John Derbyshire in his book [PRIME OBSESSION](http://tomlr.free.fr/Math%E9matiques/Fichiers%20Claude/Nombres/Derbyshire%20-%20Prime%20Obsession%20-%20Bernhard%20Riemann%20and%20the%20Greatest%20Unsolved%20Problem%20in%20MathematicsAAA.pdf) says on page 343: 



> 
> "I’ll round off with a complete calculation of $\pi(1,000,000)$, the
>  number of primes up to one million, using Riemann’s formula -- not for
>  the fun of it, though it is of course great fun, but to make some
>  important points about the error term."
> 
> 
> 


He finds that secondary terms contribute an error of -29.37378.  

**My question:** 
How many zeta zeros are needed to find it?



How many zeta zeros are needed to accurately calculate five digits for π(1000000), where π(x) is the prime counting function?
[Monte-Carlo Approximation of the Prime Counting Function](https://wstein.org/edu/2010/414/projects/stueve.pdf)


This is a numerical study of the dependence of the error $\delta\pi(x)$ in the prime counting function when $N$ zeros $\rho$ of the zeta function are used in the sum $\sum\_\rho {\rm Li}\,(x^\rho)$. 


Figure 3 shows that for $x=10^{12}$ the error $\delta\pi(x)$ is about 1100 for $N=10^3$ and 300 for $N=10^5$. The error drops by roughly a factor of two when the number of zeros is increased by a factor of ten. More extensive numerical data, with $x$ in the range $10^{10}$ to $10^{17}$, indicates that to reach the precise value of $\pi(x)$ one needs to include about $x^{1.38}$ zeros.


The value $x=10^6$ in the OP is well below this range, but as a first estimate I would conclude that to obtain $\pi(10^6)=78498$ accurate to the last of its five digits one would need $N\simeq 10^8$ zeros.


