Formulas for the Spinor Representation Product Decompositions $2^{[frac{N-1}{2}]} otimes 2^{[frac{N-1}{2}]}=?$ and ...

We know that given the dimension $N$, we can construct the corresponding spinors for the $Spin(N)$ group (which has $Spin(N)/mathbb{Z}_2=SO(N)$ so $Spin(N)$ is a double cocver of the spatial rotational group $SO(N)$).

Now, for a general $N$, I believe there is a
$$2^{[frac{N-1}{2}]}$$-dimensional irreducible Spinor representation (irrep) of the $Spin(N)$ group. Here $[frac{N-1}{2}]$ is the step function takes the maximum integer value but smaller than $frac{N-1}{2}$. For example, $N=2n$ even, we have
$$2^{[frac{N-1}{2}]}=2^{[frac{2n-1}{2}]}=2^{{n}-1}=2^{frac{N}{2}-1}.
$$
For example, $N=2n+1$ odd, we have
$$2^{[frac{N-1}{2}]}=2^{n}=2^{frac{N-1}{2}}.
$$

My question is that are there general simple formulas that can decompose the product of the spinor representations $2^{[frac{N-1}{2}]}$ and their complex conjugation representations $overline{2^{[frac{N-1}{2}]}}$, say for the even $N$:
$$
2^{[frac{N-1}{2}]} otimes 2^{[frac{N-1}{2}]} = N oplus dots ?
$$
$$
2^{[frac{N-1}{2}]} otimes overline{2^{[frac{N-1}{2}]}} = 1 oplus frac{N(N-1)}{2} oplus dots?
$$
For the odd $N$:
$$
2^{[frac{N-1}{2}]} otimes 2^{[frac{N-1}{2}]} =
1 oplus N oplus frac{N(N-1)}{2} oplus dots ?
$$

For example, when $N=3$, we have:
$$
2 otimes 2 =2 otimes overline{2} = 1 oplus 3.
$$
When $N=4$, we have:
$$
2 otimes 2 = 4.
$$
$$
2 otimes overline{2} = 1 oplus 3.
$$
For example, when $N=5$,
$$
4 otimes 4 = 1 oplus 5 oplus 10.
$$
$N=6$, we have:
$$
4 otimes 4 = 6 oplus 10.
$$
$$
4 otimes overline{4} = 1 oplus 15.
$$
$N=7$:
$$
8 otimes 8 = 1 oplus 7 oplus 21 oplus 35.
$$

Do we have a formula for general $N$?

Ref/s are welcome!