$$U_{1236}(x, y) = \frac{\left(1 - \sqrt{1 - 4 y}\right)^{2} \left(\frac{x \left(1 - \sqrt{1 - 4 y}\right)^{2}}{4 y^{2}} + 1\right)^{3}}{4 y^{2}}$$

$$T_{1236}(n, m, k) = \frac{\left(2 k + 2 n\right) {\binom{3 k}{n}} {\binom{2 k + 2 m + 2 n - 1}{m}}}{2 k + m + 2 n}$$