Timeline for Happy New Prime Year!
Current License: CC BY-SA 2.5
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 3, 2011 at 9:12 | comment | added | Gerhard Paseman | Here is a different "argument" with hopefully an easier path toward justification. Since n often has log(log(n)) distinct factors, it would be highly unlikely that all of the k numbers in the sequence would deviate from that. Thus one of the k numbers n must have size at least somewhere around exp(exp(k/2)), give or take a few orders of orders of magnitude. How about that? Gerhard "What's a Googleplex Between Colleagues?" Paseman, 2011.01.02 | |
| Dec 30, 2010 at 8:03 | comment | added | Wadim Zudilin | Thanks, S-in-B, for the newer edition and especially for the link to De Koninck (is it available online?). | |
| Dec 30, 2010 at 2:10 | comment | added | Gerhard Paseman | An interesting "argument" goes as follows: for large k, of the k(k+1)/2 factors in the run of k numbers, about k/2 are 2, about k/3 are 3, and so on for the first few primes. Counting all the factors suggests summing {1/p_i} up to (k+1)/2, which needs the first n primes where log(log(n)) would roughly be about k/2, thus examples of such runs should be roughly exp(exp(k/2)). The biggest problem with this argument is that it seems to work for small k. Perhaps someone knows a good argument with a similar conclusion? Gerhard "Ask Me About System Design" Paseman, 2010.12.29 | |
| Dec 29, 2010 at 20:18 | history | edited | sleepless in beantown | CC BY-SA 2.5 |
added OEIS reference to sequence A086560
|
| Dec 29, 2010 at 19:50 | history | edited | sleepless in beantown | CC BY-SA 2.5 |
change past to future tense
|
| Dec 29, 2010 at 19:41 | comment | added | sleepless in beantown | @A-Rex asaurus, the $n+1$ being prime was my original conjecture which I did not edit out when I found the first counter example for $k=3$, $n=63$, with $n+1=64=2^6$, $n+2=65=5\cdot13$, $n+3=66=2\cdot3\cdot11$. So $n+1$ only has to be prime (or a power of a prime) if $n$ is even; if $n$ is odd, $n+1$ can be a power of an even prime thus it must be a power of $2$. | |
| Dec 29, 2010 at 18:48 | comment | added | aorq | Maybe I'm missing something, but why does $n+1$ have to be prime? | |
| Dec 29, 2010 at 8:30 | comment | added | Wadim Zudilin | Thanks, S-in-B. Yes, I would prefer to be back to 1867 or 1999 just because they give lengthier intervals... On the other hand, I hope I don't make you sleepless. | |
| Dec 29, 2010 at 8:05 | comment | added | sleepless in beantown | @Gerhard-Paseman: Perlify it into a one-liner, and I'll see your code and raise it to APL! Or maybe get out the punch-cards and do some FORTRAN-WATFOR. | |
| Dec 29, 2010 at 8:00 | comment | added | sleepless in beantown | There are 185 $n$ values in the range 1-100000 (with values for n ranging from 1866 to 99906) for $k=4$. | |
| Dec 29, 2010 at 7:56 | comment | added | Gerhard Paseman | I would tell you to go to sleep, but that might come across as an insult. In the meantime, I'll see about replicating your results with awk. Gerhard "You Only Need One Tool" Paseman, 2010.12.28 | |
| Dec 29, 2010 at 7:48 | comment | added | sleepless in beantown | wow, i'm tired. Yes, my correction has an error in it! Yes, n+2 has to be divisble by 2, but I really meant to say n+3 is divible by 3 and if it isn't then n+5 is divisble by 3. (I'll try to explain that again more coherently later. But please see computational results for interesting "prime years" for up to 4 factors. | |
| Dec 29, 2010 at 7:43 | comment | added | Gerhard Paseman | If by n+2 or n+4 you mean n+2 or n+3, then I agree. Gerhard Paseman, 2010.12.28 | |
| Dec 29, 2010 at 7:40 | comment | added | sleepless in beantown | @Gerhard-Paseman, oops, I messed that up. n+1 has to be prime, n+2 has to be divisible by 2, n+3 does not have to be divisible by 3. But either n+2 or n+4 has to be divisble by 3. | |
| Dec 29, 2010 at 7:35 | history | edited | sleepless in beantown | CC BY-SA 2.5 |
added examples from computation
|
| Dec 29, 2010 at 7:23 | comment | added | Gerhard Paseman | n+3 does not have to be a multiple of 3. 383 is prime, and 385=5x7x11. Gerhard "Ask Me About Smooth Numbers" Paseman, 2010.12.28 | |
| Dec 29, 2010 at 4:51 | comment | added | sleepless in beantown | and as you say in your question, such a sequence requires that the base number be larger than the product of the first $k$ primes. | |
| Dec 29, 2010 at 4:45 | comment | added | sleepless in beantown | I meant to say that either "(n+2) or (n+4)" would be divisible by 4 while the other would be divisible by 2 but not by 4; not "s(n+2) or s(n+4)" | |
| Dec 29, 2010 at 4:39 | comment | added | sleepless in beantown | $n+3$ has to be divisible by $3$ and must have two other prime factors not including $2$. | |
| Dec 29, 2010 at 4:36 | history | edited | sleepless in beantown | CC BY-SA 2.5 |
s(n+k) can be a multiple of the product of $k$ prime numbers
|
| Dec 29, 2010 at 4:31 | history | answered | sleepless in beantown | CC BY-SA 2.5 |