Divisible by $54$

I will start by reducing the divisibility problem's difficulty. Notice that $54 = 2 \times 27 = 2 \times 3^3$ . And $2^{2n+1} - 9n^2 + 3n - 2 \equiv 0 - 1 + 1 - 0 \equiv 0 \mod 2$ . This means that the only difficult part is proving the numbers are always divisible by $27$ . Later I will reduce this to divisibility by $9$ but for that we need a little algebra: $2^{2n+1} - 9n^2 + 3n - 2 = 2^{2n+1} - 2 - 9n^2 + 3n = 2(2^{2n} - 1) - 9n^2 + 3n = 2(4^n - 1) - 9n^2 + 3n = 2 \times 3 \times \sum_{k=0}^{n-1} 4^k - 9n^2 + 3n = \\ 3 \left( 2 \sum_{k=0}^{n-1} 4^k - 3n^2 + n \right)$ .

It is clear that this number if divisible by 27 if and only if $2 \sum_{k=0}^{n-1} 4^k - 3n^2 + n$ is divisible by 9. To prove that this number is indeed divisible by $9$ I will find a formula for $\sum_{k=0}^{n-1} 4^k \mod 9$ . Notice that:

$\sum_{k=0}^{n-1} 4^k= \sum_{k=0}^{n-1} (3+1)^k = \sum_{k=0}^{n-1} \sum_{j=0}^k \binom{k}{j} 3^j$ . And we can actually write this as: $1 + 4 + \sum_{k=2}^{n-1} \sum_{j=0}^k \binom{k}{j} 3^j = 5 + \sum_{k=2}^{n-1} \sum_{j=0}^k \binom{k}{j} 3^j \equiv 5 + \sum_{k=2}^{n-1} \sum_{j=0}^2 \binom{k}{j} 3^j \equiv 5 + \sum_{k=2}^{n-1} \left( 1 + 3k \right) \equiv 5 + \frac{3n^2}{2} - \frac{n}{2} - 5 \equiv \frac{3n^2}{2} - \frac{n}{2} \mod 9$

So to conclude: $2 \sum_{k=0}^{n-1} 4^k - 3n^2 + n \equiv 3n^2 - n - 3n^2 + n \equiv 0 \mod 9$