# Does this follow from the definition of the LambertW function?

The LambertW function $W(s)$ also called ProductLog seems to satisfy this relation:

$$-W(s) = \underbrace{-s e^{-s e^{\cdot^{\cdot^{-s e^{-W(s)}}}}}}_n$$

Or truncated:

$$-W(s) = -se^{-s e^{-s e^{-s e^{-s e^{-s e^{-s e^{-s e^{-s e^{-s e^{-s e^{-s e^{-s e^{-W(s)}}}}}}}}}}}}}$$

for some complex number $s$.

Does this follow from the definition of the LambertW function?

Mathematica:

s = ZetaZero[1]
N[-s*Exp[-s*
Exp[-s*Exp[-s*
Exp[-s*Exp[-s*
Exp[-s*Exp[-s*
Exp[-s*Exp[-s*
Exp[-s*
Exp[-s*Exp[-s*Exp[-ProductLog[s]]]]]]]]]]]]]], 30]
N[-ProductLog[s], 30]


which outputs:

-1.88420341024267232833723733662 - 1.03367078404796759147577622553 I

and

-1.88420341024267232833723733662 - 1.03367078404796759147577622553 I

Excel spreadsheet formula (European dot-comma) apparently giving $\frac{W(n)}{n}$:

by using the relation above.

Excel output: 0 | 0,56714329 | 0,426302751 | 0,349969501 | 0,300547296 | 0,26530177

which is the same as the Mathematica command:

N[Table[LambertW[n]/n, {n, 1, 12}]]


Yup, it follows from $W(s)e^{W(s)} = s$. Rearranging yields
$$W(s) = se^{-W(s)} = se^{-se^{-W(s)}} = \cdots$$