5

Let $0<a<m$ be integers with $\gcd(a,m)=1$. For $s\in\mathbb{C}$ with $\Re s>1$, define $$\zeta(m,a;s)=\prod_{\substack{p\text{ is prime}\\p\equiv a\pmod m}}(1-p^{-s})^{-1}.$$

I'm looking for an efficient (arbitrary precision) computation of $\zeta(m,a;s)$. Assume that we have such a computation of Dirichet $L$-functions.

This paper by H. Cohen (which motivated me the most) covers various sums and products over primes, as well as Dirichlet $L$-functions and more, but not the above.

I found a satisfactory solution for some specific cases.

The simplest are $m\in\{3,4,6\}$, with a single nontrivial character $\chi$ modulo $m$. Let \begin{align*}f_1(s)&=\log\zeta(3,1;s),&g_1(s)&=\log[(1-3^{-s})\zeta(s)],\\f_2(s)&=\log\zeta(3,2;s),&g_2(s)&=\log L(s,\chi),\end{align*} then $g_1(s)=f_1(s)+f_2(s)$ and $g_2(s)=f_1(s)+f_2(2s)-f_2(s)$. This gives $$f_2(s)=\sum_{n=0}^\infty 2^{-n-1}g(2^n s),\qquad g(s)=g_1(s)-g_2(s),$$ a series that converges very quickly (its term is $O(2^{-2^n\sigma})$ with $\sigma=\Re s$). Thus, we have an efficient (enough) way to compute $f_2(s)$ and then $f_1(s)=g_1(s)-f_2(s)$.

The case $m=8$ is just a bit harder. Take the following table of characters:

$a$ $1$ $3$ $5$ $7$
$\chi_1(a)$ $1$ $1$ $1$ $1$
$\chi_2(a)$ $1$ $1$ $-1$ $-1$
$\chi_3(a)$ $1$ $-1$ $-1$ $1$
$\chi_4(a)$ $1$ $-1$ $1$ $-1$

Then, if we put $g_k(s)=\log L(s,\chi_k)$, we obtain (for $a=3$, and similarly for other values) $$\log\zeta(8,3;s)=\sum_{n=0}^\infty 2^{-n-2}g(2^n s),\qquad g(s)=g_1(s)+g_2(s)-g_3(s)-g_4(s).$$

But already the case $m=5$ gets me confused completely.

metamorphy
  • 43,591
  • $\log\prod_{p\equiv a\bmod m} (1-p^{-s})^{-1}$ is a series of $\log L(ks,\chi)$ with $\chi\bmod m$ but the coefficients might be a bit messy – reuns Apr 19 '22 at 22:05
  • 1
    I get $\sum_p \chi(p) p^{-s}= \sum_{k\ge 1} \frac{\mu(k)}{k} \log L(ks,\chi^k)$ then your function is $\exp(\sum_{k\ge 1} \frac1k\sum_{p\equiv a\bmod m} p^{-sk}) = \exp(\sum_{k\ge 1} \frac1k\sum_p p^{-sk}\frac1{\phi(m)}\sum_{\chi \bmod m}\overline{\chi}(a)\chi(p))$ – reuns Apr 25 '23 at 05:44

1 Answers1

1

A partial (and a bit disappointing) answer following the comment by @reuns.

Things look good if $a\equiv\pm1\pmod m$ and not as good otherwise. We get $$ \varphi(m)\log\zeta(m,a;s)=\sum_{\chi\bmod m}\overline{\chi}(a)F(s,\chi), \\ F(s,\chi)=\sum_{n>0}\frac1n\sum_{d\ |\ n}\mu(d)\log L(ns,\chi^d). $$ The inner sum can be simplified, but let's feed it to PARI/GP as is:

experiment(m, a, n) = {
    my (g = znstar(m, 1));
    my (w = [Mod(x, polcyclo(g.no)), g.no]);
    my (t = matrix(m, n));
    for (i = 1, m, if (gcd(i, m) == 1,
        my (c = 1 / chareval(g, i, a, w));
        for (j = 1, n, fordiv(j, d,
            my (k = lift(Mod(i, m) ^ d));
            t[k, j] += c * moebius(d)
        ))
    ));
    my (r = matrix(g.no, n), j = 0);
    for (i = 1, m, if (gcd(i, m) == 1,
        j += 1; r[j,] = lift(t[i,])
    ));
    return (r)
};

