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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.

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.

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    $\begingroup$ Should be 0.55. Truncate to the first n digits to get an answer within $10^{-n}$. Then use the prime number theorem for arithmetic progressions to show that all sequences of n digits with the first odd appear with equal probability (with error going to 0 as $x \rightarrow \infty$). Now let $n \rightarrow \infty$. $\endgroup$
    – Frank Thorne
    Commented Jan 12, 2012 at 4:45
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    $\begingroup$ First digit odd and not equal 5, you mean (though as it happens this doesn't change the average). Your comment appeared while I was writing much the same thing as an answer... $\endgroup$
    – Noam D. Elkies
    Commented Jan 12, 2012 at 4:53
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    $\begingroup$ I think we can do this without the prime number theorem since the mean for $v_{2n+1}$ is 0.55. Hence in general any sequence $v_{a_n}$ the mean should depend only the set of different possible last (unit's) digit that $a_n$ can take. $\endgroup$
    – user20174
    Commented Jan 12, 2012 at 6:16
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    $\begingroup$ well, $\varphi(b)$ is even for $b>2$, and $(-1,b)=1$, so you pair off $r+0.5$ with $b-r+0.5$ when you average. $\endgroup$
    – Timothy Foo
    Commented Jan 12, 2012 at 6:49
  • $\begingroup$ wonder what other variations can be invented for this question? $\endgroup$
    – Timothy Foo
    Commented Jan 12, 2012 at 7:49

1 Answer 1

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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.

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    $\begingroup$ And I guess that in base $b$, the mean would be $(b+1)/2b$. $\endgroup$
    – Timothy Foo
    Commented Jan 12, 2012 at 5:51

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