S_n^3 = \frac{\left(n+1\right)^4 - 6S_n^2 - 4S_n^1 - S_n^0 -1}{4} = \frac{\left(n+1\right)^4 - n\left(n+1\right)\left(2n+1\right) - 2n\left(n+1\right) - \left(n+1\right)}{4} = \frac{\left(n+1\right)\left(n^3+3n^2+3n+1-2n^2-n-2n-1\right)}{4} =