Let $ g(n) = \frac{f(n)}{2^{n(n-1)}}$. Verify that $ g(n+1) = \frac{f(n+1)}{2^{n(n-1)} 4^{n}}$ $ g(n+2) = \frac{f(n+2)}{2^{n(n-1)} 16^{n+1}}$. Dividing the given by $ 2^{n(n-1)} 16^{n+1} = 2^{n^2 + 3n + 4}$ therefore gives $ g(n+2) = g(n+1) - g(n) + n \cdot 2^{-7n - 8}$. We now work $ \bmod 13$ (noting that $ g(n)$ is well-defined $ \bmod 13$ because $ 2$ is invertible). We can quickly compute that $ 2^6 \equiv -1 \implies 2^7 \equiv -2 \implies 2^{-7} \equiv 6 \bmod 13$ and $ 2^{-8} \equiv 3 \bmod 13$. Hence $ \bmod 13$ we can write $ g(n+2) - g(n+1) + g(n) = 3n \cdot 6^n \bmod 13$. The rest is computation (all done $ \bmod 13$). Write $ g(n) = h(n) + (an + b) 6^n$ where $ h(n+2) - h(n+1) + h(n) = 0$. Then $ (an + 2a + b) 36 - (an + a + b) 6 + (an + b) = 3n \implies$ $ 5an + (a + 5b) = 3n \implies$ $ a = 11, b = 3$. $ h(n)$ has characteristic polynomial $ \Phi_6(x)$ and so is periodic with period $ 6$. It remains to compute its six values, which is tedious.

orl wrote: Let $ f(0) = f(1) = 0$ and \[ f(n + 2) = 4^{n + 2} \cdot f(n + 1) - 16^{n + 1} \cdot f(n) + n \cdot 2^{n^2}, \quad n = 0, 1, 2, \ldots\] Show that the numbers $ f(1989), f(1990), f(1991)$ are divisible by $ 13.$ Wow ! I really like this solution Let $ g(n) = \frac {f(n)}{2^{n^2}} - \frac {1}{15^3} \cdot \frac{1}{2^{4(n-1)}}(15n+2)$ Then we have $ g(n + 2) = 2g(n + 1) - g(n)$, and hence $ g(n + 2),g(n + 1),g(n)$ form an arithmetic progression, and therefore $ g(0),g(1),g(2),..$ is an arithmetic progression. $ g(0) = -\frac{32}{15^3},g(1) = -\frac {17}{15^3}$, gives $ g(n) = \frac{1}{15^3}(15n-32)$. Hence: $ f(n) = \frac{1}{15^3} 2^{(n-2)^2} \left( 2^{4(n-1)}(15n-32) + (15n+2) \right )$. From here it is easy to prove that $ 13 \mid f(1989),f(1990),f(1991)$: $ 1989 \equiv 9 \bmod 12, 1989 \equiv 0 \bmod 13$ so $ f(1989) \equiv 15^{-3}2^{7^2}(2^{32} \cdot (-32) + 2) \equiv 0 \bmod 13$, since $ 2^{32} \equiv 2^8 \equiv 9 \bmod 13$,so $ 13 \mid f(1989)$ $ 1990 \equiv 10 \bmod 12, 1990 \equiv 1 \bmod 13$ so $ f(1990) \equiv 15^{-3}2^{8^2}(2^{36} \cdot (-17) + 17) \equiv 0 \bmod 13$, since $ 2^{36} \equiv 1 \bmod 13$, so $ 13 \mid f(1990)$. And $ f(1991) = 4^{1991}f(1990)-16^{1990}f(1989)+1989 \cdot 2^{1989^2} \equiv 0 \bmod 13$ since $ 13 \mid f(1989),f(1990),1989$, so $ 13 \mid f(1991)$. Hence the conclusion: $ 13 \mid f(1989),f(1990),f(1991)$ and we are done