Number Theory - Differences of powers of 2

Let $n=2k$ where $k$ is an integer. Then

$2^n - 2^{n-2} = 2^{2k} - 2^{2k-2} = 2^{2k-2}(4-1) = 3 \cdot 2^{2k-2}$. Thus taking a difference of consecutive even powers of $2$ yields a product of $3$ with the lower power of $2$ that you used.

If you take successive differences, you are introducing more $3$ factors and continuing to reduce the power of $2$ by two. By the time you've reached $0$ as the power on $2$, you've only accumulated powers of $3$.

The original sequence is $a_n = 2^{2n} = 4^n$. The first number in the second row is $4^1-4^0$, the first in the third row is $4^2-2\cdot 4^1 + 4^0$, etc. In general the first number of the $k^{th}$ row is

$$4^k - \binom{k}{1} 4^{k-1} + \binom{k}{2} 4^{k-2} - \cdots + \binom{k}{k} 4^0 (-1)^k = (4-1)^k=3^k$$ via the binomial theorem.

Using the forward shift operator it's more clear: the $k^{th}$ difference operator is just $$(\mathbb E -1)^k = \sum_{i=1}^k \binom{k}{i} (-1)^{k-i} \mathbb E^i $$ Therefore $$(\mathbb E -1)^k a_0 = \sum_{i=1}^k \binom{k}{i} (-1)^{k-i} \mathbb E^i a_0 = \sum_{i=1}^k \binom{k}{i} (-1)^{k-i} a_i \\= \sum_{i=1}^k \binom{k}{i} (-1)^{k-i} 4^i = (4-1)^k$$