What is the generating function of $z^3 + 2z^5 + 5z^7 + 14 z^9+\dots$?

Question

What is the generating function of $z^3 + 2z^5 + 5z^7 + 14 z^9+\dots$ ?

The generating function can be written as follows: $$A(z)=\sum_{i>2}^{\infty} a_i z^{2i+1},\text{where } a_i \text{ is the Catalan numbers} $$

Answer 1

The Catalan numbers go, $$1,1,2,5,14,42,132,429, \dots$$ The generating series you are after is, $$z^3+2z^5+5z^7+14z^9+42z^{11}+\dots$$ As Lord Shark the Unknown points out the generating function for the Catalan numbers is well known, $$\frac{1-\sqrt{1-4z}}{2z}=1+z^1+2z^2+5z^3+14z^4+42z^5+...$$ To insert a single interlacing zero between the terms you need to replace $z$ with $z^2$, $$\frac{1-\sqrt{1-4(z^2)}}{2(z^2)}=1+(z^2)^1+2(z^2)^2+5(z^2)^3+14(z^2)^4+42(z^2)^5+\dots$$ Now multiply both sides by z, $$\frac{1-\sqrt{1-4z^2}}{2z}=z+z^3+2z^5+5z^7+14z^9+42z^{11}+\dots$$

To remove the unwanted initial z, simply subtract it from both sides and so the generating function you are after is, $$\frac{1-\sqrt{1-4z^2}}{2z}-z=z^3+2z^5+5z^7+14z^9+42z^{11}+...$$ There is a marvellous online tool you can use to quickly check what generating series a generating function expands into. It's the Taylor Series Expansion Calculator at https://www.numberempire.com/taylorseriesexpansion.php

Answer 2

Let $C(z)=\sum_{n=0}^\infty C_nz^n$ be the generating function for the Catalan numbers. Then your series is the generating function of the odd terms minus the first term i.e. given by $$ \frac{C(z)-C(-z)}{2}-z $$