are there infinitely many triples of consecutive square-free integers? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T17:46:33Z http://mathoverflow.net/feeds/question/59741 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/59741/are-there-infinitely-many-triples-of-consecutive-square-free-integers are there infinitely many triples of consecutive square-free integers? Ewan Delanoy 2011-03-27T17:11:52Z 2011-03-28T10:32:10Z <p>The title says it all ... Obviously, any such triple must be of the form $(4a+1,4a+2,4a+3)$ where $a$ is an integer. Has this problem already been studied before ? The result would follow from Dickson's conjecture on prime patterns, which implies that there are infinitely many integers $b$ such that $4(9b)+1,2(9b)+1$ and $4(3b)+1$ are all prime (take $a=9b$). </p> <p>A related question : <a href="http://mathoverflow.net/questions/32412/question-on-consecutive-integers-with-similar-prime-factorizations" rel="nofollow">http://mathoverflow.net/questions/32412/question-on-consecutive-integers-with-similar-prime-factorizations</a></p> http://mathoverflow.net/questions/59741/are-there-infinitely-many-triples-of-consecutive-square-free-integers/59742#59742 Answer by Aaron Meyerowitz for are there infinitely many triples of consecutive square-free integers? Aaron Meyerowitz 2011-03-27T17:16:09Z 2011-03-27T17:16:09Z <p>too hasty, never mind</p> http://mathoverflow.net/questions/59741/are-there-infinitely-many-triples-of-consecutive-square-free-integers/59746#59746 Answer by Zander for are there infinitely many triples of consecutive square-free integers? Zander 2011-03-27T17:55:37Z 2011-03-27T17:55:37Z <p>I found <a href="http://mathforum.org/kb/message.jspa?messageID=347645&amp;tstart=0" rel="nofollow">these answers by Erick Wong</a>. The simpler version is that if the answer was no, then at least two of every $4a,4a+1,4a+2,4a+3$ must not be squarefree for $a$ large enough, so the density of squarefree numbers would be limited by $1/2$. But it is $6/\pi^2>1/2$ so there must be infinitely many consecutive triples of squarefree numbers.</p> <p>By the way, I don't think Dickson's conjecture applies, since one of $4p+1,2p+1,4p+3$ is divisible by 3.</p> http://mathoverflow.net/questions/59741/are-there-infinitely-many-triples-of-consecutive-square-free-integers/59751#59751 Answer by GH for are there infinitely many triples of consecutive square-free integers? GH 2011-03-27T19:02:26Z 2011-03-27T19:18:38Z <p>The answer is yes. More precisely, George Lowther's heuristic is right, i.e. the density of $n$'s such that $n-1$, $n$, $n+1$ are square-free is $\prod_p(1-3/p^2)$ over <em>all</em> primes $p$. </p> <p>To see this, let $P$ be fixed but large. As $x\to\infty$, the number of $n\leq x$ such that none of $n-1$, $n$, $n+1$ is divisible by the square of some prime $p\leq P$ is $x\prod_{p \leq P}(1-3/p^2)+o(x)$ by the Chinese Remainder Theorem. The number of $n\leq x$ such that one of $n-1$, $n$, $n+1$ is divisible by the square of some prime $p>P$ is at most $\sum_{P &lt; p\leq \sqrt{x+1}} 3\left(\frac{x}{p^2}+1\right)\ll\frac{x}{P}+O(\sqrt{x})$. Hence the number of $n\leq x$ in question is $x\prod_{p \leq P}(1-3/p^2)+O\left(\frac{x}{P}\right)+o(x)$. Letting $P\to\infty$ proves the claim.</p> <p>A more careful count, e.g. the choice $P:=(\log x)/10$, reveals that the number of $n\leq x$ in question is $x\prod_p(1-3/p^2) + O(x/\log x)$.