This (ugly) script computes the coefficients $c_{n,\chi}$ (assuming $m,a$ fixed) in $$ \varphi(m)\log\zeta(m,a;s)=\sum_{n>0}\frac1n\sum_{\chi\bmod m}c_{n,\chi}\log L(ns,\chi) $$ For $m=5$, and the characters as given here, experiment(5, 1, 16) yields

$\chi$ $c_{1,\chi}$ $c_{2,\chi}$ $c_{4,\chi}$ $c_{8,\chi}$ $c_{16,\chi}$
$\chi_{5,1}$ $1$ $-1$ $0$ $-1$ $0$ $0$ $0$ $-1$ $0$ $0$ $0$ $0$ $0$ $0$ $0$ $-1$
$\chi_{5,2}$ $1$ $1$ $0$ $1$ $0$ $0$ $0$ $1$ $0$ $0$ $0$ $0$ $0$ $0$ $0$ $1$
$\chi_{5,3}$ $1$ $1$ $0$ $1$ $0$ $0$ $0$ $1$ $0$ $0$ $0$ $0$ $0$ $0$ $0$ $1$
$\chi_{5,4}$ $1$ $-1$ $0$ $-1$ $0$ $0$ $0$ $-1$ $0$ $0$ $0$ $0$ $0$ $0$ $0$ $-1$

and experiment(5, 4, 16) yields

$\chi$ $c_{1,\chi}$ $c_{2,\chi}$ $c_{4,\chi}$ $c_{8,\chi}$ $c_{16,\chi}$
$\chi_{5,1}$ $1$ $-1$ $0$ $-1$ $0$ $0$ $0$ $-1$ $0$ $0$ $0$ $0$ $0$ $0$ $0$ $-1$
$\chi_{5,2}$ $-1$ $-1$ $0$ $-1$ $0$ $0$ $0$ $-1$ $0$ $0$ $0$ $0$ $0$ $0$ $0$ $-1$
$\chi_{5,3}$ $-1$ $-1$ $0$ $-1$ $0$ $0$ $0$ $-1$ $0$ $0$ $0$ $0$ $0$ $0$ $0$ $-1$
$\chi_{5,4}$ $1$ $3$ $0$ $3$ $0$ $0$ $0$ $3$ $0$ $0$ $0$ $0$ $0$ $0$ $0$ $3$

but, unfortunately, experiment(5, 2, 16) yields

$\chi$ $c_{1,\chi}$ $c_{2,\chi}$ $c_{4,\chi}$ $c_{8,\chi}$ $c_{16,\chi}$
$\chi_{5,1}$ $1$ $1$ $0$ $1$ $0$ $0$ $0$ $1$ $0$ $0$ $0$ $0$ $0$ $0$ $0$ $1$
$\chi_{5,2}$ $-i$ $-i$ $-2i$ $-i$ $0$ $-2i$ $-2i$ $-i$ $-2i$ $0$ $-2i$ $-2i$ $0$ $-2i$ $0$ $-i$
$\chi_{5,3}$ $i$ $i$ $2i$ $i$ $0$ $2i$ $2i$ $i$ $2i$ $0$ $2i$ $2i$ $0$ $2i$ $0$ $i$
$\chi_{5,4}$ $-1$ $-1$ $0$ $-1$ $0$ $0$ $0$ $-1$ $0$ $0$ $0$ $0$ $0$ $0$ $0$ $-1$

where, say, the row for $\chi_{5,3}$ is ($i$ times) the sequence A374366 in OEIS.

If $a\equiv\pm1\pmod m$, then $c_{n,\chi}=0$ if there is a prime that divides $n$ but not $\varphi(m)$.

metamorphy
  • 43,591