I'm curious about this pattern that I saw while adding many powers of two together, and then taking the prime factorisation of each result, and I'm curious as to why this occurs.
Here is the pattern:
$2^n$ will have the prime factors of $2^n$
For example, $4$ has $2×2$ as prime factors.
$2^n + 2^{(n+1)}$ will have prime factors of $2^n* 3$
For example, $12$ $(4+8)$ has $2x2x3$ as prime factors.
$2^n + 2^{(n+2)}$ will have prime factors of $2^n* 5$
For example, $20$ $(4+16)$ has $2x2x5$ as prime factors.
$2^n + 2^{(n+1)} + 2^{(n+2)}$ will have prime factors of $2^n* 7$
But this is where it changes:
$2^n + 2^{(n+3)}$ will have prime factors of $2^n* 3^2$
For example, $36$ $(4+32)$ has $2x2x3x3$
$2^n + 2^{(n+1)} + 2^{(n+3)}$ will then return to prime factors of $2^n* 11$
Iterating through every combination of powers we can sum up, I have found that we will get every prime number excluding $2$ (as it would factor into $2^n$) and all possible combinations of prime numbers excluding $2$ like so in ascending order for the prime factors:
$3, 5, 7, 3*3, 11, 13, 3*5, 17, 19, 3*7, 23, 5*5, 3*3*3, 29…$
$2$ and combinations with two are of course excluded because it would just be a prime factor of $2^n$.
A similar sequence begins with powers of $3$:
$2*2, 5, 7, 2*2*2, 11, 13, 2*2*2*2, 17…$
Removing all the numbers that sometimes disappear that are larger than the base m in m^n:
$2, 3, 5, 7, 11, 13, 17, 23, 29…$
We have the prime numbers.
I am curious as to how and why this works, and if it can be generalised.
Best Answer
Note that $$2^n+2^{n+1}=3\cdot 2^n\quad \quad 3=11_2\\ 2^n+2^{n+2}=5\cdot 2^n\quad \quad 5=101_2\\2^n+2^{n+1}+2^{n+2}=7\cdot 2^n\quad \quad 7=111_2$$ You can express your odd prime in binary and multiply by $2^n$ to get the pattern you are seeing.