Answer is given by the coefficient of $z^n$ in $$(z+z^2+\dots+z^6)^k = \left(\frac{1-z^6}{1-z}\right)^k left(z\frac{1-z^6}{1-z}\right)^k = z^k (1-z^6)^k(1-z)^{-k}.$$ An explicit formula for this coefficient is: $$\sum_{i=0}^{\lfloor n/6\rfloor} \sum_{i=0}^{\min(k,\lfloor (n-k)/6\rfloor)} (-1)^{n+i} \binom{k}{i} \binom{-k}{n-6i} binom{-k}{n-k-6i} = \sum_{i=0}^{\lfloor n/6\rfloor} sum_{i=0}^{\min(k,\lfloor (n-k)/6\rfloor)} (-1)^i \binom{k}{i} \binom{k+n-6i-1}{k-1}.$$binom{n-6i-1}{k-1}.$$1 Answer is given by the coefficient of z^n in$$(z+z^2+\dots+z^6)^k = \left(\frac{1-z^6}{1-z}\right)^k = (1-z^6)^k(1-z)^{-k}.$$An explicit formula for this coefficient is:$$\sum_{i=0}^{\lfloor n/6\rfloor} (-1)^{n+i} \binom{k}{i} \binom{-k}{n-6i} = \sum_{i=0}^{\lfloor n/6\rfloor} (-1)^i \binom{k}{i} \binom{k+n-6i-1}{k-1}.