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I'm an amateur mathematician and I found a property that I've never seen anyone mention before, I think I managed to demonstrate it below. I confess that I don't know if it's new or not, as I haven't found anything like it anywhere.

Theorem: Let $r > 0$, $c > 0$, and $n > 2$ be all integers, the sum $S_{n}$ of the first $n$ terms of an arithmetic progression that has $r*c$ as the first term and r as ratio, always results in a composite number.

Theorem 2: If a number $k$ is composite and is not a power of 2, then there are $c > 0$, $r > 0$, $n > 2$ all integers such that the sum of the first $n$ terms of an arithmetic progression that starts at $r*c$ is equal to $k$.

Proof of Theorem 1:

This proof seems easy to me since the formula for the sum of the terms of an AP itself can be separated as being the result of the multiplication of two integers:

$$S_{n}(r*c, n, r) = r*c + (r*c + r) + (r*c + 2*r) + ... + (r*c + (n-1)*r)$$ $$S_{n}(r*c, n, r) = \frac{n*(r*c + (r*c + r*c + (n-1)*r))}{2}$$ $$S_{n}(r*c, n, r) = \frac{n*r*(2*c + n - 1)}{2}$$

Note that the number resulting from this specific sum always results in a composite number, since $n > 2$ and therefore $(2*c + n - 1) > 1$ as well.

Proof of Theorem 2:

Suppose we have a positive composite number k that is different from a power of 2. By the Fundamental Theorem of Arithmetic:

$k = p_{1}*p_{2}*p_{3}*...*p_{n}$ , where all p's are primes in crescent order.

As the number is not a power of two, it therefore has an odd prime number that is smaller than or equal to the other odd factors that it may or may not have. Let's call this number $p_{m}$.

Then,

$$k = p_{1}*p_{2}*p_{3}*...*p_{m}*...*p_{n}$$ $$k = \frac{p_{m}*p_{1}*p_{2}*p_{3}*...*p_{m-1}*(2*(p_{m+1}*...*p_{n}))}{2}$$ $$k = \frac{p_{m}*p_{1}*p_{2}*p_{3}*...*p_{m-1}*(2*(p_{m+1}*...*p_{n}) - p_{m} + 1 + p_{m} - 1)}{2}$$ $$k = \frac{p_{m}*p_{1}*p_{2}*p_{3}*...*p_{m-1}*(2*\frac{(2*(p_{m+1}*...*p_{n}) - p_{m} + 1)}{2} + p_{m} - 1 )}{2}$$ $$k = S_{n}(\frac{2*(p_{m+1}*...*p_{n}) - p_{m} + 1)}{2} * p_{1}*p_{2}*p_{3}*...*p_{m-1}, p_{m}, p_{1}*p_{2}*p_{3}*...*p_{m-1})$$

I tried to figure something out with this property, but didn't have much success. I also tried to create factorers and primality tests, but they were relatively inefficient.

Do any of you see any potential uses for this property? (Or perhaps, an error in my demonstrations?) Thanks.

Bill Dubuque
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    Regarding your theorem 2, it is shown here expressing integers as sums of consecutive integers. – Arturo Magidin Mar 18 '24 at 01:05
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    "Is this useful" is not an on-topic question for this site. If you have a specific mathematical question regarding the above then you should ask that, else the question will likely be closed as off-topic (or a duplicate). – Bill Dubuque Mar 18 '24 at 02:51
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    An arithmetic sequence doesn't have a "ratio". It has a "common difference". – David K Mar 18 '24 at 04:32
  • You can get a vertically centred ellipsis with proper spacing to surrounding binary operators like $+$ using \cdots. – joriki Mar 18 '24 at 04:54

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