Property of Krawtchouk polynomials involving $K_s(x)$ and $K_s(x+y)$ Fix $n$ and consider the Krawtchouk polynomials
$$
K_s(\ell) = \sum_{k=0}^s(-1)^k\binom{\ell}{k}\binom{n-\ell}{s-k} \quad \text{defined for} \quad 0 \leq s \leq n
$$
Are you aware of any property linking the evaluations in some points $x$ and $x+y$, i.e. between $K_s(x)$ and $K_s(x+y)$ ?
 A: Not quite what was asked for, but maybe still useful?
$$K_{s,n}(x+y)=[z^s](1-z)^{x+y}(1+z)^{n-x-y}$$
$$K_{s,n}(x+y)=[z^s](1-z)^x(1+z)^{n-x}\big(\frac{1-z}{1+z}\big)^y$$
$$K_{s,n}(x+y)=\sum_{j=0}^s[z^j](1-z)^x(1+z)^{n-x}[z^{s-j}]\big(\frac{1-z}{1+z}\big)^y$$
$$K_{s,n}(x+y)=\sum_{j=0}^s K_{j,n}(x)F_{s-j}(y)$$
where $F_{m}(y)$ is a polynomial function of $y$ which is the coefficient of $z^m$ in the $z$-power expansion of $ \big(\frac{1-z}{1+z}\big)^y$.
$F_{0}(y)=1$
$F_{1}(y)=-2y$
$F_{2}(y)=2y^2$
$F_{3}(y)=-\frac{2}{3}(y + 2 y^3)$  etc.
These polynomial are probably known and likely have an explicit expression.
See D.E Knuth Convolution Polynomials, where it is shown how to compute these kinds of polynomials.
Edit:  If I am not wrong in the computation, a general expression for $F_m(y)$ is :
$$ F_m(y)= (-1)^m \sum \prod_{i=1}^m\frac{(2 y)^{k_i}}{k_i!i^{k_i}}$$
where the $\sum$ is over all partitions of $m= \sum i\cdot k_i$ but with odd summands $i$ only.
Further edit:  since $K_{s,n}(0)={n\choose s}$, we have (with the convention that the $K_{j,n}$ are zero for negative $j$ and $j\gt n$)
$$K_{s,n}(y)=\sum_{j=0}^n {n\choose j}F_{s-j}(y)=\sum_{j=0}^n {n\choose j}F_{s+j-n}(y)$$
then by binomial inversion we have
$$\sum_{u=0}^j  (-1)^{j-u}{j\choose u}K_{s,u}(y)=F_{s-j}(y) $$
And then
$$K_{s,n}(x+y)=\sum_{j=0}^s K_{j,n,q}(x)\sum_{u=0}^j(-1)^{j-u}{j\choose u} K_{s,u}(y)$$
$$ \color{red}{K_{s,n}(x+y)=\sum_{0\le u \le j \le s} (-1)^{j-u}{j\choose u} K_{j,n}(x)K_{s,u}(y)}$$
If true (please verify), this would be a better looking addition formula for the Krawtchouk polynomials.
