The Hardy-Littlewood Conjecture F [1] involves the infinite product $$\prod\left(1-\frac{1}{\varpi-1}\left(\frac D\varpi\right)\right)$$ where $\varpi$ ranges over the odd primes and $\left(\frac D\varpi\right)$ is the Legendre symbol.

Is there a good way to calculate this? The product converges very slowly, and none of the standard methods (Cohen-Villegas-Zagier, Wynn, etc.) seem to work because of the unpredictable sign changes.

Given that D is fixed, it suffices to calculate the partial products in various congruence classes; I don't know if this is a viable approach.

Another possibility: I've seen almost magical series acceleration with the zeta function, it may work here.

[1] G. H. Hardy, J. E. Littlewood. "Some of the problems of partitio numerorum III: On the expression of a large number as a sum of primes". Acta Mathematica 44 (1923), pp. 1-70.


This problem was studied by a few, and the ideas involve too much latex to write here. Mainly there are ideas of transforming to crazy weighted sums and then use ERH to bound errors from crazier integrals. It suffices to say that the culmination of this research is the freely available paper:

New Quadratic Polynomials With High Densities Of Prime Values, Jacobson and Williams 1999

In that year, they used their ideas to compute the constant for discriminants of up to 72 digits! (assuming ERH) The bottleneck of the process seems to be the calculation of the algebraic invariants class number and regulator. Over a decade has passed, the technology today should be able use the same methods to get up to 100 digits, within reasonable time, and up to 110 digits with a bit more time (apparently, two weeks on a cluster):

Practical Improvements to Class Group and Regulator Computation of Real Quadratic Fields, Biasse and Jacobson 2010

Improvements in the Computation of Ideal Class Groups of Imaginary Quadratic Number Fields, Biasse 2009

| cite | improve this answer | |
  • 1
    $\begingroup$ The paper of "Biasse and Michael" is really "Biasse and Jacobson". $\endgroup$ – KConrad Feb 8 at 6:27

I don't know what "the standard methods" means. Have you considered comparing this with a more conventional Euler product? First, I will write $p$ instead of $\varpi$ since I don't like $\varpi$ (looks too much like $\overline{\omega}$ for my tastes).

Expanding the $p$-th factor as a geometric series, $$ 1-\frac{1}{p-1}\left(\frac Dp\right) = 1-\frac{1/p}{1-1/p}\left(\frac Dp\right) = 1-\frac{1}{p}\left(\frac Dp\right) - \frac{1}{p^2}\left(\frac Dp\right) - \cdots, $$ and the first two terms have a product that is known: $\prod_{p} (1 - (D|p)/p) = 1/L(1,\chi_D)$, where $\chi_D(p) = (\frac{D}{p})$ for odd $p$. (I'm being a bit sloppy about the Euler factor at 2.) Therefore if we divide the $p$-th factor in the Hardy--Littlewood constant by $1 - (D|p)/p$ then we'll have the $p$-th term be $1 + O(1/p^2)$, which has better convergence (not ultra-fast, but better than what you had before.) We write $$ \prod_{p} \left(1 - \frac{(D|p)}{p-1}\right) = \prod_{p} \frac{1 - (D|p)/(p-1)}{1 - (D|p)/p} \cdot \left(1 - \frac{(D|p)}{p}\right)= \prod_{p} \frac{1 - (D|p)/(p-1)}{1 - (D|p)/p} \cdot \prod_{p} \left(1 - \frac{(D|p)}{p}\right). $$ This last expression is $$ \frac{1}{L(1,\chi_D)}\prod_{p} \frac{1 - (D|p)/(p-1)}{1 - (D|p)/p} = \frac{1}{L(1,\chi_D)}\prod_{p} \left(1 + O\left(\frac{1}{p^2}\right)\right). $$

There are ways of accelerating these quasi-Euler products further. See Pieter Moree, Approximation of singular series and automata, Manuscripta Math. 101 (2000), 385--399.

Last comment: yes, when $\chi$ is nontrivial the Euler product of $L(s,\chi)$ at $s = 1$ does equal the $L$-function at $s = 1$. That requires an argument, since $s=1$ is not in the range of absolute convergence of the Euler product.

| cite | improve this answer | |
  • $\begingroup$ springerlink.com/content/mkqmp6q58j00v271 $\endgroup$ – Charles Jul 9 '10 at 4:07
  • $\begingroup$ I really wish I could accept two answers. Your comment has been very helpful, especially since Jacobson & Williams don't discuss solutions for -4 <= delta <= 0. $\endgroup$ – Charles Jul 16 '10 at 6:25

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.