Let's go full Rambo on this. We're going to prove by example with prime power pairs. So let's write \(m = p_1^{r_1}p_2^{r_2}\):
\[
q(m) = \sum_{n\le m} \chi_m(n) = \sum_{n \le p_1^{r_1}p_2^{r_2}} \chi_{p_1^{r_1}}(p^{\nu_{p_1}(n)})\chi_{p_2^{r_2}}(p^{\nu_{p_2}(n)}) [\frac{r_1}{\nu_{p_1}(n)} = \frac{r_2}{\nu_{p_2}(n)}]\\
\]
But we are only summing over \(0 \le t_1 < r_1\) and \(0 \le t_2 < r_2\) here. So we can break this into:
\[
\sum_{t_1 \le r_1} \sum_{t_2 \le r_2} \chi_{p_1^{r_1}}(p^{t_1})\chi_{p_2^{r_2}}(p^{t_2})[\frac{r_1}{t_1} = \frac{r_2}{t_2}]\\
\]
This is ALMOST actually a product sum:
\[
\left( \sum_{t_1 \le r_1} \chi_{p_1^{r_1}}(p^{t_1})\right) \left(\sum_{t_2 \le r_2}\chi_{p_2^{r_2}}(p^{t_2})\right)\\
\]
Where our solution is less than this sum, as the indicator will eliminate some of the values...
But each of these are just the divisor sum:
\[
\sigma(r_1) \sigma(r_2)\\
\]
Where, again, \(\sigma(r_1)\) is how many divisors \(r_1\) has...
\[
\sigma(r ) = \sum_{d \mid r} 1\\
\]
AND THERE YOU FUCKING HAVE IT!!! We can find a non trivial bound of the power indicator function!
\[
\begin{align}
m &= p_1^{r_1} p_2^{r_2}\cdots p_c^{r_c}\\
\sum_{n\le m} \chi_m(n) &\le \sigma(r_1)\sigma(r_2)\cdots\sigma(r_c) = \Pi(m)\\
\end{align}
\]
I KNEW I'D SEEN THIS BEFORE!!!!!!!!! The function \(\Pi(m)\) is actually pretty well understood, so no problem there. I believe it grows about \(\sqrt{m}\log(m)\) or something like that. So now when we write:
\[
g(m) = (-1)^{m+1}\sum_{n\le m} \frac{\log(n)^{\frac{\log m}{\log n}}}{\frac{\log m}{\log n}!} \chi_m(n)\\
\]
For :
\[
T(x,2) = \sum_{m=1}^\infty g(m) m^{-x}\\
\]
We can bound this much better. I just have to go over my Abel summation bound technique notes I have. I'm lazy, but we should expect something like \(g(m) \le \log(m) \Pi(m)\) right off the bat!!!!!! Which is a much better bound than I got initially. It's probably the best naive bound we can get.......
Abel's tricks with arithmetic functions/integrals/zeta functions will probably afford us much more. If you're anxious, I suggest looking for expressions of the form:
\[
A(m) = \sum_{n\le m} f(n) g(m/n)\\
\]
which is Abel's convolution of \(f\) and \(g\). Gottfried's weighted counting function is ALMOST of this form. I'm trying to reduce it to this form to ignore problems. The way I have reduced it is by creating:
\[
\Pi(m) = \prod_{j=1}^\infty \sigma(\nu_{p_j}(m))\\
\]
And showing this is a pretty fucking tight bound on the growth of \(q(m)\). It just ignores the Iverson bracket part of my deduction...
The cleaner version of a bound, which I can call off my head is a bit more difficult to describe. But we can get a better bound than \(g(m) = O(\Pi(m)\log(m))\). We can probably bound a good amount tighter..... To do this, we would have to look at:
\[
y(m) = \sum_{d \mid m} \frac{\log(d)^{m/d}}{m/d!}\\
\]
And guess how this grows. Which is just guessing how fast Gottfried's weights grow.... This should lower the \(\log(m)\) though, this definitely grows slower than \(\log(m)\)........ I think an ambitious guess would be something like \(\sqrt{\log(m)}\) or something like that...
