Timeline for Counting powerful integers. Lower bounds
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 31, 2022 at 8:48 | comment | added | GH from MO | In my response below, I recorded the numerical upper and lower bounds of Golomb (1970). | |
| Oct 31, 2022 at 8:31 | answer | added | GH from MO | timeline score: 4 | |
| Oct 31, 2022 at 1:32 | comment | added | Wlod AA | @GerryMyerson, I do NOT whine! (And I never said that you've down-voted my post above). | |
| Oct 31, 2022 at 1:28 | comment | added | Gerry Myerson | I thought you were disputing the $x^2y^3$ definition, on the grounds that you couldn't get anything less than $72$ using it. Also, you asked for "more" definitions, without telling anyone which ones you already knew – was I meant to read your mind? And could you please stop whining about downvotes? It's not very becoming. (And, no, I haven't downvoted.) | |
| Oct 31, 2022 at 1:15 | comment | added | Wlod AA | @GerryMyerson, the $\ p\to p^2\ $ and $\ x^2\cdot y^3\ $ are the classical, common definitions. Here, I have added two more (and each for a good reason). BTW, after my last definition, another ugly guy had to commit his filthy deed and has cowardly down-voted my post. | |
| Oct 31, 2022 at 0:58 | comment | added | Gerry Myerson | $4=2^21^3$. $8=1^22^3$. So, yes, the $x^2y^3$ definition is correct. Another equivalent definition is that for all primes $p$, if $p$ divides $n$, then $p^2$ divides $n$. | |
| Oct 30, 2022 at 21:44 | history | edited | Wlod AA | CC BY-SA 4.0 |
Another eq. def. of powerful integers
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| Oct 30, 2022 at 16:25 | comment | added | Wlod AA | @JohnOmielan, many thanks for your comments (and to Staanley Yao Xiao too). I'll try to digest them. | |
| Oct 30, 2022 at 15:00 | comment | added | John Omielan | (cont.) I assume your comment reply to StanleyYaoXiao indicated you realized the issue but, if not, please let me know. Also, I believe my comment's referenced paper of On a theorem of Erdös and Szekeres gives about a tight a bound, both lower & upper, i.e., of $o(x^{1/6})$, as you can reasonably expect to get on how many powerful numbers there are up to $x$. | |
| Oct 30, 2022 at 14:59 | comment | added | John Omielan | I don't know what you're referring to. The Wikipedia article that StanleyYaoXiao indicated (which mentions the $x^2\cdot y^3$ form) has a link to OEIS A001694 (that lists the powerful numbers), from which I got my link to OEIS A118896. This other page gives just how many powerful numbers are $\le 10^n$. Looking at A001694, I see it matches, with $1$ for $1$ (i.e., $1$), $4$ for $10$ (i.e., $3$ more of $4$, $8$ and $9$) and $14$ for $100$ (i.e., with $10$ more of $16$, $25$, $27$, $32$, $36$, $49$, $64$, $72$, $81$ and $100$). ... | |
| Oct 30, 2022 at 8:13 | comment | added | Wlod AA | @StanleyYaoXiao, in. my comment I meant OEIS only, I knew about wikipedia. However, something has jinxed me since I got nowhere with OEIS. And, indeed, I overlooked a ref. to OEIS in the wikipedia. After your link, everything became "normal", thank you, I do see that seq. there and over. Once again, thank you. | |
| Oct 30, 2022 at 7:57 | comment | added | Stanley Yao Xiao | en.wikipedia.org/wiki/Powerful_number | |
| Oct 30, 2022 at 5:32 | answer | added | Wlod AA | timeline score: 1 | |
| Oct 30, 2022 at 4:51 | history | edited | Wlod AA | CC BY-SA 4.0 |
a mth typo -- |...| missing
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| Oct 30, 2022 at 4:32 | comment | added | Wlod AA | @JohnOmielan, these integers are different from $\ x^2\cdot y^3.\ $ I was not able to find powerful numbers as: $$ 1\ 4\ 8\ 9\ 16\ 25\ 27\ 36\ 49\ 64\ 72 \ldots $$ -- I tried in several different ways $\ (72=2^3\cdot3^2\ $ is the first "generic" term). | |
| Oct 30, 2022 at 3:47 | comment | added | John Omielan | In OEIS's A118896 (Number of powerful numbers <= 10^n), the second comment in the Comments section states "Bateman & Grosswald proved that the number of powerful numbers up to x is zeta(3/2)/zeta(3) * x^1/2 + zeta(2/3)/zeta(2) * x^1/3 + o(x^1/6). This approximates the series very closely: up to a(24), all absolute errors are less than 75." Thus, it seems that a reasonably good lower bound, especially for larger $x$ values, would be $\frac{\zeta(3/2)x^{1/2}}{\zeta(3)}+\frac{\zeta(2/3)x^{1/3}}{\zeta(2)}-cx^{1/6}$ for some relatively small constant $c \gt 0$. | |
| Oct 30, 2022 at 3:36 | comment | added | Wlod AA | Actually, outside the above definition of powerful numbers, I know only two classical definitions; do you know more of them -- perhaps they can be mentioned here in the comments. | |
| Oct 30, 2022 at 3:26 | history | asked | Wlod AA | CC BY-SA 4.0 |