Well, I'm bored again, so let's write some more about this, as this is an interesting problem. Let's see what happens for co-prime pairs in a certain manner.
So let \(\chi_m(n) = 1\) if there exists a \(k\) where \(m = n^k\), and zero other wise. We want to decompose this function into a simpler form than this. A good way to do this is to work with prime powers initially; in which we set \(m=p^r\) for some \(p\), then:
\[
\chi_{m}(n) = 1\,\,\text{if and only if}\,\, \log(n)/p \mid r\\
\]
Where \(a \mid b\) means \(a\) divides \(b\).
So for prime powers we are given:
\[
\sum_{n\le p^r} \chi_{m}(n) = \sum_{d \mid r} 1 = \sigma( r)\\
\]
Where \(\sigma\) is the divisor function. Then Gottfried's counting function for prime powers, is precisely:
\[
\sum_{n\le p^r} \frac{\log(n)^{r \log(p)/\log n}}{(r \log(p)/\log n !)}\chi_{m}(n) = \sum_{d \mid r} \frac{\log(d)^{\frac{r}{d}}}{\frac{r}{d}!}\\
\]
Now let's see what happens when we have two primes, rather than one. So let's write \(m = p_1^{r_1} p_2^{r_2}\). And let's see if we can relate this to the single prime case.
\[
\chi_{m}(n) = 1\,\,\text{if and only if}\,\, n = p_1^{t_1} p_2^{t_2}\,\,\text{where}\,\,kt_1 = r_1\,\,kt_2 = r_2\\
\]
This is a little trickier, because now we are testing two ratios, and seeing if they equal. To simplify this discussion, I am going to introduce the \(\nu\) character. This can be defined as:
\[
\begin{align}
m &= p_1^{a_1}p_2^{a_2} \cdots p_c^{a_c}\\
\nu_{p_j}(m) &= a_j
\end{align}
\]
This states that, in the Fundamental theorem of arithmetic--in the decomposition of \(m\), \(\nu_p\) counts how many times the prime \(p\) occurs. Nothing more nothing less. And from this we can write:
\[
\chi_{p_1^{r_1}p_2^{r_2}}(n) = 1\,\,\text{if and only if}\,\, \frac{r_1}{\nu_{p_1}(n)} = \frac{r_2}{\nu_{p_2}(n)}\,\,\text{and}\,\, \nu_{p_1}(n) \mid r_1,\,\,\nu_{p_2}(n)\mid r_2\\
\]
Now the average mathematician might gawk that this is useless and pointless and looks ugly as fuck. I don't like that line of reasoning. It just means we are looking at it wrong. Instead. we are going to look at the expression:
\[
A(n) = \chi_{p_1^{r_1}}(p_1^{\nu_{p_1}(n)}) \chi_{p_2^{r_2}}(p_2^{\nu_{p_2}(n)})[\frac{r_1}{\nu_{p_1}(n)} = \frac{r_2}{\nu_{p_2}(n)}] \\
\]
Where \([a=b]\) is an Iverson bracket which produces \(1\) if true, and \(0\) if false...
We want to prove that \(A(n) = \chi_{p_1^{r_1}p_2^{r_2}}(n)\). We can do this by doing the first implication, by proving that \(A(n) = 1\) if \(n^k = p_1^{r_1}p_2^{r_2}\) for some \(k\). So take such an \(n\). Then since we've projected into the first prime number, and we know that \(\nu_{p_1}(n) k = r_1\) we have a \(1\). Multiply this by the \(1\) you get for the projection into \(p_2\). And we get \(1\). This proves the implication one way. To prove the opposite direction we have to pay close attention to \(k\). Suppose that \(A(n) = 1\), but \(\chi_{m}(n) = 0\). (Literally checking for false posiives). For this we'll note that the first part of our expression equals \(1\) if \(k_1 \nu_{p_1}(n) = r_1\) and \(k_2 \nu_{p_2}(n) = r_2\) where \(k_1 \neq k_2\). Which is the entrance of the Iverson bracket; which tests if the exponents are the same.
So........................................................
