# s=s... again?

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we remember that $\sum_{n=1}^\infty\dfrac{Ch(n)}{n^s}=-2\zeta(2s)+\zeta(s)$ consider $Cs(n)=(Ch*1)(n)=\sum_{d|n}Ch(d)$ so, if d is a perfect square it is -1, 1 otherwise. if there are k perfect square divisors, there are $$\tau(n)-k$$ non-square divisor. add them and it is $$\tau(n)-2k$$.

let k be a function of n. It is multiplicative. So we can compute for prime powers $k(p^a)=\sum_{d^2|p^a} d^{0}=1^0+(p^2)^0+(p^4)^0+.... \left(p^{2\lfloor \dfrac{a}{2}\rfloor}\right)^0=\lfloor \dfrac{a}{2}\rfloor+1$ we have $\sum_{n=1}^\infty\dfrac{Ch(n)}{n^s}\sum_{n=1}^\infty\dfrac{1}{n^s}=\sum_{n=1}^\infty\dfrac{\tau(n)}{n^s}-2\sum_{n=1}^\infty\dfrac{k(n)}{n^s}$ $-2\zeta(s)\zeta(2s)+\zeta^2(s)=\zeta^2(s)-2\sum_{n=1}^\infty\dfrac{k(n)}{n^s}$ $\sum_{n=1}^\infty\dfrac{k(n)}{n^s}=\zeta(s)\zeta(2s)$ Lets try to compute the summation using the fact it is multpicative $\prod_{p=prime} \sum_{a≥0}\dfrac{k(p^a)}{p^{as}}=\prod_{p=prime} \sum_{a≥0}\dfrac{\lfloor \dfrac{a}{2}\rfloor+1}{p^{as}}=\prod_{p=prime} \dfrac{p^{3s}}{(p^s-1)^2(p^s+1)}\\=\prod_p \dfrac{1}{(1-p^{-s})^2}\dfrac{1-p^ {-s}}{1-p^{-2s}}=\zeta^2(s)\dfrac{\zeta(2s)}{\zeta(s)}=\zeta(s)\zeta(2s)$ so we get $\boxed{\large{s=s}}$ again!!!

Note by Aareyan Manzoor
2 years, 4 months ago

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The combo part was not found by me but @Andrew Ellinor so he gets the credit for that.

He was not sure wether or not he was correct but this proves that $k(n)=\prod_{i=1}^{\omega(n)} (\lfloor \dfrac{k_i}{2}\rfloor +1)$ if $$n=p_1^{k_1}p_2^{k_2}.....p_{\omega(n)}^{k_{\omega(n)}}$$

- 2 years, 4 months ago

Well he is correct.

- 2 years, 3 months ago