For $$m, n > 0$$, let $$\alpha(m, n) = \int_0^2 t^m(1 + 3t)^n dt$$. If $$11\alpha(10, 6) + 18\alpha(11, 5) = p \cdot 14^6$$, then $$p$$ is equal to _______

We have $$\alpha(m, n) = \int_0^2 t^m(1 + 3t)^n \, dt$$.

Consider the derivative:

$$\frac{d}{dt}\left[t^{11}(1+3t)^6\right] = 11t^{10}(1+3t)^6 + t^{11} \cdot 18(1+3t)^5$$

Integrating both sides from 0 to 2:

$$\left[t^{11}(1+3t)^6\right]_0^2 = 11\int_0^2 t^{10}(1+3t)^6 \, dt + 18\int_0^2 t^{11}(1+3t)^5 \, dt$$

$$= 11\alpha(10, 6) + 18\alpha(11, 5)$$

Evaluating the left side:

$$2^{11}(1 + 6)^6 - 0 = 2^{11} \cdot 7^6 = 2048 \cdot 7^6$$

Given that $$11\alpha(10, 6) + 18\alpha(11, 5) = p \cdot 14^6 = p \cdot 2^6 \cdot 7^6$$:

$$2048 \cdot 7^6 = p \cdot 64 \cdot 7^6$$

$$p = \frac{2048}{64} = 32$$