So when arguably having a lot of free time and try to enumerate the prime factorisation of the numbers 0-100
\begin{align}
& 0\\
& 10\\
& 01\\
& 0010\\
& 020\\
& 00010\\
& \vdots\\
& 0410\\
& LONG\\
& 010020\\
& 0020010\\
& 02020
\end{align}
where LONG means it is something of the form $000000…010$, which are the prime numbers and the numerical places are read from left to right as the exponents of $\{1,2,3,5,7,11,13,17,19,…\}$ in that order (assuming no typos for the larger numbers). Thus for example $0410$ reads $1^0\times 2^4 \times 3^1 \times 5^0 \times \cdots = 48$ and anything not written on the right are all zeros.
By checking every even number in the list, ($0,01,020,…$, corresponding to the numbers $0,2,4,…$) the following pattern is observed for the exponents of $2$, that is the value of the 2nd digit from the left:
$$0,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,6,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,4,1,2,…$$
There seemed to be apparent patterns on the spacing between two exponents of the same value such as any number whose exponent of $2$ is $2$ is exactly 4 even numbers away from each other, and every 2nd even number (counting $0$ as the 0th even number hence $2$ is the 1st even number, this means every odd-th even number) always have the exponent of $2$ to be $1$, and similarly for $3$ as well.
There is also an apparent pattern of $4$s that breaks down after the 3rd one
To further investigate, I use some prime number calculator and randomly choose a sequence of big even numbers for example $12000-12020$:
which gives the list:
\begin{align}
& 05130\\
& 010000010…10\\
& 010…10\\
& 012000000110\\
& 0300000010…10\\
& 01010…10\\
& 02101110\\
& 010…10\\
& 040…10\\
& 0110…10\\
& 02010…10
\end{align}
The entry $010…10$, which corresponds the number $12004$ showed the pattern of $2$ actually breaks down here before seemly picking up again at $12012$ ($02101110$). The pattern of $1$s however seemed to still holds as fine.
Unconvinced, I then try some ridiculously huge number such as $1214654365170$ and found the same pattern of 1s for something like $1214654365170 + 4n$ for $n=1,2,3,4$.
What theorem explains that every 2nd even number, the exponent of $2$ is only one, or is it really just an apparent pattern that will break down somewhere higher, similar to the pattern of $2$s?
Best Answer
You are looking at numbers of the form $2n$.
Looking at your enumeration, the most obvious thing is that every second number is divisible by $2$ only once. That this happens is clear, as these are the numbers where $n$ from above is odd.
The second pattern you note is that every $3+4k$th term is divisible by $2$ only twice, and these are the only numbers divisible by $2$ twice. In order to make this more clear, remove all the numbers of the form $2(2k+1)$ from the list to get
$$0,2,3,2,4,2,...$$
These $2$'s correspond to numbers of the form $2(2 (2k+1))=8k+4$, which gives you every number divisible by $2$ exactly twice.
If we strip these two obvious patterns, what you write is
$$3,4,3,5,3,4,3,6,3,4,3,4$$
and here there is an error, the last term should not be a $4$ but at $5$ (it corresponds to $96=2^5\cdot 3^1$).
So we actually find the same pattern as in the other two instances, just that everything got a $+2$.
What happens is this: You are removing from the sequence all odd terms and then you recover the same sequence but with a $+1$. Thus the pattern you see is an artefact of the odd numbers having no $2$ divisors and the rule "remove every second term and you receive the same list but with $+1$ to everything".