Number Theory - Differences of powers of 2 I'm not really familiar with number theory, or mathy stuff for that matter.
However I recently noticed that if you take the even powers of 2 ( $2^n$ where $n$ is even)
and write the difference of each two consecutive numbers ( $2^4 - 2^2$ which is 12),
and turn those differences into a sequence and repeat, like so:
1, 4, 16, 64,256
   3, 12, 48,192
       9, 36,144
          27,108
              81 (let's call this 'm')

the final numbers is a power of 3.
Why does this happen? Is there any way to prove this? Is it a proven thing?
 A: 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 $3$.
A: 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
$$(\mathbb E -1)^k a_0 = \sum_{i=1}^k \binom{k}{i} (-1)^{k-i} \mathbb E^i a_0 = \sum_{i=1}^k \binom{k}{i} (-1)^{k-i} a_i \\= \sum_{i=1}^k \binom{k}{i} (-1)^{k-i} 4^i = (4-1)^k$$
