7
$\begingroup$

Lehmer 1988 and Keiper 1992 made major progress on evaluating the series:

$$\sigma_r = \sum_{k=1}^{\infty} \left( \frac{1}{\rho_k^r} + \frac{1}{(1-\rho_k)^r}\right) \quad r \in \mathbb{N}$$

where $\rho_k$ is the $k$-th non-trivial zero of the Riemann $\zeta$-function.

The function can be expressed in terms of Stieltjes-constants $\gamma_x$:

$\sigma_1= 1 - \frac12\ln(4\pi) + \frac{\gamma}{2}$

$\sigma_2= 1 - \,\frac34\zeta(2) \,\,\,\,+ \gamma^2+\frac{2}{1}\gamma^0\gamma_1$

$\sigma_3= 1 - \,\frac78\zeta(3) \,\,\,\,+ \gamma^3+\frac{3}{1}\gamma^1\gamma_1+ \frac32\gamma^0\gamma_2 $

$\sigma_4= 1 - \frac{15}{16}\zeta(4) \,\,\,+ \gamma^4+\frac{4}{1}\gamma^2\gamma_1+\frac{4}{2}\gamma^1\gamma_2+\frac46\gamma^0\gamma_3 \qquad\qquad\ \ + \frac{4}{2}\gamma^0\gamma_1^2$

$\sigma_5= 1 - \frac{31}{32}\zeta(5) \,\,\,+ \gamma^5+ \frac{5}{1}\gamma^3\gamma_1+\frac{5}{2}\gamma^2\gamma_2+\frac{5}{6}\gamma^1\gamma_3+\frac{5}{24}\gamma^0\gamma_4 \,\,+ \frac{5}{1}\gamma^1\gamma_1^2+\frac{5}{2}\gamma^0\gamma_1\gamma_2$

$...$

$\displaystyle\sigma_r=1-\left(1-\frac{1}{2^r} \right )\zeta(r)+\frac{\gamma\,\gamma_{r-2}}{\Gamma(r-1)}+\frac{r\,\gamma_{r-1}}{\Gamma(r)}-\sum_{j=1}^{r-2}\frac{\gamma_{j-1}}{\Gamma(j)}\,\left( 1-\left(1-\frac{1}{2^{r-j}}\right)\zeta(r-j)-\sigma_{r-j}\right)$

a recurrence relation valid for $r>1$.

Let's define: $\sigma_r^* = \sigma_r-1+\left(1-\frac{1}{2^r} \right )\zeta(r)$, i.e. excluding the first two terms of $\sigma_r$ that represent the pole at $s=1$ and the trivial zeros of $\zeta(s)$, so that $\sigma_r^*$ only has terms with Stieltjes-constants. For a fixed $r$, the number of these terms turns out to be equal to the partition number $p_r$ (verified up to $r=30$).

My first thought was that this would be a simple consequence of the similarity in recurrence relations that both functions possess, however these turn out to be quite different.

Q1: What could be a logical explanation for this phenomenon?

ADDED 1:

Numerical evidence up to $r=999$, suggests that:

  1. all $\sigma_r^*$ are positive (contrary to $\sigma_r$ and the Stieltjes constants);
  2. $\displaystyle\lim_{r\rightarrow \infty} \frac{\sigma_r^*}{\sigma_{r-1}^*}=\frac13$.

Q2: Are these two observations trivial? Could they be proven?

ADDED 2 & 3:

A few additional attributes of $\sigma_r^*$ that I found:

$\qquad \displaystyle (s-1)\,\zeta(s) = \prod_{r=1}^\infty \exp\left(-\frac{\sigma_r^*}{r}\,(1-s)^r\right)$, with $\sigma_1^*=\gamma$.

or stated differently:

$\qquad \displaystyle \log\big((s-1)\,\zeta(s)\big) = -\sum_{r=1}^\infty \frac{\sigma_r^*}{r}\,(1-s)^r$

which gives the logarithmic derivative:

$\qquad \displaystyle \frac{\zeta'(s)}{\zeta(s)} = \sum_{r=0}^\infty \sigma_r^*\,(1-s)^{r-1}$, with $\sigma_0^*= 1$

all with a certain radius of convergence.

Also found that:

$\qquad \displaystyle \sum_{r=0}^\infty \sigma_r^* = \log(2\,\pi)$

and that the sum over the non-trivial and the trivial zeros:

$\qquad z(r)=\displaystyle \sum_{k=1}^{\infty} \left( \frac{1}{\rho_k^r} + \frac{1}{(1-\rho_k)^r}\right)+\sum_{k=1}^{\infty} \frac{1}{(-2k)^r} = \sigma_r + \frac{1}{(-2)^r}\zeta(r)$

yields for r = even:

$\qquad \sigma_r^* = z(r) -1 + \eta(r)$

and for r = odd:

$\qquad \sigma_r^* = z(r) -1 + \zeta(r)$

where $\eta(s)$ is the Dirichlet $\eta$-function.

P.S.: list of $\sigma_r$ for $r=1..999$.

$\endgroup$
4
  • $\begingroup$ If you write $\sigma_r$ in terms of $\delta_j:=\frac{\gamma_{j-1}}{\Gamma(j)}$ and put $\delta_1:=\gamma$, the recursion formula should simply become $\sigma^*_r= r \delta_{r}+ \delta_1\delta_{r-1}+\sum_{j=1}^{r-2}\delta_j\sigma^*_{r-j} $, thus positive integer coefficients for each partition of $r$. Maybe writing $\frac42$ in $\sigma_4$ in my edit was misleading... $\endgroup$
    – Wolfgang
    Nov 3, 2020 at 9:36
  • $\begingroup$ If we put $\sigma^*_1:=\delta_1= \gamma$ (instead of $\frac{\gamma }{2}$), we just have $\sigma^*_r= r \delta_{r}+ \sum_{j=1}^{r-\color{red}1}\delta_j\sigma^*_{r-j} $. Further note that it's all "homogenous of height $r$", whence the natural appearance of the partitions of $r$. $\endgroup$
    – Wolfgang
    Nov 3, 2020 at 9:50
  • $\begingroup$ @wolfgang Merci, this is a nice simplification. I had also found that $\sigma^*_0= (\gamma^0)=1$ and $\sigma^*_1=\gamma^1$. The attributes of $\sigma_r^*$ seem more 'elegant' than those of $\sigma_r$ (simpler recurrence formula only using Stieltjes constants, a conjectured lower bound of $\frac{\gamma}{3^r}$, $\sum_{r=0}^\infty \sigma_r^* = \log(2\,\pi) $). Since $\sigma_r$ can be related directly to the Keiper/Li-constants $\lambda_r$, whose non-negativity implies the RH (Li's-criterion), I also wonder whether the conjectured non-negativity of $\sigma_r^*$ would have the same consequence. $\endgroup$
    – Agno
    Nov 3, 2020 at 12:15
  • $\begingroup$ The following program in Mathematica 8.0.1 also gives the partition numbers as the number of terms in a expansion: Clear[a, f, s, n, nn]; nn = 20; a = CoefficientList[Series[1/f[x], {x, 0, nn}], x]; Table[Length[Expand[a[[n]]]], {n, 1, Length[a]}] $\endgroup$ Jan 3, 2021 at 19:46

0

Your Answer

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

Browse other questions tagged or ask your own question.