I'm recently studying some topics in analytic number theory and I have encountered results involving the infinite product $$C=\prod_{p}\left(1-\frac{1}{p(p+1)}\right)$$ where $p$ denotes calculating the product over all prime numbers.

Or more intuitively,$$C=\left(1-\frac{1}{2\cdot3}\right)\cdot\left(1-\frac{1}{3\cdot4}\right)\cdot\left(1-\frac{1}{5\cdot6}\right)\cdot\left(1-\frac{1}{7\cdot8}\right)\cdot\left(1-\frac{1}{11\cdot12}\right)\cdots$$ Mathematica can show that $C\approx0.704442.$

However, I need a high precision result, roughly accurate to at least 20 digits, so that I could use the Inverse Symbolic Calculator to make an educated guess of any possible analytic expression for $C$ in forms like $\dfrac{\zeta(2)\zeta(3)}{\zeta(6)}$.

When I use NProduct, I get

NProduct[1 - 1/(Prime[k] (Prime[k] + 1)), {k, 1, Infinity},
         AccuracyGoal -> 20, PrecisionGoal -> 20]

During evaluation of In[14]:= Prime::intpp: Positive integer argument expected in Prime[15.]. >>
During evaluation of In[14]:= Prime::intpp: Positive integer argument expected in Prime[14.]. >>
During evaluation of In[14]:= Prime::intpp: Positive integer argument expected in Prime[13.]. >>
During evaluation of In[14]:= General::stop: Further output of Prime::intpp will be suppressed during this calculation. >>

However, when I use N[Product[]], the calculation was so slow that I had to abort the calculation after minutes of calculating.

N[Product[1 - 1/(Prime[k] (Prime[k] + 1)), {k, 1, 100000}], 50]


N[Product[1 - 1/(Prime[k] (Prime[k] + 1)), {k, 1, 1000000}], 50]


N[Product[1 - 1/(Prime[k] (Prime[k] + 1)), {k, 1, 10000000}], 50]


Is there any way to calculate products like $C$ to high precision in Mathematica? Thanks.

  • 1
    $\begingroup$ Page 11 of arxiv.org/pdf/0903.2514v2.pdf gives the expansion, as $Q^{(1)}_1$. It's the "carefree constant", according to Wikipedia. (Thanks to [OEIS] for this.) [OEIS]: oeis.org/A065463 $\endgroup$ – Patrick Stevens Aug 15 '15 at 12:22
  • 1
    $\begingroup$ Did not have any luck with the inverse symbolic calculator though. $\endgroup$ – bobbym Aug 15 '15 at 15:34
  • $\begingroup$ Thanks, @PatrickStevens! @bbgodfrey, $\endgroup$ – Zhenhua Liu Aug 16 '15 at 0:32
  • $\begingroup$ @bbgodfrey, I'll try my best to help, though I'm still a novice in using Mathematica. $\endgroup$ – Zhenhua Liu Aug 16 '15 at 0:41
  • $\begingroup$ @bobbym, Thanks. It seems no analytic expression is available. $\endgroup$ – Zhenhua Liu Aug 16 '15 at 0:41

Using the formula given in the arXiv preprint Patrick linked to for the "carefree constant" gives:

Exp[NSum[(-1)^k PrimeZetaP[k] (1 - LucasL[k])/k, {k, 2, ∞}, Compiled -> False,
         Method -> "AlternatingSigns", NSumTerms -> 20, WorkingPrecision -> 30]]

Note that this agrees with the result in the OEIS up to twenty digits. The only speed-limiting part of this is the calculation of the prime zeta function.

  • $\begingroup$ Thanks a lot! That's great help to me. $\endgroup$ – Zhenhua Liu Aug 16 '15 at 0:42

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