Ivic writes, at the beginning of chapter 13 of his The Riemann Zeta Function, about a method of expressing the principal terms of the Dirichlet Divisor Problem as polynomials of $log\\ n $ with coefficients built from arrangements of the Stieltjes Constants. He gives these coefficients for the first 3 divisor problems, mentions that general formulas exist for generating the coefficients for other divisor problems, and then moves on.

From further research, I've tracked down that this problem was explored in this paper by A.F. Lavrik.

Unfortunately, I'm having a really hard time following how these coefficients are generated. Is there anywhere online that has these polynomials elaborated more fully, or any other works that cover this topic? Or can someone explain this process to me more clearly?

Also, is there a way to translate the general approach here to versions of the divisor functions that exclude 1 and the number itself? (I believe Iwaniec, here, referred to this as the strict divisor function.) I know these two styles of divisor functions can be expressed in terms of each other combinatorially, but I'm curious if there are other issues to thinking about these different functions analytically...


So, here is the context for my fervent desire for more information about this. I apologize if what follows is too hand wavy - I'm trying to go fast here. Let me know if anything needs better clarification.

The summary is, different ways of approximating divisor problems might lead to interesting different ways of approximating the number of primes.

Yuri Linnik proved the following combinatorial identity

$\frac{\Lambda(n)}{log\\ n} = \sum\limits_{k=1} \frac{-1^{k+1}}{k}d_k'(n)$

where $\Lambda(n)$ is the Mangoldt function,

$d_1'(n) = 1$ if $\\ n > 1$


$d_k'(n) = \sum\limits_{1 < j < n, j|n} d_{k-1}'(j)$


$\Pi(n) = \pi(n) + \frac{1}{2}\pi(n^\frac{1}{2})+ \frac{1}{3}\pi(n^\frac{1}{3})+ \frac{1}{4}\pi(n^\frac{1}{4}) + ...$

with $\pi(n)$ the prime counting function, then, by summing up Linnik's identity from 1 to n (which takes us from a count of divisor function starting at 2 to its summatory equivalent), we have

$\Pi(n) = \displaystyle\sum\limits_{j=2}^n 1 - \frac{1}{2}\sum\limits_{j=2}^n\sum\limits_{k=2}^\frac{n}{j} 1 + \frac{1}{3}\sum\limits_{j=2}^n\sum\limits_{k=2}^\frac{n}{j}\sum\limits_{l=2}^\frac{n}{j k} 1- \frac{1}{4}\sum\limits_{j=2}^n\sum\limits_{k=2}^\frac{n}{j}\sum\limits_{l=2}^\frac{n}{j k}\sum\limits_{m=2}^\frac{n}{j k l} 1 \\ +\\ ...$

One way, but only one way, to approximate this series of sums is

$\displaystyle\int_1^n dx - \frac{1}{2}\int_1^n\int_1^\frac{n}{x}dy\\ dx + \frac{1}{3}\int_1^n\int_1^\frac{n}{x}\int_1^\frac{n}{x y} dz\\ dy\\ dx \\ - \frac{1}{4}\int_1^n\int_1^\frac{n}{x}\int_1^\frac{n}{x y}\int_1^\frac{n}{x y z} dw\\ dz\\ dy\\ dx \\ + ... $

which, when evaluated, comes out to

$n - 1 - \frac{1}{2}(n log n - n + 1) + \frac{1}{3}( \frac{1}{2}n(log\\ n)^2 - n log n + n - 1)$

$ \\ \\ \\ \\ \\ \\ \\ \\ - \frac{1}{4}( \frac{1}{6}n(log\\ n)^3 - \frac{1}{2}n(log\\ n)^2 + n log n - n + 1) + ... $

This sum appears to be exactly equal to

$li(n) - log\\ log\\ n - \gamma$

where $li(n)$ is the logarithmic integral and $\gamma$ is the Euler-Mascheroni constant.

As I say, though, this is only one way to approximate that sequence of divisor summatory functions.

It seems like relying on alternative ways of approximating these divisor summatory functions, like the work Lavrik is doing with their principal terms, could result in some other potentially interesting approximations to $\Pi(n)$, right?

Unfortunately, pursuing this line of inquiry outstrips my own uneven analytic abilities.


I assume you're asking how you can explicitly calculate the polynomials $P_{k-1}(x)$. The answer is in equation 13.4: $$ P_{k-1}(\log x) = \mathop{\rm Res}_{s=1} \big( x^{s-1} \zeta(s)^k s^{-1} \big). $$ To calculate this residue, expand everything as a Laurent series at $s=1$: $$ x^{s-1} \zeta(s)^k s^{-1} = \bigg( \sum _{j=0}^\infty \frac{((s-1)\log x)^j}{j!} \bigg) \bigg( \frac1{s-1} + \gamma + \sum _{j=1}^\infty \gamma_j (s-1)^j \bigg)^k \bigg( \sum _{j=0}^\infty (-1)^j (s-1)^j \bigg), $$ multiply it all out, and extract the coefficient of $(s-1)^{-1}$. Note that you only need to retain the power series to their $j=k-1$ terms to successfully calculate the final coefficient of $(s-1)^{-1}$. This is where the polynomial of degree $k-1$ in $\log x$ comes from.

  • $\begingroup$ Greg: First off, a huge thanks for taking the time to respond... This is the core of what I'm asking, and it really helps me. Does something like this work for the version of the divisor function that doesn't include 1 or n as divisors? Or do I have to use the standard divisor functions here and then use combinatorial identities to get the non-1, non-n version? For approximating the prime power function with Linnik's identity, that's the divisor function I need. Anyway, once again, a huge, huge, thanks. Back to work for me... $\endgroup$ – Nathan McKenzie Sep 8 '11 at 17:10
  • $\begingroup$ Let's see. If $f(n)$ is a function with corresponding Dirichlet series $F(s) = \sum _{n=1}^\infty f(n)n^{-s}$, and $g(n) = \sum _{d\mid n} f(d)$, then its D. series $G(s) = \sum _{n=1}^\infty g(n)n^{-s}$ satisfies $G(s) = \zeta(s) F(s)$. You want $g'(n) = \big( \sum _{d\mid n} f(d) \big) - f(1) - f(n)$, whose corresponding D. series thus satisfies $G'(s) = \zeta(s) \big( F(s) - f(1) \big) - F(s)$. If you start with $d_1'(n)$, then $D_1'(s) = \zeta(s)-1$; you can iterate this identity to compute each $D_k'(s)$ as a polynomial in $\zeta(s)$. Then insert that in place of $\zeta(s)^k$ in (13.4). $\endgroup$ – Greg Martin Sep 9 '11 at 7:25
  • $\begingroup$ Okay - I follow that, I think. Thanks again. $\endgroup$ – Nathan McKenzie Sep 10 '11 at 9:25
  • $\begingroup$ @GregMartin Hi, do you know if it is obvious that for $p$ prime and $n \le 2^p$ : $\mu(n) \equiv d_{p-1}(n) \bmod p$ and $\text{Mertens}(N) \equiv D_{p-1}(N) \bmod p$, or does it mean something ? $\endgroup$ – reuns Jul 24 '16 at 1:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.