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Let's consider generating function $$F(x) = (1+x)^r = \sum_{n=0} \binom{r}{n} x^n$$ And another generating function $$G(x) = (1+x^2)^r = \sum_{n=0} \binom{r}{n}x^{2n}$$ Please note those 2 functions differs only on $x^k$

After that let's take definition of binomial $$\binom{r}{n} = \frac{r^{\underline{n}}}{n!}$$ and apply it to our function $F(x)$ and $G(x)$ first.

$$F(x)=(1+x)^r=\sum_{n=0}^\infty r^{\underline n} \frac{x^n}{n!}$$ $$G(x)=(1+x^2)^r=\sum_{n=0}^\infty r^{\underline n} \frac{x^{2n}}{n!}$$

Whole question is about multiplying those functions.

To multiply $F(x)$ and $G(x)$ we have to have two function of form $\sum_{n=0}^\infty a_n x^n$, but our G(x) has form $\sum_{n=0}^\infty a_n x^{2n}$ So idea here is to make it right form, obviously.

Here i have doubts: should i treat this function as exponential generating function or ordinary generating function?

Because in first case it's seems like i'm iterating over sequence $\langle r^{\underline n} \rangle$ And in second case, if i want to make "correct" ordinary generating function i have to take $x=\sqrt{x}$ and apply it to function, it will give me obviously $G(\sqrt{x})=G_2(x)=(1+x)^r=\sum_{n=0}^\infty x^n$, but now $F(x)G(x) \neq F(x)G_2(x)$, so it doesnt seem like a good solution.

On the side note, $$F(x)G(x) = \sum_{n=0}^\infty ( \sum_{k=0}^n \binom{r}{k} \binom{r}{n-2k}) x^n$$

So my question is: how to work with such generating function? How to multiply them and how to change them to "right" form?

(Second side question : can we multiply our functions if both are in form $\sum_{n=0}^\infty a_n x^{kn}$ for some k being an integer/real number?)

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Whether you are contemplating OGFs or EGFs at this point is irrelevant -- multiplication here is simply as formal power series, without regards to interpretation.

There are a couple of ways that you can think about this; here's one way.

Note that when you multiply to power series $\sum a_nx^n$ and $\sum b_nx^n$, the result is $$ \sum_{n=0}^{\infty}c_nx^n,\qquad c_n:=\sum_{k=0}^{n}a_kb_{n-k}. $$ In this case, you can think of $$ a_n=\begin{cases}\binom{r}{n/2} & \text{if $n$ is even}\\0 & \text{else}\end{cases},\qquad b_n=\binom{r}{n}. $$ This means that your $c_n$ in this case will be $$ \sum_{k=0}^{n}a_kb_{n-k}=\sum_{j=0}^{\lfloor n/2\rfloor}a_{j}b_{n-2j}=\sum_{j=0}^{\lfloor n/2\rfloor}\binom{r}{j}\binom{r}{n-2j} $$ where we have noticed that all terms with odd $k$ are $0$, and reindexed using $k=2j$.

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  • $\begingroup$ Thanks for your solution, it's very nice, but i won't accept your question yet, cause i'm interested at another question too :) and maybe other methods of it to... $\endgroup$ – Krzysztof Lewko Apr 3 '14 at 13:24
  • $\begingroup$ Been thinking about this next day and i think it's the best solution out there. Thanks :) $\endgroup$ – Krzysztof Lewko Apr 5 '14 at 8:37

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