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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 $2x2$ 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.

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  • $\begingroup$ Welcome to MSE. Please use MathJax to format both, your questions and your answers :) $\endgroup$ – mrtaurho Oct 15 '18 at 20:50
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    $\begingroup$ @mrtaurho thank you for this link, I'll start formatting it now. $\endgroup$ – user10176969 Oct 15 '18 at 20:51
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    $\begingroup$ Use $\cdot$ instead of $*$. $\endgroup$ – Shine On You Crazy Diamond Oct 15 '18 at 21:08
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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.

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  • $\begingroup$ Ah, that makes a lot of sense. So, it take it that instead of generating the primes, I am just making all numbers excluding $2^n$ as I am including it in all equations, hence all numbers end with 1 in binary. Thanks, this makes a lot more sense now! $\endgroup$ – user10176969 Oct 15 '18 at 21:00

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