I found a notation this notation in this result:
(The reverse involution) If $x=\sum x_{k}\left(x_{k} \in \wedge^{k} E \right)$ is a multivector, we define $\tilde{x}=\sum(-1)^{k // 2} x_{k}(k / / 2=\lfloor q / 2\rfloor) .$ The linear map $\wedge E \rightarrow \wedge E, x \mapsto \tilde{x},$ is an involution that is called the reverse involution of $\wedge E$. Since it satisfies $\widehat{x \wedge y}=\tilde{y} \wedge \tilde{x}$ for all $x, y \in \wedge E,$ we say that it is an algebra antiautomorphism of $\wedge E.$
Proof The key point is that if $x=x_{1} \wedge \cdots \wedge x_{k}$ is a $k$ -blade, then $\tilde{x}=x_{k} \wedge \cdots \wedge x_{1}$. Indeed, since the exterior product is skew-symmetric, $x_{k} \wedge \cdots \wedge x_{1}=(-1)^{\left(\begin{array}{c}k \\ 2\end{array}\right)} x$ and $(-1)^{\left(\begin{array}{c}k \\ 2\end{array}\right)}=(-1)^{k // 2}$ because $\left(\begin{array}{c}k \\ 2\end{array}\right)$ has the same parity as $k / / 2$. With this, the proof is easily completed.
By googling the notation I found pragramming explanations of the same notation and not one in the context of algebra.
From what's written on the proof $(k / / 2=\lfloor q / 2\rfloor) $ what I guessed didn't make sense, I also thought there was a mistake because $k$ is a natural number, $q$ was noted as a metric or a bilinear form all along, I don't see why it would be the case, but if k//2 meant the floor of $k/2$ instead of $q/2$ then $\left(\begin{array}{c}k \\ 2\end{array}\right)$ would be of the same parity as the floor of the result of the division of $k$ by 2, I tried a few examples, it sounds to be the case , if so, how ?
Thank you.