Timeline for Remainders $\quad 1\quad 2\quad $ only
Current License: CC BY-SA 3.0
29 events
| when toggle format | what | by | license | comment | |
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| Jul 1, 2013 at 6:44 | vote | accept | Włodzimierz Holsztyński | ||
| Jul 1, 2013 at 6:41 | comment | added | Włodzimierz Holsztyński | Nice format is important to me. I understand though "different strokes for different folks"--why, some prefer "de gustibus non est disputandum". | |
| Jun 29, 2013 at 12:28 | comment | added | Nate Eldredge |
Are all the extra spaces (\quad, \ , ) really necessary?
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| Jun 27, 2013 at 20:17 | answer | added | Włodzimierz Holsztyński | timeline score: 1 | |
| Jun 27, 2013 at 14:20 | comment | added | Yoav Kallus | Also here -- primepuzzles.net/conjectures/conj_018.htm -- we have the note: "Erdös conjectured that min (X-Y)=1 for only n=1, 2, 3, 4 & 7. This has been verified recently by Chris Nash up to n=600000. As far as John knows his conjecture has never been posted before." | |
| Jun 27, 2013 at 7:34 | comment | added | Gerry Myerson | In 1993-94, I got David Bailey interested in the question of whether the product of the primes up to $n$, $n\ge19$, could be a product of two consecutive integers, and he searched up to $n=23,000$. Later, Peter Montgomery took the search up to $n=50,000$. No examples were found. I don't know whether anything got published. | |
| Jun 27, 2013 at 7:15 | comment | added | Gerry Myerson | Daniel Berend, On the roots of certain sequences of congruences, Acta Arith 67 (1994) 97-104, proves that if $P_k$ is the product of the first $k$ primes, and $x_k$ is the smallest positive solution of $x(x+1)\equiv0\pmod{P_k}$, then $x_k/P_k\to0$ as $k\to\infty$. | |
| Jun 27, 2013 at 7:03 | comment | added | Włodzimierz Holsztyński | @Gary: I imagine that many people (naive and sophisticated mathematically) considered such pairs and similar after encountering Euclid infinitude of primes; one would replace Euclid's $P+1$ with $R+S$ or $R-S$, where $P$ and $R\cdot S$ are products of a finite number of the initial primes. More generally, $R\cdot S$ can be a product of powers of such primes, with $\gcd(R\ S)=1$. These numbers are most of the time huge or $1$, and the case of $1$ makes it hard to get a new prime as small as possible. | |
| Jun 27, 2013 at 6:14 | comment | added | Gerry Myerson | @Yoav, this was discussed in Nelson, Penney, and Pomerance, 714 and 715, J Rec Math 7 (1974) 87-89. They went up to $k=3049$ without finding another example. | |
| Jun 27, 2013 at 3:09 | comment | added | Yoav Kallus | I mean, we also have $2\times3=P(3)$, $5\times6=P(5)$, $14\times 15=P(7)$, but are there any instances with $k\ge19$? | |
| Jun 27, 2013 at 2:54 | comment | added | Yoav Kallus | So $714\times 715=P(17)$. Is there another pair of consecutive integers so that $n(n+1)=P(k)$ for some $k$? Or is this impossible? | |
| Jun 27, 2013 at 2:41 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
typo
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| Jun 27, 2013 at 2:38 | comment | added | Włodzimierz Holsztyński | @Greg: close. After your "therefore" you lost a factor of $\sqrt{2}$ though. | |
| Jun 27, 2013 at 2:34 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
a proof of the prezented theorem
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| Jun 27, 2013 at 2:06 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
First part of the proof
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| Jun 26, 2013 at 17:24 | comment | added | Greg Martin | Presumably the proof of the theorem goes something like this: given $x$, we can write $P(x) = 2P_1(x)P_2(x)$ where $n(x) \equiv 1\pmod {P_1(x)}$ and $n(x) \equiv 2\pmod {P_2(x)}$; therefore $n(x)$ exceeds the larger of $P_1(x)$ and $P_2(x)$. | |
| Jun 26, 2013 at 12:07 | comment | added | Gerry Myerson | What's going on with 716 reminds me of the "Ruth-Aaron pair", 714 and 715; $714=2x3x7x17$, $715=5x11x13$ (and $2+3+7+17=5+11+13$, although that's not relevant here). Also brings to mind Stormer's Theorem, about consecutive smooth numbers. | |
| Jun 26, 2013 at 10:22 | answer | added | Waldemar | timeline score: 4 | |
| Jun 26, 2013 at 10:17 | comment | added | Stefan Kohl♦ | So far I haven't found an instance of $n(p) = n(q) = n(r)$, at least. | |
| Jun 26, 2013 at 10:14 | answer | added | Stefan Kohl♦ | timeline score: 6 | |
| Jun 26, 2013 at 10:11 | comment | added | Włodzimierz Holsztyński | @Stefan: you have discovered an interesting case of primes $p<q$ such that $n(p)=n(q)$. On one hand it should not happen too often (just naive probability), and on the other hand somehow it does not surprize me (after the fact) that there are pairs of primes like this. Actually one should talk about intervals of primes (not just pairs). What would be the longest interval? How big (if any) would be three primes $p<q<r$ such that $n(p)=n(q)=n(r)$? | |
| Jun 26, 2013 at 10:01 | comment | added | Włodzimierz Holsztyński | How nice--thank you, @Stefan. Also, I am curious, can't help it, if you used a computer? | |
| Jun 26, 2013 at 9:51 | comment | added | Stefan Kohl♦ | We have $n(13) = n(17) = 716$, while $\sqrt{P(17)} \approx 714$. -- Thus for $x = 17$ your inequality is pretty sharp. | |
| Jun 26, 2013 at 9:34 | comment | added | Włodzimierz Holsztyński | Continuation. Since $11=1 \mod 5$ we need $k=1 \mod 5$ or $5|k$ or $k=4 \mod 5$. Is it beneath one's honor to perform the reduced amount of calculation by hand (by the brute force of one's calculating power)? | |
| Jun 26, 2013 at 9:20 | comment | added | Włodzimierz Holsztyński | To double-check Gerry's nearly-calculation one needs to verify only integers $11\cdot k+1$ and $11\cdot k+2$ in the range $49\ldots 210$, which amounts to less than $30$ candidate integers. It can be patiently computed by hand like this: $$7|56\quad 3|57\quad _7 67_4\quad _7 68_5\ \ldots$$ except that it's a little bit too embarrassing :-) (we already see that $n(11) \ge 78$, ok, that $n(11)\ge 79$, etc.). | |
| Jun 26, 2013 at 8:25 | comment | added | Włodzimierz Holsztyński | @Gerry: I can see that the remainders are right hence $n(11)\le 211$. And now I should write a small program or improve my small theorem, which only assures us that $n(11)\ge 49$, which already feels awfully weak (mea culpa :-). Thank you Gerry. | |
| Jun 26, 2013 at 1:21 | comment | added | Gerry Myerson | $n(11)=211$, I think. | |
| Jun 25, 2013 at 23:35 | review | Close votes | |||
| Jun 27, 2013 at 23:13 | |||||
| Jun 25, 2013 at 23:08 | history | asked | Włodzimierz Holsztyński | CC BY-SA 3.0 |