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however, for some sequences it makes more sense to count from $1$, giving the description \begin{equation}  G_a(z)=a(1)z^1 + a(2)z^2 + \cdots= \sum_{n=1}^\infty a(n)z^n,  \end{equation}  the latter is the description we will use.Next we introduce the inverse Z-transform of a function $F(z)$, which is given by the contour integral   \begin{equation}  \mathcal{Z}^{-1}[F(z)](n)=\frac{1}{2\pi i}\oint F(z)z^{n-1}\;dz  \end{equation}  We can take the inverse Z-transform of the generating function with reciprocal argument (i.e. $G(\frac{1}{z})$) to produce the expression for the sequence.  \begin{equation}  \mathcal{Z}^{-1}[G_a(\frac{1}{z})](n)=a(n)  \end{equation}  We can see that the generating function for the natural numbers $1,2,3,\cdots$ (which are also the 2-rough numbers) is then \begin{equation}  G_2(z)=\frac{z}{(z-1)^2}=z+2z^2+3z^3+\cdots\\  a_2(n)=n