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!