Timeline for Sieving the values of an arithmetic sequence which is infinitely many times $1$
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| May 5, 2018 at 13:04 | comment | added | The Number Theorist | A possibility that might work in certain cases is to consider the following identity: $$\sum_{\substack{n\leq x \\ a_n>1 \\ (a_n, P)=1}}1= -\sum_{\substack{1<d\ll X}}\mu(d)\sum_{\substack{n\leq x \\ d\mid a_n}}\sum_{e\mid (a_n, P)}\mu(e)=-\sum_{\substack{1<d\ll X}}\mu(d)\sum_{e|P}\mu(e)\sum_{\substack{n\leq x \\ [d,e]\mid a_n}}1,$$ where $X$ is an upper bound for the sequence $a_n$ when $n\leq x$. | |
| May 1, 2018 at 17:19 | comment | added | Michael Hardy | Notice the use of \text{} and \mid in my edits to the question. Both affect spacing. | |
| May 1, 2018 at 17:18 | history | edited | Michael Hardy | CC BY-SA 3.0 |
deleted 20 characters in body; edited title
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| Apr 30, 2018 at 19:30 | comment | added | The Number Theorist | Actually, what I expect (and what I would like to prove) is that the set $\{n : a_n=1\}$ has positive natural density. But in order to reach this, I need to estimate the cardinality stated above. | |
| Apr 30, 2018 at 19:18 | comment | added | Denis Chaperon de Lauzières | If you really have no information, it could be that $a_n$ is always $1$, and then you can't do better than the trivial bound. | |
| Apr 30, 2018 at 17:33 | comment | added | The Number Theorist | Yes, you are right. The problem is that I don't have any information about the number of integers $n\leq x$ such that $a_n>1$. Therefore, I don't see how can I find a bound using your formula. | |
| Apr 30, 2018 at 16:08 | comment | added | GH from MO | I don't see how the condition (or the lack of it) that infinitely many $a_n$'s are equal to $1$ changes anything. The idea of sieve theory is to upper bound and lower bound the divisor sum $\sum_{d\mid m}\mu(d)$ by replacing $\mu(d)$ in it by other functions. Starting from such bounds, you can estimate your quantity, which is $\sum_{\substack{n\leq x\\a_n>1}}\sum_{d\mid(a_n,\mathcal{P})}\mu(d)$. | |
| Apr 30, 2018 at 14:05 | history | asked | The Number Theorist | CC BY-SA 3.0 |