most recent 30 from http://mathoverflow.net 2013-05-25T01:32:19Z http://mathoverflow.net/feeds/question/85468 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/85468/analogue-of-van-der-corput-sequence-for-prime-numbers Xaus 2012-01-12T04:22:02Z 2012-01-12T15:57:31Z

<p>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. <a href="http://en.wikipedia.org/wiki/Van_der_Corput_sequence" rel="nofollow">Read more</a>. </p> <p>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</p> <p>$$ 0.2, 0.3, 0.5, 0.7, 0.11, 0.31, 0.71, 0.91, 0.32, 0.92, \ldots $$</p> <p>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. </p> <p>Q1. What is the mean value of the sequence $v_p$? In other words does the following limit exist?</p> <p>$$ \lim_{x \rightarrow \infty}\frac{1}{x}\sum_{p \le x}v_p. $$</p> <p><strong>Edit</strong> (<em>Adding Timothy's guess as a question</em>)</p> <p>Q2. What would be the mean value of base $b$?</p> <p>Q3. <strong>Edit</strong> (<em>@ one more variation for this problem</em>)</p> <p>Find a closed form representation of </p> <p>$$ \lim_{x\to\infty}\frac{1}{x}\sum_{p\le x}f(v_p) $$</p> <p>where $f$ is any function Riemann integrable in $(0,1)$?</p> <p>Considering the analogy with the equidistribution theorem, I think this should be possible. </p>

http://mathoverflow.net/questions/85468/analogue-of-van-der-corput-sequence-for-prime-numbers/85470#85470 Noam D. Elkies 2012-01-12T04:50:16Z 2012-01-12T07:08:02Z

<p>While $\lbrace v_p \rbrace$ is clearly not equidistributed in $(0,1)$, it <em>is</em> 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.</p> <p>[I see that Frank Thorne wrote much the same thing in a comment as I was editing this...]</p> <p>[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.</p> <p>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.</p>