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I[G(x)]=\frac{1}{1-G(x)}-1  \end{equation}  Let $I^{(n)}[G(x)]$ denote a repeated application of the invert transform $n$ times. Then \begin{equation}  I^{(2)}[G(x)]=\frac{1}{1-\frac{1}{1-G(x)}-1}-1\\  I^{(3)}[G(x)]=\frac{1}{1-\frac{1}{1-\frac{1}{1-G(x)}-1}-1}-1 I^{(2)}[G(x)]=\frac{1}{1-(\frac{1}{1-G(x)}-1)}-1\\  I^{(3)}[G(x)]=\frac{1}{1-(\frac{1}{1-(\frac{1}{1-G(x)}-1)}-1)}-1  \end{equation}  We then have \begin{equation}  I^{(0)}[P(x)]=2x+3x^2+5x^3+7x^4+11x^5+\cdots\\