Is there an explicit expression for Chebyshev polynomials modulo $x^r-1$?

There's a rapid algorithm to compute $T_n(x)$ modulo $(n,x^r-1)$. Note that $$ \pmatrix{T_n(x) \\ T_{n-1}(x)} = \pmatrix { 2x & -1 \\ 1&0} \pmatrix{T_{n-1}(x) \\ T_{n-2}(x)} = \pmatrix { 2x & -1 \\ 1&0}^{n-1} \pmatrix{ x\\ 1}. $$ Now you can compute these matrix powers all modulo $(n, x^{r}-1)$ rapidly by repeated squaring. Clearly $O(\log n)$ multiplications (of $2\times 2$ matrices) are required, and the matrices have entries that are polynomials of degree at most $r$ and coefficients bounded by $n$. So the complexity is a polynomial in $r$ and $\log n$.

The coefficient of $x^j$ in $(T_n(x)\bmod (x^r-1))$ equals the coefficient of $t^{n+r-j-1}$ in $$\frac{(1+t^2)^{r-j}}{2^{r-j}} \frac{((1+t^2)^{r-1}t - 2^{r-1}t^{r-1})}{((1+t^2)^r - 2^rt^r)}.$$ This coefficient can be explicitly computed as $$\sum_{k\geq 0} 2^{rk-r+j} \left( \binom{r-1-j-rk}{\frac{n+r-j-2-rk}{2}} - 2^{r-1}\binom{-j-rk}{\frac{n-j-rk}{2}}\right).$$ (here the binomial coefficients are zero whenever their lower indices are noninteger or negative)