How do we show that

$$\prod\limits_{k=1}^{n} \tau(k) = 2^{n (\log \log n + C) + \phi(n)},$$

where $\tau(k)$ is the number of divisors of $k$, the constant $C$ is given by $$C = \gamma + \sum_{\nu = 2}^{\infty} \left\{ \log_2 \left(1 + \frac{1}{\nu}\right) - \frac{1}{\nu} \right\} \left(\sum\limits_{p \textrm{ prime }} \frac{1}{p^{\nu}} \right),$$ and $$\begin{eqnarray*} \frac{\phi(n)}{n} = \frac{\gamma -1}{\log n} + \frac{1!}{(\log n)^2} (\gamma + \gamma_1 -1 ) + \frac{2!}{(\log n)^3} (\gamma + \gamma_1 + \gamma_2 - 1) + \ldots + \frac{(r-1)!}{(\log n)^r} (\gamma + \gamma_1 + \gamma_2 + \ldots + \gamma_{r-1} - 1) + O \left\{\frac{1}{(\log n)^{r+1}} \right\}? \end{eqnarray*}$$

Reference: equation number (10) in the link.


1 Answer 1


The proof relies on the following identity, which is straightforward to verify: $$ \sum_{k=1}^\infty\frac{\log_2\tau(k)}{k^s}=\zeta(s)\sum_p\sum_{\nu=1}^\infty\frac{\log_2\left(1+\frac{1}{\nu}\right)}{p^{\nu s}},\qquad\Re s>1.$$ Here $p$ runs through the prime numbers. This implies $$ \sum_{k=1}^\infty\frac{\log_2\tau(k)}{k^s}=\zeta(s)\{\log\zeta(s)+g(s)\},\qquad\Re s>1, $$ where $$g(s):=\sum_p\sum_{\nu=2}^\infty\frac{\log_2\left(1+\frac{1}{\nu}\right)-\frac{1}{\nu}}{p^{\nu s}}$$ is analytic in the half-plane $\Re s>1/2$. From here we can proceed as in Section 3 of Diaconis (1976). The result follows upon noting that $C=\gamma+g(1)$.

  • $\begingroup$ (+1) many thanks!! I need some time to read through Diaconis' paper .. $\endgroup$
    – r9m
    Jun 9, 2015 at 16:22
  • 1
    $\begingroup$ @r9m: Diaconis decomposes the genarating function into two parts: here $\zeta(s)\log\zeta(s)$ and $\zeta(s)g(s)$. For both parts, the deformed contour is decomposed into 7 parts, of which 6 parts contribute to the error term only. The paper is very nicely written, I did not know about it before (nor did I know about the online Ramanujan archive). So I thank you, too! $\endgroup$
    – GH from MO
    Jun 9, 2015 at 16:25

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