Using a similar method to my previous answer...

43

Since

$439204 = 2^{(2^3)}\cdot 5\cdot 7^3+2\cdot 3^4+2$

(phi formula incoming...)

Using a similar method to my answer to the previous question...

 43  42

Since

$N = \frac{\phi + \phi + ...}{\phi}$
and
$439204 = 2^{(2^3)}\cdot 5\cdot 7^3+2\cdot (3^4+1)$

$\frac{\phi + \phi}{\phi}^{(\frac{\phi + \phi}{\phi}^\frac{\phi + \phi + \phi}{\phi})}\cdot \frac{\phi + \phi + \phi + \phi + \phi}{\phi} \cdot \frac{\phi + \phi + \phi + \phi + \phi + \phi + \phi}{\phi} ^\frac{\phi + \phi + \phi}{\phi} + \frac{\phi + \phi}{\phi}\cdot (\frac{\phi + \phi + \phi}{\phi} ^\frac{\phi + \phi + \phi + \phi}{\phi} + \frac{\phi}{\phi})$