$$\bigoplus_{k \ge 0} t^k Sym^{k} V = \frac{1}{(1−tV)(1−t^2 \mathbf{g})(1−t^3 V)(1−t^4)(1−t^4 V_2)},$$ where $V = \wedge^3 \mathbb{C}^6 = [0,0,1,0,0], V_2 = [0,1,0,1,0]$, and $\mathbf{g} = [1,0,0,0,1].$

See section 6 of "Series of Lie Groups" by Landsberg and Manivel.

$$\bigoplus_{k \ge 0} t^k Sym^{k} V = \frac{1}{(1−tV)(1−t^2 \mathbf{g})(1−t^3 V)(1−t^4)(1−t^4 V_2)},$$ where $V = \wedge^3 \mathbb{C}^6 = [0,0,1,0,0], V_2 = [0,1,0,1,0]$, and $\mathbf{g} = [1,0,0,0,1].$

See section 6 of "Series of Lie Groups" by Landsberg and Manivel.

$$\bigoplus_{k \ge 0} Sym^{\bullet} V = \frac{1}{(1−tV)^{−1}(1−t^2 \mathbf{g})^{−1}(1−t^3 V)^{−1}(1−t^4)^{−1}(1−t^4 V_2)^{−1}},$$ where $V = \wedge^3 \mathbb{C}^6 = [0,0,1,0,0], V_2 = [0,1,0,1,0]$, and $\mathbf{g} = [1,0,0,0,1].$

See section 6 of "Series of Lie Groups" by Landsberg and Manivel.

$$\bigoplus_{k \ge 0} t^k Sym^{k} V = \frac{1}{(1−tV)(1−t^2 \mathbf{g})(1−t^3 V)(1−t^4)(1−t^4 V_2)},$$ where $V = \wedge^3 \mathbb{C}^6 = [0,0,1,0,0], V_2 = [0,1,0,1,0]$, and $\mathbf{g} = [1,0,0,0,1].$

See section 6 of "Series of Lie Groups" by Landsberg and Manivel.