5

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. >>
0.705024
```

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]
0.70444223707595873775750824971600704569326374380066
N[Product[1 - 1/(Prime[k] (Prime[k] + 1)), {k, 1, 1000000}], 50]
0.70444220359808164073982578152558851839613830115938
N[Product[1 - 1/(Prime[k] (Prime[k] + 1)), {k, 1, 10000000}], 50]
$Aborted
```

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

1

Page 11 of http://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]: https://oeis.org/A065463

– Patrick Stevens – 2015-08-15T12:22:32.6771Did not have any luck with the inverse symbolic calculator though. – bobbym – 2015-08-15T15:34:40.027

Thanks, @PatrickStevens! @bbgodfrey, – Zhenhua Liu – 2015-08-16T00:32:14.303

@bbgodfrey, I'll try my best to help, though I'm still a novice in using Mathematica. – Zhenhua Liu – 2015-08-16T00:41:15.150

@bobbym, Thanks. It seems no analytic expression is available. – Zhenhua Liu – 2015-08-16T00:41:21.707