r). $$ Subsequent approximations are defined recursively by $$ \tilde W^{n + 1} (x,r) = \frac{1}{r}\ln \bigg[\frac{{\tilde W^n (1 + \tilde W^n )}}{x}\varphi (x,r)\bigg]. $$

Example. For $x=2000$, even $r$ as low as $80$ gives $$ \tilde W^5 (x,r) \approx 5.83673149492073 $$ and $$ \tilde W^6 (x,r) \approx 5.836731494908671, $$ while the exact solution (according to Wims Function Calculator, and as already noted) is $$ W(x) = 5.836731494908178747.... $$
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The approximation described below is original, explicit (in an appropriate sense), and extremely accurate. It is closely related to this question (second paragraph).

So, you want to approximate the solution $W(x)$ of $x=W(x)e^{W(x)}$, for large $x$ (the order of $x$ plays no significant role in what follows). First, define $$ \varphi (x,r) = 1 + \sum\limits_{k = 1}^{\left\lceil r \right\rceil } {\frac{{x^k [r - (k - 1)]^k }}{{k!}}} . $$ Now, consider the following series of approximations, where $r$ is assumed sufficiently large. The first one is $$ \tilde W^1 (x,r) = \frac{1}{r}\ln \varphi (x,r). $$ Subsequent approximations are defined recursively by $$ \tilde W^{n + 1} (x,r) = \frac{1}{r}\ln \bigg[\frac{{\tilde W^n (1 + \tilde W^n )}}{x}\varphi (x,r)\bigg]. $$

Example. For $x=2000$, even $r$ as low as $80$ gives $$ \tilde W^5 (x,r) \approx 5.83673149492073 $$ and $$ \tilde W^6 (x,r) \approx 5.836731494908671, $$ while the exact solution (according to Wims Function Calculator) is $$ W(x) = 5.836731494908178747.... $$