During my try of proofing something else, I found the following equality:
$\binom{r+2n-1}{n-1} = \binom{2n-1}{n-1} + \sum\limits_{k=1}^{n-1} \binom{2k-1}{k} \binom{r+2(n-k)-1}{r+n-k}\frac{r}{n-k}; ~~~r \in \mathbb{R}, n \in \mathbb{N_0}, k \in \mathbb{N}$
I am not a mathematician. I just got to this equality by coincidence, which should be true. But I can't proof it. I am stuck at this point. I tried already by induction, but I can't figure out the induction step.
Can somebody help me?
Starting from the claim (we treat the case $r$ a positive integer)
$${r+2n-1\choose n-1} - {2n-1\choose n-1} = S = \sum_{k=1}^{n-1} {2k-1\choose k} {r+2(n-k)-1\choose r+n-k} \frac{r}{n-k} \\ = \sum_{k=1}^{n-1} {2n-2k-1\choose n-k} {r+2k-1\choose r+k} \frac{r}{k} \\ = \sum_{k=1}^{n-1} {2n-2k-1\choose n-k} {r+2k-1\choose k-1} \frac{r}{k}$$
we use the fact that
$${r+2k-1\choose k-1} \frac{r}{k} = {r+2k-1\choose k} - {r+2k-1\choose k-1}$$
to get two pieces, call them $S_1$ and $S_2$ where $S=S_1-S_2$ and
$$S_1 = \sum_{k=1}^{n-1} {2n-2k-1\choose n-k} {r+2k-1\choose k} $$
and
$$S_2 = \sum_{k=1}^{n-1} {2n-2k-1\choose n-k} {r+2k-1\choose k-1}.$$
We find for $S_1$
$$\underset{w}{\mathrm{res}}\; (1+w)^{r-1} \sum_{k=1}^{n-1} {2n-2k-1\choose n-k} \frac{(1+w)^{2k}}{w^{k+1}} \\ = \underset{w}{\mathrm{res}}\; \frac{(1+w)^{r-1}}{w} [z^n] (1+z)^{2n-1} \sum_{k=1}^{n-1} z^k (1+z)^{-2k} \frac{(1+w)^{2k}}{w^{k}}.$$
Including the term at $k=0$ and compensating
$$-{2n-1\choose n-1} + \underset{w}{\mathrm{res}}\; \frac{(1+w)^{r-1}}{w} [z^n] (1+z)^{2n-1} \sum_{k=0}^{n-1} z^k (1+z)^{-2k} \frac{(1+w)^{2k}}{w^{k}}.$$
Including the term at $k=n$ and again compensating
$$-{2n-1\choose n-1} - {r+2n-1\choose n} \\ + \underset{w}{\mathrm{res}}\; \frac{(1+w)^{r-1}}{w} [z^n] (1+z)^{2n-1} \sum_{k=0}^{n} z^k (1+z)^{-2k} \frac{(1+w)^{2k}}{w^{k}}.$$
Now we may extend $k$ beyond $n$ owing to the coefficient extractor $[z^n]$ to get
$$-{2n-1\choose n-1} - {r+2n-1\choose n} \\ + \underset{w}{\mathrm{res}}\; \frac{(1+w)^{r-1}}{w} [z^n] (1+z)^{2n-1} \frac{1}{1-z(1+w)^2/w/(1+z)^2} \\ = -{2n-1\choose n-1} - {r+2n-1\choose n} \\ + \underset{w}{\mathrm{res}}\; (1+w)^{r-1} [z^n] (1+z)^{2n+1} \frac{1}{w(1+z)^2-z(1+w)^2}.$$
We get for $S_2$
$$\underset{w}{\mathrm{res}}\; (1+w)^{r-1} [z^n] (1+z)^{2n-1} \sum_{k=1}^{n-1} z^k (1+z)^{-2k} \frac{(1+w)^{2k}}{w^{k}}.$$
The term $k=0$ contributes zero. Compensating for $k=n$ we find
$$-{r+2n-1\choose n-1} + \underset{w}{\mathrm{res}}\; (1+w)^{r-1} [z^n] (1+z)^{2n-1} \sum_{k\ge 0} z^k (1+z)^{-2k} \frac{(1+w)^{2k}}{w^{k}} \\ = -{r+2n-1\choose n-1} + \underset{w}{\mathrm{res}}\; w (1+w)^{r-1} [z^n] (1+z)^{2n+1} \frac{1}{w(1+z)^2-z(1+w)^2}.$$
We therefore have
$$S = S_1-S_2 = - {2n-1\choose n-1} - {r+2n-1\choose n} + {r+2n-1\choose n-1} \\ + \underset{w}{\mathrm{res}}\; (1+w)^{r-1} [z^n] (1+z)^{2n} \frac{(1-w)(1+z)}{w(1+z)^2-z(1+w)^2}.$$
Working with the remaining residue we note that
$$\frac{(1-w)(1+z)}{w(1+z)^2-z(1+w)^2} = \frac{1}{w} \frac{1}{1-z/w} - \frac{1}{1-zw}.$$
We see on substituting into the residue that we get no contribution from the second term. This leaves
$$\underset{w}{\mathrm{res}}\; \frac{1}{w} (1+w)^{r-1} [z^n] (1+z)^{2n} \frac{1}{1-z/w} \\ = \underset{w}{\mathrm{res}}\; \frac{1}{w} (1+w)^{r-1} \sum_{q=0}^n {2n\choose n-q} w^{-q} \\ = \sum_{q=0}^n {2n\choose n-q} {r-1\choose q} = [z^n] (1+z)^{2n} \sum_{q=0}^n {r-1\choose q} z^q.$$
The coefficient extractor once more enforces the range and we find
$$[z^n] (1+z)^{2n} \sum_{q\ge 0} {r-1\choose q} z^q \\ = [z^n] (1+z)^{2n} (1+z)^{r-1} = [z^n] (1+z)^{r+2n-1} = {r+2n-1\choose n}.$$
Collecting all four pieces yields
$$S = S_1-S_2 = - {2n-1\choose n-1} - {r+2n-1\choose n} + {r+2n-1\choose n-1} + {r+2n-1\choose n} \\ = {r+2n-1\choose n-1} - {2n-1\choose n-1}$$
which is the claim.
Remark. The next-to-last step may also be done as follows:
$$\underset{w}{\mathrm{res}}\; \frac{1}{w} (1+w)^{r-1} [z^n] (1+z)^{2n} \frac{1}{1-z/w} \\ = \underset{w}{\mathrm{res}}\; \frac{1}{w} \sum_{q=0}^{r-1} {r-1\choose q} w^q [z^n] (1+z)^{2n} \frac{1}{1-z/w} \\ = [z^n] (1+z)^{2n} \sum_{q=0}^{r-1} {r-1\choose q} z^q = [z^n] (1+z)^{2n} (1+z)^{r-1} = {r+2n-1\choose n}.$$