To relate this to Gottfried's \(T(x,2)\) is actually pretty standard. If you can control the growth of the zeta function coefficients, you can control the growth of the zeta function itself. They are both related through a Fourier inversion....
\[
q(m) = \sum_{n\le m} \chi_m(n) = \sum_{n \le p_1^{r_1}p_2^{r_2}} \chi_{p_1^{r_1}}(p^{\nu_{p_1}(n)})\chi_{p_2^{r_2}}(p^{\nu_{p_2}(n)}) [\frac{r_1}{\nu_{p_1}(n)} = \frac{r_2}{\nu_{p_2}(n)}]\\
\]
But we are only summing over \(0 \le t_1 < r_1\) and \(0 \le t_2 < r_2\) here. So we can break this into:
\[
\sum_{t_1 \le r_1} \sum_{t_2 \le r_2} \chi_{p_1^{r_1}}(p^{t_1})\chi_{p_2^{r_2}}(p^{t_2})[\frac{r_1}{t_1} = \frac{r_2}{t_2}]\\
\]
This is ALMOST actually a product sum:
\[
\left( \sum_{t_1 \le r_1} \chi_{p_1^{r_1}}(p^{t_1})\right) \left(\sum_{t_2 \le r_2}\chi_{p_2^{r_2}}(p^{t_2})\right)\\
\]
Where our solution is less than this sum, as the indicator will eliminate some of the values...
But each of these are just the divisor sum:
\[
\sigma(r_1) \sigma(r_2)\\
\]
Where, again, \(\sigma(r_1)\) is how many divisors \(r_1\) has...
\[
\sigma(r ) = \sum_{d \mid r} 1\\
\]
AND THERE YOU FUCKING HAVE IT!!! We can find a non trivial bound of the power indicator function!
\[
\begin{align}
m &= p_1^{r_1} p_2^{r_2}\cdots p_c^{r_c}\\
\sum_{n\le m} \chi_m(n) &\le \sigma(r_1)\sigma(r_2)\cdots\sigma(r_c) = \Pi(m)\\
\end{align}
\]
I KNEW I'D SEEN THIS BEFORE!!!!!!!!! The function \(\Pi(m)\) is actually pretty well understood, so no problem there. I believe it grows about \(\sqrt{m}\log(m)\) or something like that. So now when we write:
\[
g(m) = (-1)^{m+1}\sum_{n\le m} \frac{\log(n)^{\frac{\log m}{\log n}}}{\frac{\log m}{\log n}!} \chi_m(n)\\
\]
For :
\[
T(x,2) = \sum_{m=1}^\infty g(m) m^{-x}\\
\]
We can bound this much better. I just have to go over my Abel summation bound technique notes I have. I'm lazy, but we should expect something like \(g(m) \le \log(m) \Pi(m)\) right off the bat!!!!!! Which is a much better bound than I got initially. It's probably the best naive bound we can get.......
Abel's tricks with arithmetic functions/integrals/zeta functions will probably afford us much more. If you're anxious, I suggest looking for expressions of the form:
\[
A(m) = \sum_{n\le m} f(n) g(m/n)\\
\]
which is Abel's convolution of \(f\) and \(g\). Gottfried's weighted counting function is ALMOST of this form. I'm trying to reduce it to this form to ignore problems. The way I have reduced it is by creating:
\[
\Pi(m) = \prod_{j=1}^\infty \sigma(\nu_{p_j}(m))\\
\]
And showing this is a pretty fucking tight bound on the growth of \(q(m)\). It just ignores the Iverson bracket part of my deduction...
The cleaner version of a bound, which I can call off my head is a bit more difficult to describe. But we can get a better bound than \(g(m) = O(\Pi(m)\log(m))\). We can probably bound a good amount tighter..... To do this, we would have to look at:
\[
y(m) = \sum_{d \mid m} \frac{\log(d)^{m/d}}{m/d!}\\
\]
And guess how this grows. Which is just guessing how fast Gottfried's weights grow.... This should lower the \(\log(m)\) though, this definitely grows slower than \(\log(m)\)........ I think an ambitious guess would be something like \(\sqrt{\log(m)}\) or something like that...
To relate this to Gottfried's \(T(x,2)\) is actually pretty standard. If you can control the growth of the zeta function coefficients, you can control the growth of the zeta function itself. They are both related through a Fourier inversion....
