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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?

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    $\begingroup$ The 1s are even numbers which are not divisible by 4, and the 2s are the multiples of 4 which are not multiples of 8. In fact, you must have made an error in calculating the factorization of 12004, since it is also divisible by 4, so it should have a 2 instead of a 1. $\endgroup$ – Rahul Oct 28 '18 at 13:12
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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".

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  • $\begingroup$ Also you can notice that the first time a number appears, it is the center of a palindrome (ie if you go left or right you get the same sequence). This is due to such numbers being of the form $2^n$ and $2^n+k$ having the same divisibility by $2$ property as $2^n-k$ provided $k<2^n$. $\endgroup$ – s.harp Nov 9 '18 at 10:38

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