I’ve come across the following identity which seems to be related to the Lucas polynomials $\sum_{j=0}^{\lfloor{\frac{n}{2}}\rfloor}\frac{n}{n-j}\binom{n-j}{j}x^{n-2j}$:
$${\sum_{j=0}^{\lfloor{\frac{n}{2}}\rfloor}\binom{i+j+x}{i-j+1}\frac{n}{n-j}\binom{n-j}{j}}=\binom{x+n+i}{i+1}$$ for $i\ge0$ and $n\ge{i+2}$. Is there a simple proof?
We seek to prove that with $n\ge m+2$
$$\sum_{j=0}^{\lfloor n/2 \rfloor} {m+j+k\choose m-j+1} \frac{n}{n-j} {n-j\choose j} = {n+k+m\choose m+1}.$$
This is
$${m+k\choose m+1} + \sum_{j=1}^{\lfloor n/2 \rfloor} {m+j+k\choose m-j+1} \frac{n}{n-j} {n-j\choose j} = {n+k+m\choose m+1}$$
or
$${m+k\choose m+1} + \sum_{j=1}^{\lfloor n/2 \rfloor} {m+j+k\choose m-j+1} \frac{n}{j} {n-j-1\choose j-1} = {n+k+m\choose m+1}$$
Now observe that
$${n-j-1\choose j} = \frac{n-2j}{j} {n-j-1\choose j-1} = \frac{n}{j} {n-j-1\choose j-1} - 2 {n-j-1\choose j-1}.$$
We thus get two terms:
$${m+k\choose m+1} + \sum_{j=1}^{\lfloor n/2 \rfloor} {m+j+k\choose m-j+1} {n-j-1\choose j} = \sum_{j=0}^{\lfloor n/2 \rfloor} {m+j+k\choose m-j+1} {n-j-1\choose j}$$
and
$$2\sum_{j=1}^{\lfloor n/2 \rfloor} {m+j+k\choose m-j+1} {n-j-1\choose j-1}.$$
For the first one we have
$$\sum_{j=0}^{\lfloor n/2 \rfloor} {m+j+k\choose m-j+1} {n-j-1\choose n-2j-1} \\ = [z^{n-1}] (1+z)^{n-1} \sum_{j=0}^{\lfloor n/2 \rfloor} {m+j+k\choose m-j+1} (1+z)^{-j} z^{2j}.$$
We may extend $j$ to infinity because of the coefficient extractor in front (note that the following representation in the variable $w$ will produce a correct value of zero in the remaining binomial coefficient when $j\gt m+1$):
$$[z^{n-1}] (1+z)^{n-1} [w^{m+1}] (1+w)^{m+k} \sum_{j\ge 0} (1+z)^{-j} z^{2j} (1+w)^j w^j \\ = [z^{n-1}] (1+z)^{n-1} [w^{m+1}] (1+w)^{m+k} \frac{1}{1-z^2 w (1+w)/(1+z)} \\ = [z^{n-1}] (1+z)^{n} [w^{m+1}] (1+w)^{m+k} \frac{1}{1+z-z^2 w (1+w)} \\ = - [z^{n-1}] (1+z)^{n} [w^{m+2}] (1+w)^{m+k-1} \frac{1}{(z-1/w)(z+1/(1+w))}.$$
Extracting $[z^{n-1}]$ first we get
$$\mathrm{Res}_{z=0} \frac{1}{z^n} (1+z)^n \frac{1}{(z-1/w)(z+1/(1+w))}.$$
We see that the residue at infinity is zero. Residues sum to zero and we get for the residue at $z=1/w$
$$w^n \frac{(1+w)^n}{w^n} \frac{1}{1/w+1/(1+w)} = w \frac{(1+w)^{n+1}}{1+2w}.$$
For the residue at $z=-1/(1+w)$ we find
$$- (-1)^n (1+w)^n \frac{w^n}{(1+w)^n} \frac{1}{1/(1+w)+1/w} = - (-1)^n w^{n+1} (1+w) \frac{1}{1+2w}.$$
Now the coefficient extractor is $[w^{m+2}]$ but we have $n\ge m+2$ so the contribution from this is zero.
It follows that the first sum is given by
$$[w^{m+1}] \frac{(1+w)^{n+k+m}}{1+2w}.$$
Continuing with the second sum we find
$$2\sum_{j=1}^{\lfloor n/2 \rfloor} {m+j+k\choose m-j+1} {n-j-1\choose n-2j} \\ = 2[z^{n}] (1+z)^{n-1} \sum_{j=1}^{\lfloor n/2 \rfloor} {m+j+k\choose m-j+1} (1+z)^{-j} z^{2j}.$$
We may include $j=0$ here because
$$2[z^n] (1+z)^{n-1} {m+k\choose m+1} = 0,$$
getting
$$2[z^{n}] (1+z)^{n-1} \sum_{j=0}^{\lfloor n/2 \rfloor} {m+j+k\choose m-j+1} (1+z)^{-j} z^{2j}.$$
We skip forward to the residue computation since the intermediate steps are the same as before. We get for the residue at $z=1/w$
$$2 w^{n+1} \frac{(1+w)^n}{w^n} \frac{1}{1/w+1/(1+w)} = 2 w^2 \frac{(1+w)^{n+1}}{1+2w}.$$
For the residue at $z=-1/(1+w)$ we find
$$- (-1)^{n+1} (1+w)^{n+1} \frac{w^n}{(1+w)^n} \frac{1}{1/(1+w)+1/w} = (-1)^n w^{n+1} (1+w)^2 \frac{1}{1+2w}.$$
We note once more that the coefficient extractor is $[w^{m+2}]$ but we have $n\ge m+2$ so the contribution from this is zero. It follows that the second sum is given by
$$[w^{m+1}] 2 w \frac{(1+w)^{n+k+m}}{1+2w}.$$
Adding the two sums we obtain at last
$$[w^{m+1}] \frac{(1+w)^{n+k+m}}{1+2w} + [w^{m+1}] 2 w \frac{(1+w)^{n+k+m}}{1+2w} = [w^{m+1}] (1+w)^{n+k+m}.$$
or
$$\bbox[5px,border:2px solid #00A000]{ {n+k+m\choose m+1}.}$$