Timeline for Squareful values of polynomials
Current License: CC BY-SA 3.0
14 events
when toggle format | what | by | license | comment | |
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May 21, 2017 at 11:57 | comment | added | Daniel Loughran | This $\ll$ is Vinogradov notation, which is standard in analytic number theory. | |
May 21, 2017 at 10:32 | comment | added | Yaakov Baruch | Thank you. I think I get confused by the "<<" notation, which I must have learnt as "much less than" a long time ago. | |
May 21, 2017 at 7:44 | comment | added | Daniel Loughran | There are $cX^{1/2}$ squareful integers of absolute value up to $X$, for some $c>0$. This is very easy to prove; see e.g. Powerful Numbers S. W. Golomb The American Mathematical Monthly Vol. 77, No. 8 (Oct., 1970), pp. 848-852 | |
May 21, 2017 at 5:09 | comment | added | Yaakov Baruch | @DanielLoughran: for the density of squarefuls, do you mean $\gg X^{1/2}$, or $\ll X^{5/6}$? | |
May 21, 2017 at 0:21 | answer | added | Dr. Pi | timeline score: 4 | |
May 8, 2017 at 15:43 | vote | accept | Daniel Loughran | ||
May 8, 2017 at 10:54 | answer | added | joro | timeline score: 5 | |
May 7, 2017 at 20:03 | comment | added | Lucia | Miscalculation on my part, sorry! | |
May 7, 2017 at 19:52 | comment | added | Daniel Loughran | @Lucia:The problem is as follows.(though if you are able to fix this I would be very interested). Suppose we are using the large sieve to count the number of squareful integers $n$ and sieving modulo $p^2$ as you suggest. Note that units modulo $p^2$ are squareful. So we are only excluding the $p-1$ residue classes given by $p,\ldots,(p-1)p$. The local $p$-adic density of squareful numbers is then roughly given by $(p^2 - (p - 1))/p^2 = 1 - 1/p + 1/p^2$, which decays like $1/(\log X)$, as claimed. | |
May 7, 2017 at 19:32 | comment | added | Lucia | Won't the large sieve give something like $X^{3/4+\epsilon}$ by just arguing with moduli of the form $p^2$? For most primes $p$, it seems like $n$ would have to live in a bounded number of residue classes $\pmod {p^2}$ and then the exponential sums at $a/p^2$ for $(a,p)=1$ must be big ... . | |
May 7, 2017 at 19:15 | comment | added | Daniel Loughran | Yes good point thanks. I added the assumption that $f$ is separable. | |
May 7, 2017 at 19:14 | history | edited | Daniel Loughran | CC BY-SA 3.0 |
added 14 characters in body
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May 7, 2017 at 19:13 | comment | added | Vesselin Dimitrov | I think the one necessary condition should be that the polynomial itself (over $\mathbb{C}$) should not be squareful. Apart from the constants you have to exclude the trivial examples such as $f(x) = x^2 (x+1)^3$. | |
May 7, 2017 at 19:04 | history | asked | Daniel Loughran | CC BY-SA 3.0 |