4t-2j-1}.$$ By Theorem \[thm-z-3\] (iii), we have $$|C(v,S_0\cup S_1)|=2^{4m-2}+(-1)^m2^{2m-1}.$$

It completes the proof.

Note that, by Lemma \[lem-z-7\] (a), $|S_0\cup S_1|=\sum_{i=1}^m{4m\choose 4i}+\sum_{i=0}^m{4m\choose 4i+1}=\sum_{i=0}^m\left({4m\choose 4i}+{4m\choose 4i+1}\right)-1=\sum_{i=0}^m{4m+1\choose 4i+1}-1=2^{4m-1}+(-1)^m2^{2m-1}-1$. Combining Lemmas \[lem-z-9\], \[lem-z-10\] and Theorem \[cor-z-3\], we have the following result.

\[thm-z-5\] The orbit Cayley graph $Cay(Z_2^{4m},S_0\cup S_1)$ is strongly regular with parameter $$\left(2^{4m},2^{4m-1}+(-1)^m2^{2m-1}-1,2^{4m-2}+(-1)^m2^{2m-1}-2,2^{4m-2}+(-1)^m2^{2m-1}\right)$$ for any positive integer $m$.

Note that the complement of a non-trivial $(n,r,\lambda,\mu)$-strongly regular graph is an $(n,n-1-r,n-2-2r+\mu,n-2r+\lambda)$-strongly regular [@Godsil]. Since $\overline{Cay(Z_2^{4m},S_0\cup S_1)}=Cay(Z_2^{4m},S_2\cup S_3)$, we get the following result.

\[cor-z-4\] The orbit Cayley graph $Cay(Z_2^{4m},S_2\cup S_3)$ is strongly regular with parameter $$\left(2^{4m},2^{4m-1}-(-1)^m2^{2m-1},2^{4m-2}-(-1)^m2^{2m-1},2^{4m-2}-(-1)^m2^{2m-1}\right)$$ for any positive integer $m$.

By similar methods, we consider the orbit Cayley graphs $Cay(Z_2^{4m+2},S_0\cup S_1)$ and $Cay(Z_2^{4m+2},S_0\cup S_3)$.

\[lem-z-11\] Let $v$ be an element of $Z_2^{4m+2}$. If $v\in S_0\cup S_1$, then $$|C(v,S_0\cup S_1)|=2^{4m}+(-1)^m2^{2m}-2.$$

As similar as the proof of Lemma \[lem-z-9\], if $v\in S_0$, one can verify that there are $\sum_{t=0}^m\sum_{j=0}^{2k}{4k\choose 2j}{4m-4k+2\choose 4t-2j}-2$ pairs of $(x,y)\in C(v,S_0\cup S_1)$ such that $x,y\in S_0$ and there are $\sum_{t=0}^m\sum_{j=0}^{2k}{4k\choose 2j}{4m-4k+2\choose 4t+1-2j}$ pairs of $(x,y)\in C(v,S_0\cup S_1)$ such that $x,y\in S_1$. Therefore, from Theorem (\[thm-z-3\]) (iv), we have $$\begin{array}{lll}
|C(v,S_0\cup S_1)|&=&\displaystyle\sum_{t=0}^m\sum_{j=0}^{2k}{4k\choose 2j}{4m-4k+2\choose 4t-2j}+\sum_{t=0}^m\sum_{j=0}^{2k}{4k\choose 2j}{4m-4k+2\choose 4t+1-2j}-2\\
&=&\displaystyle\sum_{t=0}^m\sum_{j=0}^{2k}{4k\choose 2j}{4m-4k+3\choose 4t-2j+1}-2\\
&=&2^{4m}+(-1)^m2^{2m}-2.\end{array}$$

If $v\in S_1$, there are $\sum_{t=0}^m\sum_{j=0}^{2k}{4k+1\choose 2j}{4m-4k+1\choose 4t-2j}-1$ pairs of $(x,y)\in 