# 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?

• You'll notice that every number along the left hand side of your pyramid is a power of 3. – Duncan Ramage Jan 7 at 23:18
• Try it with powers of 5 across the top :-) – Joffan Jan 7 at 23:19
• wow this was fast! Thanks. – kasra Jan 7 at 23:24
• Nice game. Thank you! – Gottfried Helms Jan 8 at 0:31

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 $$(\mathbb E -1)^k a_0 = \sum_{i=1}^k \binom{k}{i} (-1)^{k-i} \mathbb E^i a_0 = \sum_{i=1}^k \binom{k}{i} (-1)^{k-i} a_i \\= \sum_{i=1}^k \binom{k}{i} (-1)^{k-i} 4^i = (4-1)^k$$