For $q,a\in \mathbb N$ write $d=(q,a)$. Why might be $$\sum _{h|q}\frac {c_h(a)\log (d/h)}{h}=-\frac {q'}{\phi (q')}\sum _{h|q}\frac {c_h(a)}{h}\sum _{h'|q'}\frac {\mu (h')\log h'}{h'}+\sum _{h|d}\frac {\phi (q/h)}{q/h}\log h?$$ I can see that this is true for $d=q$ or $d=1$ but in general, don't know.

My motivation for the question is quite long - if you really want it see below. But my question as it stands is just that above.

Here $c_h(a)$ is Ramanujan's sum \[ c_h(a)=\sum _{n=1\atop {(n,h)=1}}^he^{2\pi ina/h}=\frac {\mu (h/(h,a))\phi (h)}{\phi (h/(h,a))}.\]

By https://www.impan.pl/en/publishing-house/journals-and-series/acta-arithmetica/all/168/4/82214/the-divisor-function-on-residue-classes-i (stated in the abstract, no access necessary) we have $$\sum _{n\leq X\atop {n\equiv a(q)}}d(n)\approx \frac {X}{q}\sum _{h|q}\frac {c_h(a)}{h}\Big (\log (X/h^2)+2\gamma -1\Big ).$$ On the other hand the main term is provided by a residue which I think should be, writing $d=(q,a)$ and $q'=q/d$, $$\sum _{n\leq X\atop {n\equiv a(q)}}d(n)\approx \frac {1}{\phi (q')}Res\left \{ \frac {(X/d)^{s}}{s}\sum _{n=1\atop {(n,q')=1}}^\infty \frac {d(nd)}{n^s}\right \} =\frac {1}{\phi (q')}Res\left \{ \frac {X^{s}}{s}\sum _{n=1\atop {(n,q)=d}}^\infty \frac {d(n)}{n^s}\right \} =:\frac {\mathcal R}{\phi (q')}$$ and as a plausability check it's possible to compare with Theorem 1 of http://matwbn.icm.edu.pl/ksiazki/aa/aa47/aa4713.pdf (there it's stated for $d_3(n)$ but I'm hoping it should also be the case for $d(n)$). So these two main terms should be equal, i.e. we should have $$\frac {\mathcal R}{\phi (q')}=\frac {X}{q}\sum _{h|q}\frac {c_h(a)}{h}\Big (\log (X/h^2)+2\gamma -1\Big )\hspace {10mm}(1)$$ and my issue is I can't see this. Here's my reasoning: As $$\sum _{n=1\atop {(n,q)=1}}^\infty \frac {1}{n^s}=\prod _{p\not |q}\left (1-p^{-s}\right )^{-1}=\zeta (s)\prod _{p|q}(1-p^{-s})=\zeta (s)\sum _{h|q}\frac {\mu (h)}{h^s}$$ the series in $\mathcal R$ is $$\sum _{n,m=1\atop {(nm,q)=d}}^\infty \frac {1}{(nm)^s}=\sum _{\delta |d}\sum _{n,m=1\atop {d|nm\atop {(nm,q)=d\atop {(n,d)=\delta }}}}^\infty \frac {1}{(nm)^s}=\sum _{\delta |d}\frac {1}{\delta ^s}\sum _{n,m=1\atop {d/\delta |m\atop {(n\delta m,q)=d\atop {(n,d/\delta )=1}}}}^\infty \frac {1}{(nm)^s}=\frac {1}{d^s}\sum _{\delta |d}\sum _{n,m=1\atop {(ndm,q)=d\atop {(n,d/\delta )=1}}}^\infty \frac {1}{(nm)^s}=\frac {1}{d^s}\left (\sum _{m=1\atop {(m,q')=1}}^\infty \frac {1}{m^s}\right )\left (\sum _{\delta |d}\sum _{n=1\atop {(n,q/\delta )=1}}^\infty \frac {1}{n^s}\right )=\frac {\zeta (s)^2}{d^s}\left (\sum _{h|q'}\frac {\mu (h)}{h^s}\right )\left (\sum _{\delta |d\atop {h|q/\delta }}\frac {\mu (h)}{h^s}\right )=:\frac {\zeta (s)^2}{d^s}\Delta (s)$$ so \[ \mathcal R=Res_{s=1}\Bigg \{ \frac {(X/d)^{s}}{s}\zeta (s)^2\Delta (s)\Bigg \} =:Res\Bigg \{ \lambda (s)\zeta (s)^2\Delta (s)\Bigg \} .\] And here I get stuck - I don't see how the $\log $'s in the residue can reduce to something so simple as in (1). Below is calculating the residue explicitly.