To deconstruct the power function:
\[
\begin{align}
m &= p_1^{r_1} p_2^{r_2}\cdots p_c^{r_c}\\
\chi_m(n) &= \prod_{j=1}^c \chi_{p_j^{r_j}}(p_j^{\nu_{p_j}(n)}) \cdot [\frac{r_1}{\nu_{p_1}(n)} = \frac{r_2}{\nu_{p_2}(n)} = ... = \frac{r_c}{\nu_{p_c}(n)}]
\end{align}
\]
Now this may seem stupid to talk about on a tetration forum, but now, we can take the sum:
\[
\sum_{n\le m} \chi_m(n)\\
\]
Which much much much more finesse. Especially because we know the prime power sum is just the divisor function.
I'm going to read more Abel summation shit, because I haven't even really introduced this yet. I'm just fuzzy on some things I remember. Also, in this instance the Iverson bracket is an arithmetical function. You can encode it using basic divisor sums! So the Iverson bracket looks ugly, but it's actually entirely plain mathematical operations to find it. A benefit of the bracket on natural numbers. We can test the equality of two natural numbers by summing numbers! (Fourier/orthogonality relationships abound).
So the root of Gottfried's problem \(\chi_m(n)\), is almost a multiplicative function, upto this stupid Iverson Bracket, lol.
I'll post more as I work through it. I have a sneaking suspicion this zeta function should have an Euler expansion--or something like it. I have to refresh myself on decomposing Euler sums.
******
The Iverson bracket in our case can be written as:
\[
[a = b] = \frac{1}{m} \sum_{j=1}^m e^{2 \pi i j \frac{a-b}{m}}\\
\]
Since all the terms in our bracket are less than \(m\). And if \(a-b\) = 0, then we are summing \(\sum_{j=1}^m 1\). Otherwise, we are summing all the m roots of unity, which is famously the value zero. The function:
\[
[a_1 = a_2 = a_3 =...=a_c]\\
\]
Is a permutation problem; and we get:
\[
\frac{1}{m^c}\prod_{i < l \le c} \left(\sum_{j=1}^m e^{2 \pi i j \frac{a_{i}-a_{l}}{m}}\right)\\
\]
So let \(\chi_m(n) = 1\) if there exists a \(k\) where \(m = n^k\), and zero other wise. We want to decompose this function into a simpler form than this. A good way to do this is to work with prime powers initially; in which we set \(m=p^r\) for some \(p\), then:
\[
\chi_{m}(n) = 1\,\,\text{if and only if}\,\, \log(n)/p \mid r\\
\]
Where \(a \mid b\) means \(a\) divides \(b\).
So for prime powers we are given:
\[
\sum_{n\le p^r} \chi_{m}(n) = \sum_{d \mid r} 1 = \sigma( r)\\
\]
Where \(\sigma\) is the divisor function. Then Gottfried's counting function for prime powers, is precisely:
\[
\sum_{n\le p^r} \frac{\log(n)^{r \log(p)/\log n}}{(r \log(p)/\log n !)}\chi_{m}(n) = \sum_{d \mid r} \frac{\log(d)^{\frac{r}{d}}}{\frac{r}{d}!}\\
\]
Now let's see what happens when we have two primes, rather than one. So let's write \(m = p_1^{r_1} p_2^{r_2}\). And let's see if we can relate this to the single prime case.
