I found this equation, results looks reasonable, but can please someone explain me where it come form ?

I'm no expert in this, in fact I'm a mathematician, not an EE, but the coefficients in your formula rang a bell, so perhaps I can help a bit.

If you have a sinusoidal input current, $$V_{peak}\cos\omega t$$ and pass it trough an ideal rectifier, you get $$V_{peak} \vert \cos\omega t \vert$$. Taking the Fourier series of the rectified voltage, you get, to the second harmonic:

$$ \frac{2V_{peak}}{\pi} + \frac{4V_{peak}}{3\pi}\cos 2\omega t + \ldots $$

If your $V_{ac.min}$ is RMS, then, the input voltage is:

$$\frac{2\sqrt{2}V_{ac.min}}{\pi} + \frac{4\sqrt{2}V_{ac.min}}{3\pi}\cos 2\omega t + \ldots $$

The amplitude of the second harmonic is clearly recognizable in your formula, and in looks to be like a duty cycle format (disregarding efficiency, and taking the peak-to-peak voltage):

$$ D = 1 - \frac{V_{in, min}}{V_o} = 1 - \frac{8\sqrt{2}V_{ac.min}}{3\pi V_o} $$

So, somehow, your formula looks like:

$$I_S = \frac{P_o}{V_{ac.min}}\sqrt{D}$$

This is the part I don't really understand. I tried the computation for a normal booster, and found something along the lines of $$I_S = \frac{P_o}{V_{ac.min}}\cdot D^2$$ (continuous mode), so I don't know where the square root comes from. I also don't know why the zero-order harmonic doesn't come to play into the duty cycle, unless your output voltage equals the zero-order rectified voltage for some reason.

Edits: LaTeX cleanup.