Here's my work in detail which gets it to the point above I'm stuck at. We have \begin{align} \mathcal R&=\lim _{s\rightarrow 1}\frac {d}{ds}\lambda (s)(s-1)^2\zeta (s)^2\Delta (s) \\ &=\lim _{s\rightarrow 1}\left (\Delta (s)\frac {d}{ds}\lambda (s)(s-1)^2\zeta (s)^2+\lambda (s)(s-1)^2\zeta (s)^2\frac {d}{ds}\Delta (s)\right ) \\ &=\Delta (1)\lim _{s\rightarrow 1}\left (\lambda (s)\frac {d}{ds}(s-1)^2\zeta (s)^2+(s-1)^2\zeta (s)^2\frac {d}{ds}\lambda (s)\right )+\lambda (1)\Delta '(1) \\ &=\Delta (1)\Big (2\gamma \lambda (1)+\lambda '(1)\Big )+\lambda (1)\Delta '(1) \\ &=\frac {X}{d}\Bigg (\left (\log (X/d)+2\gamma -1\right )\Delta (1)+\Delta '(1)\Bigg ). \end{align} Then \[ \sum _{\delta h|q\atop {\delta |d}}\frac {\mu (h)}{h}=\sum _{\delta |h|q\atop {\delta |d}}\frac {\mu (h/\delta )}{h/\delta }=\sum _{h|q}\frac {1}{h}\sum _{\delta |h,d}\mu (h/\delta )\delta =\sum _{h|q}\frac {c_h(a)}{h}\] and similarly \[ \sum _{\delta h|q\atop {\delta |d}}\frac {\mu (h)\log h}{h}=\sum _{h|q}\frac {1}{h}\sum _{\delta |h,d}\mu (h/\delta )\delta \log (h/\delta )=\sum _{h|q}\frac {c_h(a)\log h}{h}-\sum _{\delta |d\atop {\delta h|q}}\frac {\mu (h)\log \delta }{h}=\sum _{h|q}\frac {c_h(a)\log h}{h}-\sum _{\delta |d}\frac {\phi (q/\delta )}{q/\delta }\log \delta \] so \[ \Delta (1)=\frac {\phi (q')}{q'}\sum _{\delta h|q\atop {\delta |d}}\frac {\mu (h)}{h}=\frac {\phi (q')}{q'}\sum _{h|q}\frac {c_h(a)}{h}\] and \begin{align} \Delta '(1)&=\frac {d}{ds}\vline _{s=1}\left (\sum _{\delta |d\atop {h|q/\delta \atop {h'|q'}}}\frac {\mu (h)\mu (h')}{(hh')^s}\right ) \\ &=-\left (\sum _{\delta |d\atop {h|q/\delta }}\frac {\mu (h)\log h}{h}\right )\left (\sum _{h'|q'}\frac {\mu (h')}{h'}\right )-\left (\sum _{\delta |d\atop {h|q/\delta }}\frac {\mu (h)}{h}\right )\left (\sum _{h'|q'}\frac {\mu (h')\log h'}{h'}\right ) \\ &=-\frac {\phi (q')}{q'}\left (\sum _{h|q}\frac {c_h(a)\log h}{h}-\sum _{\delta |d}\frac {\phi (q/\delta )}{q/\delta }\log \delta \right )-\left (\sum _{h|q}\frac {c_h(a)}{h}\right )\left (\sum _{h'|q'}\frac {\mu (h')\log h'}{h'}\right ) \end{align} so \begin{align} \mathcal R=\frac {X}{d}\sum _{h|q}\frac {c_h(a)}{h}\left (\left (\log (X/d)+2\gamma -1\right )\frac {\phi (q')}{q'}-\frac {\phi (q')\log h}{q'}-\sum _{h'|q'}\frac {\mu (h')\log h'}{h'}\right ) \\ +\frac {X\phi (q')}{dq'}\sum _{\delta |d}\frac {\phi (q/\delta )}{q/\delta }\log \delta \end{align} so, comparing with (1) we should have \begin{align} \sum _{h|q}\frac {c_h(a)}{h}\Big (\log (X/h^2)+2\gamma -1\Big )&=\sum _{h|q}\frac {c_h(a)}{h} \\ &\left (\left (\log (X/d)+2\gamma -1\right )-\log h-\frac {q'}{\phi (q')}\left (\sum _{h'|q'}\frac {\mu (h')\log h'}{h'}\right )\right ) \\ &+\sum _{\delta |d}\frac {\phi (q/\delta )}{q/\delta }\log \delta \end{align} and therefore $$\sum _{h|q}\frac {c_h(a)\log (d/h)}{h}=-\frac {q'}{\phi (q')}\sum _{h|q}\frac {c_h(a)}{h}\sum _{h'|q'}\frac {\mu (h')\log h'}{h'}+\sum _{\delta |d}\frac {\phi (q/\delta )}{q/\delta }\log \delta $$ and that's the identity I want to establish at the start of this post.

