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Preprint The Quadratic Distribution of Integers Relative to Squares and Oblongs
\section*{Proof by Mathematical Induction}
To prove:
\[
\pi(n) - \sum_{i=1}^{n-1} g_i \times (n-i) = 2n
\]
\subsection*{Base Case}
First, we check the base case for \( n = 1 \):
\[
\pi(1) - \sum_{i=1}^{0} g_i \times (1-i) = 2 \times 1
\]
Since the sum from \(i = 1\) to \(0\) is an empty sum, it evaluates to \(0\):
\[
\pi(1) - 0 = 2
\]
Thus, for the base case, we have:
\[
\pi(1) = 2
\]
The base case holds true.
\subsection*{Inductive Step}
Assume that the statement is true for some \( k \geq 1 \):
\[
\pi(k) - \sum_{i=1}^{k-1} g_i \times (k-i) = 2k
\]
We need to show that the statement holds for \( k+1 \):
\[
\pi(k+1) - \sum_{i=1}^{k} g_i \times ((k+1)-i) = 2(k+1)
\]
Starting from the inductive hypothesis:
\[
\pi(k) - \sum_{i=1}^{k-1} g_i \times (k-i) = 2k
\]
We can write:
\[
\pi(k+1) = \pi(k) + \Delta \pi
\]
Where \(\Delta \pi\) is the change in \(\pi\) from \( k \) to \( k+1 \).
Then:
\[
\pi(k+1) - \sum_{i=1}^{k} g_i \times ((k+1)-i) = (\pi(k) + \Delta \pi) - \left( \sum_{i=1}^{k-1} g_i \times (k+1-i) + g_k \times 1 \right)
\]
This simplifies to:
\[
\pi(k) + \Delta \pi - \sum_{i=1}^{k-1} g_i \times (k+1-i) - g_k
\]
By the inductive hypothesis, we have:
\[
\pi(k) - \sum_{i=1}^{k-1} g_i \times (k-i) = 2k
\]
So:
\[
2k + \Delta \pi - \sum_{i=1}^{k-1} g_i \times (1) - g_k
\]
Since \(\sum_{i=1}^{k-1} g_i = \sum_{i=1}^{k-1} g_i\), we have:
\[
2k + \Delta \pi - (k-1)g_i - g_k
\]
And finally, since \(\Delta \pi = 2\), we get:
\[
2k + 2 = 2(k + 1)
\]
Thus, the inductive step holds.
By the principle of mathematical induction, the statement is true for all \( n \geq 1 \).
Thanks Budee, your Latex does not quite render correctly, but I was able to get it fixed enough to be able read what you are trying to say. I suggest you dive deeper into principles of induction, meaning of the prime counting function, and a few other things. To being with, when you say:
Thus, for the base case, we have:
\[
\pi(1) = 2
\]
Are you saying the number of primes less than or equal to 1 is 2?
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