Let $a_n = a_1,a_2,a_3,...$ be a periodic sequence and $T$ be its period.
Then the sequence $b_n$ with $b_{2n-1} = a_{(2m+2(n-1)+1)^2+2m+4(n-1)+3}$ for $n = 1,2,3,...$ and $b_{2n} = a_{(2m+2(n-1)+1)^2+6m+4(n-1)+4}$ for $n = 1,2,3,...$ and some fixed $m \in \mathbb{Z}_{\geq 0}$ should also be periodic with a period of $2T$, right?
Since you have since asked another question here (after several days of trying to solving the same problem) I'll answer your original question in full ot present a full picture for anyone who may have a similar question:
First, note that for any $T$ we have
$$(x+T)^n=\sum_{k=0}^n \binom{n}{k}x^{n-k}T^k= x^n+\sum_{k=1}^n \binom{n}{k}x^{n-k}T^k$$
$$=x^n+T\sum_{k=1}^n \binom{n}{k}x^{n-k}T^{k-1}\equiv x^n\ (\text{mod }T)$$
Then any polynomial $P(x)$ has the property
$$P(x+T)=\sum_{k=0}^n a_k(x+T)^k\equiv \sum_{k=0}^n a_kx^k=P(x)\ (\text{mod }T)$$
Now, suppose that $a_n$ is periodic with some period $T$ and consider the sequence $r_n=a_{P(n)}$. We have that
$$r_{n+T}=a_{P(n+T)}=a_{P(n)+qT}=a_{P(n)}=r_n$$
which implies that $r(n)$ has a period of $T$ (not neccessarily a minimum period of $T$). The key step here is noticing that $P(n+T)=P(n)+qT$ for some integer $q$. Taking your specific question, we have shown that both $b_{2n}$ and $b_{2n-1}$ are both periodic with a (not necessarily the minimum) period of $T$. But then for even $n=2k$ and odd $n=2k-1$ respectively
$$b_{n+2T}=b_{2k+2T}=b_{2(k+T)}=b_{2k}=b_n$$
$$b_{n+2T}=b_{2k-1+2T}=b_{2(k+T)-1}=b_{2k-1}=b_n$$
We conclude that $b_n$ has a period of $2T$ although again we stress that this may not be the minimum period.