</p> <p>BTW these questions have a large literature. See e.g. some of Harald Helfgott's papers with the words "square-free sieve" or "power-free values" in the title, and the references in them.</p> http://mathoverflow.net/questions/59741/are-there-infinitely-many-triples-of-consecutive-square-free-integers/59753#59753 Answer by George Lowther for are there infinitely many triples of consecutive square-free integers? George Lowther 2011-03-27T19:21:08Z 2011-03-28T01:40:17Z <p>To expand on the answer in my comment, the proportion of integers $a$ for which $4a+1,4a+2,4a+3$ are all squarefree is $\prod_{p\not=2}(1-3/p^2)$, with the product taken over all odd primes $p$. As this product converges to a positive limit, there are infinitely many such $a$. A quick heuristic is to look modulo $p^2$. For odd prime $p$, precisely $p^2-3$ of the possible $p^2$ values of $a$ mod $p^2$ lead to $4a+1,4a+2,4a+3$ all being nonzero mod $p^2$, so this has probability $1-3/p^2$. Independence of mod $p^2$ arithmetic as $p$ runs through the primes suggests the claimed limit.</p> <p>More precisely, if $\phi(n)$ is the number of positive integers $a\le n$ with $4a+1,4a+2,4a+3$ squarefree then $$\frac{\phi(n)}{n}\to\prod_{p\not=2}(1-3/p^2).\qquad\qquad{\rm(1)}$$ It's not too hard to turn this heuristic into a rigorous argument. If we let $\phi_N(n)$ denote the number of $a\le n$ such that none of $4a+1,4a+2,4a+3$ is a multiple of $p^2$ for a prime $p &lt; N$, then the Chinese remainder theorem says that we get equality $$\frac{\phi_N(n)}{n}=\prod_{\substack{p\not=2,\\ p &lt; N}}(1-3/p^2).\qquad\qquad{\rm(2)}$$ wherever $n$ is a multiple of $\prod_{\substack{p\not=2,\\ p &lt; N}}p^2$ and, therefore, the error in (2) is of order $1/n$ for arbitrary $n$. It only needs to be shown that ignoring primes $p\ge N$ leads to an error which is vanishingly small as $N$ is made large. In fact, the number of $a \le n$ which are a multiple of $p^2$ is $\left\lfloor\frac{n}{p^2}\right\rfloor\le \frac{n}{p^2}$. The number of $a\le n$ which is a multiple of $p^2$ for some prime $p\ge N$ is bounded by $n\sum_{p\ge N}p^{-2}$. So, the proportion of $a\le n$ for which one of $4a+1,4a+2,4a+3$ is a multiple of $p^2$ for an odd prime $p\ge N$ is bounded by $3\frac{4n+3}{n}\sum_{p\ge N}p^{-2}\sim3(4+3/n)/(N\log N)$. This means that $\phi_N(n)/n\to\phi(n)/n$ <em>uniformly</em> in $n$ as $N\to\infty$, and the limit (1) follows from approximating by $\phi_N$.</p> http://mathoverflow.net/questions/59741/are-there-infinitely-many-triples-of-consecutive-square-free-integers/59817#59817 Answer by Marius Overholt for are there infinitely many triples of consecutive square-free integers? Marius Overholt 2011-03-28T10:32:10Z 2011-03-28T10:32:10Z <p>The question of the number of positive integers $n \leq x$ for which all members of an associated fixed pattern are squarefree (or r-free) was studied by Leon Mirsky:</p> <p>L. Mirsky, Note on an asymptotic formula connected with r-free integers. Quart. J. Math., Oxford Ser. 18 (1947), 178-182. </p> <p>L. Mirsky, Arithmetical pattern problems relating to divisibility by rth powers. Proc. London Math. Soc. (2) 50 (1949), 497–508.</p> <p>As I remember it, Mirsky proved that the number is $cx + O(x^{2/3})$ for patterns of squarefrees, where $c$ is a constant depending on the pattern, and is positive if the pattern is not excluded by certain necessary congruential conditions.</p>