\[
\chi_{m}(n) = 1\,\,\text{if and only if}\,\, n = p_1^{t_1} p_2^{t_2}\,\,\text{where}\,\,kt_1 = r_1\,\,kt_2 = r_2\\
\]
This is a little trickier, because now we are testing two ratios, and seeing if they equal. To simplify this discussion, I am going to introduce the \(\nu\) character. This can be defined as:
\[
\begin{align}
m &= p_1^{a_1}p_2^{a_2} \cdots p_c^{a_c}\\
\nu_{p_j}(m) &= a_j
\end{align}
\]
This states that, in the Fundamental theorem of arithmetic--in the decomposition of \(m\), \(\nu_p\) counts how many times the prime \(p\) occurs. Nothing more nothing less. And from this we can write:
\[
\chi_{p_1^{r_1}p_2^{r_2}}(n) = 1\,\,\text{if and only if}\,\, \frac{r_1}{\nu_{p_1}(n)} = \frac{r_2}{\nu_{p_2}(n)}\,\,\text{and}\,\, \nu_{p_1}(n) \mid r_1,\,\,\nu_{p_2}(n)\mid r_2\\
\]
Now the average mathematician might gawk that this is useless and pointless and looks ugly as fuck. I don't like that line of reasoning. It just means we are looking at it wrong. Instead. we are going to look at the expression:
\[
A(n) = \chi_{p_1^{r_1}}(p_1^{\nu_{p_1}(n)}) \chi_{p_2^{r_2}}(p_2^{\nu_{p_2}(n)})[\frac{r_1}{\nu_{p_1}(n)} = \frac{r_2}{\nu_{p_2}(n)}] \\
\]
Where \([a=b]\) is an Iverson bracket which produces \(1\) if true, and \(0\) if false...
We want to prove that \(A(n) = \chi_{p_1^{r_1}p_2^{r_2}}(n)\). We can do this by doing the first implication, by proving that \(A(n) = 1\) if \(n^k = p_1^{r_1}p_2^{r_2}\) for some \(k\). So take such an \(n\). Then since we've projected into the first prime number, and we know that \(\nu_{p_1}(n) k = r_1\) we have a \(1\). Multiply this by the \(1\) you get for the projection into \(p_2\). And we get \(1\). This proves the implication one way. To prove the opposite direction we have to pay close attention to \(k\). Suppose that \(A(n) = 1\), but \(\chi_{m}(n) = 0\). (Literally checking for false posiives). For this we'll note that the first part of our expression equals \(1\) if \(k_1 \nu_{p_1}(n) = r_1\) and \(k_2 \nu_{p_2}(n) = r_2\) where \(k_1 \neq k_2\). Which is the entrance of the Iverson bracket; which tests if the exponents are the same.
So........................................................
To deconstruct the power function:
\[
\begin{align}
m &= p_1^{r_1} p_2^{r_2}\cdots p_c^{r_c}\\
\chi_m(n) &= \prod_{j=1}^c \chi_{p_j^{r_j}}(p_j^{\nu_{p_j}(n)}) \cdot [\frac{r_1}{\nu_{p_1}(n)} = \frac{r_2}{\nu_{p_2}(n)} = ... = \frac{r_c}{\nu_{p_c}(n)}]
\end{align}
\]
Now this may seem stupid to talk about on a tetration forum, but now, we can take the sum:
\[
\sum_{n\le m} \chi_m(n)\\
\]
Which much much much more finesse. Especially because we know the prime power sum is just the divisor function.
I'm going to read more Abel summation shit, because I haven't even really introduced this yet. I'm just fuzzy on some things I remember. Also, in this instance the Iverson bracket is an arithmetical function. You can encode it using basic divisor sums! So the Iverson bracket looks ugly, but it's actually entirely plain mathematical operations to find it. A benefit of the bracket on natural numbers. We can test the equality of two natural numbers by summing numbers! (Fourier/orthogonality relationships abound).
So the root of Gottfried's problem \(\chi_m(n)\), is almost a multiplicative function, upto this stupid Iverson Bracket, lol.
I'll post more as I work through it. I have a sneaking suspicion this zeta function should have an Euler expansion--or something like it. I have to refresh myself on decomposing Euler sums.
******
The Iverson bracket in our case can be written as:
\[
[a = b] = \frac{1}{m} \sum_{j=1}^m e^{2 \pi i j \frac{a-b}{m}}\\
\]
Since all the terms in our bracket are less than \(m\). And if \(a-b\) = 0, then we are summing \(\sum_{j=1}^m 1\). Otherwise, we are summing all the m roots of unity, which is famously the value zero. The function:
\[
[a_1 = a_2 = a_3 =...=a_c]\\
\]
Is a permutation problem; and we get:
\[
\frac{1}{m^c}\prod_{i < l \le c} \left(\sum_{j=1}^m e^{2 \pi i j \frac{a_{i}-a_{l}}{m}}\right)\\
\]
