A van der Corput sequence is a low-discrepancy sequence over the unit interval first published in 1935 by the Dutch mathematician J. G. van der Corput. It is constructed by placing a decimal point and writing the base $n$ representation of the sequence of natural numbers in the reverse order. [Read more](http://en.wikipedia.org/wiki/Van_der_Corput_sequence). 


This question is motivated by curiosity to study the analogue of van der Corput sequence for prime numbers. Consider the sequence $v\_p$ formed by placing a decimal point and writing the digits of the sequence of prime numbers $p$ in the reverse order (in base 10). The first few terms of this sequence are


$$
0.2, 0.3, 0.5, 0.7, 0.11, 0.31, 0.71, 0.91, 0.32, 0.92, \ldots
$$


Clearly $v\_p$ is not equidistributed in the unit interval and therefore it is not a low discrepancy sequence. I am curious to know if $v\_p$ has anything interesting property in it. 


Q1. What is the mean value of the sequence $v\_p$? In other words does the following limit exist?


$$
\lim\_{x \rightarrow \infty}\frac{1}{x}\sum\_{p \le x}v\_p.
$$


**Edit** (*Adding Timothy's guess as a question*)


Q2. What would be the mean value of base $b$?


Q3. **Edit** (*@ one more variation for this problem*)


Find a closed form representation of 


$$
\lim\_{x\to\infty}\frac{1}{x}\sum\_{p\le x}f(v\_p)
$$


where $f$ is any function Riemann integrable in $(0,1)$?


Considering the analogy with the equidistribution theorem, I think this should be possible. 



Analogue of van der Corput sequence for prime numbers
While $\lbrace v\_p \rbrace$ is clearly not equidistributed in $(0,1)$, it *is* equidistributed in
$$
\Pi\_{10} := [.1,.2) \cup [.3,.4) \cup [.7,.8) \cup [.9,1)
$$
by the prime number theorem (PNT) for arithmetic progressions modulo powers of $10$. In particular, the average tends to $0.55$, the average of the midpoints $0.15$, $0.35$, $0.75$, and $0.95$ of these four intervals.


[I see that Frank Thorne wrote much the same thing in a comment as I was editing this...]


[added in reply to further inquiries:] Replacing $10$ by an arbitrary base $b>1$, we likewise use the PNT for arithmetic progressions (APs) modulo powers of $b$ to show that the sequence is equidistributed in
$$
\Pi\_b := \lbrace x \in [0,1) : \gcd(\lfloor bx \rfloor, b) = 1 \rbrace.
$$
(I made the intervals half-open in $\Pi\_{10}$ for consistency with this definition, though it doesn't change the distribution.) That's a union of $\varphi(b)$ intervals of length $1/b$ whose set of left endpoints is symmetric about $1/2$, so the average is $1/2 + 1/(2b) = (b+1)/2b$ as Timothy Foo surmised.


On further thought, not only does equidistribution in $\Pi\_b$ follow from the PNT for APs modulo every power of $b$, but the two statements are readily seen to be equivalent.


