The identity is
$$\sum_{i\ge 0}(-1)^i \binom{\frac{s^k + s^{-k} - 10}{4}}{i}\binom{\frac{s^k + s^{-k} - 10}{4}+4}{\frac{gs^k + 2g^{-1}s^{-k} - 4}{8}-i}= 0$$
where $k\gt 1$ , $s = 3+2\sqrt{2}$ and $ g = 2-\sqrt{2}$
exemples $$\sum_{i\ge 0}(-1)^i \binom{6}{i}\binom{10}{2-i}= 0$$ $$\sum_{i\ge 0}(-1)^i \binom{47}{i}\binom{51}{14-i}= 0$$ $$\sum_{i\ge 0}(-1)^i \binom{286}{i}\binom{290}{84-i}= 0$$ $$...$$ ....etc.
It comes from this question.
Are there other identities like this one, where the integers are obtained from a recursion?
A generalisation (in the neater presentation suggested by @JeanMarie) would be, for a given positive integer $p$
$$\sum_{i\ge 0}(-1)^i \binom{u_k}{i}\binom{u_k+p}{v_k-i}= 0$$
This was the case $p=4$. The cases $p\le 3$ are dealt in the previous question.
For $p=5$ we have four pairs of sequences $(u_k,v_k)$
$$u_k=18u_{k-1}-u_{k-2}+48\ \ \ v_k=18v_{k-1}-v_{k-2}+8$$ with $$u_0=2,u_1=-2\ \ \ v_0=6,v_1=0$$ $$u_0=-1,u_1=-1\ \ \ v_0=2,v_1=0$$ $$u_0=-1,u_1=-1\ \ \ v_0=3,v_1=1$$ $$u_0=-2,u_1=2\ \ \ v_0=1,v_1=3$$
For $p=6$ we have two pairs of sequences $(u_k,v_k)$
$$u_k=14u_{k-1}-u_{k-2}+42 \ \ \ v_k=14v_{k-1}-v_{k-2}+6$$ with $$u_0=-1,u_1=-2\ \ \ v_0=4,v_1=0$$ $$u_0=-2,u_1=-1\ \ \ v_0=2,v_1=0$$
For $p=8$ we have two pairs of sequences $(u_k,v_k)$
$$u_k=6u_{k-1}-u_{k-2}+18 \ \ \ v_k=6v_{k-1}-v_{k-2}+2$$ with $$u_0=-1,u_1=-2\ \ \ v_0=5,v_1=1$$ $$u_0=-2,u_1=-1\ \ \ v_0=3,v_1=1$$
Question: Find other pairs of sequences $(u_k,v_k)$, $v_k \lt u_k$ for $p=7$ or $p\ge 9$ (if any).
Two pieces of information
1) I advise you to read Chapter 5 of the marvelous book "Concrete Mathematics" (Graham, Knuth, Patashnik) Addison-Wesley, 1989, which is in the same spirit as the book "A=B" that has been recommended by @Gabriel Nivasch.
2) Your expression should be simplified in this way:
$$\sum_{i\ge 0}(-1)^i \binom{u_k}{i}\binom{u_k+4}{v_k-i}= 0$$
where $u_n$ and $v_n$ are the sequences defined resp. by the second order recurrence relationships:
$$u_{n+1}=6u_n-u_{n-1}+10 \ \ \text{with} \ \ u_0=6 \ \text{and} \ u_1=47$$
and
$$v_{n+1}=6v_n-v_{n-1}+2 \ \ \text{with} \ \ v_0=10 \ \text{and} \ v_1=51$$
In this way you have a neater presentation without spurious irrationals.