$$U_{1103}(x, y) = \frac{x \left(1 - \sqrt{1 - 4 y}\right)^{3}}{8 y^{3}} + 1$$

$$T_{1103}(n, m, k) = \begin{cases}1&\text{if n=0,m=0} ,\ \\\frac{3 n {\binom{k}{n}} {\binom{2 m + 3 n - 1}{m}}}{m + 3 n} \end{cases} $$