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As for the name, according to wikipedia the Todd genus is given by:

$$\mathrm{Td}(z)=\frac{z}{1-e^{-z}}.$$

So, $f(z)=1/\mathrm{Td}(-z)$.
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As for the name, according to wikipedia the Todd genus is given by:

$$\mathrm{Td}(z)=\frac{z}{1-e^{-z}}.$$

So, $f(z)=1/\mathrm{Td}(-z)$.