  • $\begingroup$ I suggest that at the start of your post you define the unexplained notation $c_h(a)$, called Ramanujan's sum in some places and not worth giving any special name in others: $c_h(a) = {\rm Tr}_{\mathbf Q(\zeta_h)/\mathbf Q}(\zeta_h^a) = \mu(h/(a,h))\varphi(h)/\varphi(h/(a,h))$. $\endgroup$
    – KCd
    Dec 27, 2022 at 3:06
  • $\begingroup$ sure, have put it in $\endgroup$
    – tomos
    Dec 27, 2022 at 9:25
  • $\begingroup$ You left out a denominator $h$ in the exponent in your definition of $c_h(a)$. $\endgroup$
    – KCd
    Dec 27, 2022 at 9:44
  • $\begingroup$ cheers, corrected $\endgroup$
    – tomos
    Dec 27, 2022 at 9:54

1 Answer 1


I feel I'm answering my own questions too much here but I guess I should post the answer if I posted the question and found the answer in the meantime.

Write $\Delta (n)=\phi (n)/n$ and write $\Delta =1-1/p$. Let $$R(q,d)=\sum _{h|q}\frac {c_h(d)}{h}\hspace {6mm}\text { and }\hspace {6mm}S(q,d)=\sum _{h|d}\Delta (q'h).$$ Then $$\sum _{h|q}\frac {c_h(d)\log h}{h}=R(q,d)\left (-\sum _{p|q'}\frac {\log p}{p\Delta }+\sum _{p^\alpha |d\atop {p|q'}}\frac {\log p^\alpha }{1+D}+\sum _{p^\alpha |d\atop {p\nmid q'}}\frac {\log p^\alpha \Delta }{1+\Delta D}\right )$$ and $$\sum _{h|d}\Delta (q'h)\log h=R(q,d)\left (\sum _{p^\alpha |d,q'}\frac {\log p^\alpha }{1+D}+\sum _{p^\alpha |d\atop {p\nmid q'}}\frac {\log p^\alpha \Delta }{1+\Delta D}\right ).$$ The claim in my question then follows from these claims and this question How to prove $\sum_{d\mid q}\frac{\mu(d)\log d}{d}=-\frac{\phi(q)}{q}\sum_{p\mid q}\frac{\log p}{p-1}$? Proof The first sum in question is \begin{align} \sum _{p^\alpha |q}\frac {\log p^\alpha c_{p^\alpha }(d)}{p^\alpha }\sum _{h|q\atop {p\nmid h}}\frac {c_h(d)}{h}=:\sum _{p^\alpha |q}\frac {\log p^\alpha c_{p^\alpha }(d)}{p^\alpha }R_p(q,d). \end{align} Then \begin{align} R_p(q,d)&=&\prod _{P|q\atop {P\nmid d\atop {P\not =p}}}\sum _{h|q}\frac {c_h(d)}{h}\prod _{P|d,q'\atop {P\not =p}}\sum _{h|q}\frac {c_h(d)}{h}\prod _{P|d\atop {P\nmid q'\atop {P\not =p}}}\sum _{h|q}\frac {c_h(d)}{h} \\ &=&\prod _{P|q\atop {P\nmid d\atop {P\not =p}}}\Delta \prod _{P|d,q'\atop {P\not =p}}\left (\Delta (1+D)\right )\prod _{P|d\atop {P\nmid q'\atop {P\not =p}}}\left (1+\Delta D\right ) \\ &=&R(q,d)\underbrace {\Delta ^{-1}}_{p|q'}\underbrace {(1+D)^{-1}}_{p|d,q'}\underbrace {(1+\Delta D)^{-1}}_{p|d\atop {p\not |q'}} \end{align} so our sum has the correct factor and the rest becomes \begin{align} &&\sum _{p^\alpha |q}\frac {\log p^\alpha c_{p^\alpha }(d)}{p^\alpha }\underbrace {\Delta ^{-1}}_{p|q'}\underbrace {(1+D)^{-1}}_{p|d,q'}\underbrace {(1+\Delta D)^{-1}}_{p|d\atop {p\not |q'}} \\ &&\hspace {10mm}=\hspace {4mm}\sum _{p^\alpha |q\atop {p\not |d}}\frac {\log p^\alpha \mu (p^\alpha )}{p^\alpha \Delta }+\sum _{p^\alpha |q\atop {p|d,q'}}\frac {\log p^\alpha c_{p^\alpha }(d)}{p^\alpha \Delta (1+D)}+\sum _{p^\alpha |q\atop {p|d\atop {p\not |q'}}}\frac {\log p^\alpha c_{p^\alpha }(d)}{p^\alpha (1+\Delta D)} \\ &&\hspace {10mm}=\hspace {4mm}-\sum _{p|q\atop {p\not |d}}\frac {\log p}{p\Delta }+\sum _{p^\alpha |d\atop {p|d,q'}}\frac {\log p^\alpha }{1+D}-\sum _{p|d,q'}\frac {\log p^{D+1}}{p\Delta (1+D)}+\sum _{p^\alpha |d\atop {p\not |q'}}\frac {\log p^\alpha \Delta }{1+\Delta D} \\ &&\hspace {10mm}=\hspace {4mm}-\sum _{p|q'}\frac {\log p}{p\Delta }+\sum _{p^\alpha |d\atop {p|q'}}\frac {\log p^\alpha }{1+D}+\sum _{p^\alpha |d\atop {p\nmid q'}}\frac {\log p^\alpha \Delta }{1+\Delta D} \end{align} The second sum in question is \begin{align} \sum _{p^\alpha |d}\log p^\alpha \sum _{h|d\atop {p\nmid h}}\Delta (q'hp^\alpha )=:\sum _{p^\alpha |d}\log p^\alpha S_p(q,d). \end{align} Then \begin{align} S_p(q,d)&=&\prod _{P|q\atop {P\nmid d\atop {P\not =p}}}\Delta (q')\prod _{P|d,q'\atop {P\not =p}}\sum _{h|d}\Delta (q'h)\prod _{P|d\atop {P\nmid q'\atop {P\not =p}}}\sum _{h|d}\Delta (h)\cdot \Delta (q'P) \\ &=&R(q,d)\underbrace {\Delta ^{-1}}_{p|q'}\underbrace {(1+D)^{-1}}_{p|d,q'}\underbrace {(1+\Delta D)^{-1}}_{p|d\atop {p\nmid q'}}\cdot \Delta \end{align} so our sum has the right factor and the rest is \begin{align} &&\sum _{p^\alpha |d}\log p^\alpha \underbrace {\Delta ^{-1}}_{p|q'}\underbrace {(1+D)^{-1}}_{p|d,q'}\underbrace {(1+\Delta D)^{-1}}_{p|d\atop {p\nmid q'}}\cdot \Delta \\ &&\hspace {10mm}=\hspace {4mm}\sum _{p^\alpha |d,q'}\frac {\log p^\alpha }{1+D}+\sum _{p^\alpha |d\atop {p\nmid q'}}\frac {\log p^\alpha \Delta }{1+\Delta D} \end{align}


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