Let $x,y$ be two elements of a ring satisfying $$xy=-yx.$$ Let $n \geq 0$. Then we can calculate $(x+y)^n$ as follows:
- If $n$ is even, then $$(x+y)^n = \sum_{0 \leq k \leq n,\, k \text{ even}} \binom{n/2}{k/2} x^{n-k} y^k.$$
- If $n$ is odd, then $$(x+y)^n = \sum_{0 \leq k \leq n} \binom{\lfloor n/2 \rfloor}{\lfloor k/2 \rfloor} x^{n-k} y^k.$$
The proof is an easy induction (and it is also possible to see it as a special case of $q$-binomial coefficients for $q=-1$, see the answer by mjqxxxx). I would like to know if someone knows a reference for precisely this anti-commutative binomial theorem. Please notice that I am not asking for a proof, and I am not asking for a reference for the $q$-binomial theorem, unless it explicitly deduces and states the anti-commutative binomial theorem as stated above.
In section 3 of Scurlock 2020, expressions are derived for ${{n}\choose{k}}_{-1}$, which relate to this question through the definition $$ (x+y)^{n}=\sum_{k=0}^{n}{{n}\choose{k}}_{-1}x^ky^{n-k} $$ when $x$ and $y$ anticommute. OP and the reference both effectively break this down by cases, as $$ {{n}\choose{k}}_{-1}=\begin{cases} 0 &\qquad& \text{$n$ even, $k$ odd (Thm. 3.2);} \\ {{n/2}\choose{k/2}} &\qquad& \text{$n$ even, $k$ even (Thm. 3.1);} \\ {{\lfloor{n/2}\rfloor}\choose{\lfloor{k/2}\rfloor}} &\qquad& \text{$n$ odd (Thm. 3.1)}. \\ \end{cases} $$ The latter two cases can be combined into one: when $n$ is odd or $k$ is even, ${{n}\choose{k}}_{-1}$ is given by $${{\lfloor{n/2}\rfloor}\choose{\lfloor{k/2}\rfloor}}.$$ Theorem 3.1 of the paper expresses this slightly differently: as $$ \frac{n!!_E}{k!!_E (n-k)!!_E}, $$ where the notation $n!!_E$ is defined as the product of all even natural numbers $\le n$. But note that $n!!_E = 2^{\lfloor{n/2}\rfloor} (\lfloor{n/2}\rfloor)!$, and so $$ \frac{n!!_E}{k!!_E (n-k)!!_E}=\frac{2^{\lfloor{n/2}\rfloor} (\lfloor{n/2}\rfloor)!}{2^{\lfloor{(n-k)/2}\rfloor} (\lfloor{(n-k)/2}\rfloor)! 2^{\lfloor{k/2}\rfloor} (\lfloor{k/2}\rfloor)!}={{\lfloor{n/2}\rfloor}\choose{\lfloor{k/2}\rfloor}}, $$ where we used (twice) the fact that $\lfloor{n/2}\rfloor = \lfloor{k/2}\rfloor + \lfloor{(n-k)/2}\rfloor$ when $n$ is odd or $k$ is even. I would probably argue that the right-hand expression is simpler, since it doesn't require defining a new variant of the double factorial... but clearly the two are equivalent and the paper could be used as a reference for OP's expressions.