Timeline for Happy New Prime Year!
Current License: CC BY-SA 2.5
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 3, 2011 at 8:51 | answer | added | Unknown | timeline score: 4 | |
| Jan 1, 2011 at 15:04 | comment | added | Steven Landsburg | 2011 is not just prime; it is also the sum of a prime number of consecutive primes: 2011=157+163+167+173+179+181+191+193+197+199+211 | |
| Dec 30, 2010 at 19:11 | answer | added | Ricardo Menares | timeline score: 7 | |
| Dec 30, 2010 at 16:54 | comment | added | Ross Millikan | No, but it must be a standard thing it looks for. It reports 2010 divides 29^6-1. | |
| Dec 30, 2010 at 0:28 | answer | added | Gerhard Paseman | timeline score: 3 | |
| Dec 29, 2010 at 21:31 | answer | added | Micah Milinovich | timeline score: 6 | |
| Dec 29, 2010 at 15:56 | comment | added | Joseph O'Rourke | 2011 divides $53^{10}-1$. Anyone know why (or how) Wolfram Alpha selects this fact to report? | |
| Dec 29, 2010 at 7:44 | comment | added | sleepless in beantown | Mea culpa. The first example with four factors does NOT have to be divisible by 210, nor 5 by 2310, as I show in an example below where starting with n=1866 and k=4 gets you the prime year 1867, the two-prime factor year 1868, the three-prime-factor year 1869, and the four-prime factor 1870. See answer below for factors. | |
| Dec 29, 2010 at 7:38 | comment | added | Gerhard Paseman | But then, that may be why you say "(infinitely many integers)" instead of "infinitely many integers". Gerhard "Reading A Second Time Helps" Paseman, 2010.12.28 | |
| Dec 29, 2010 at 7:34 | comment | added | Gerhard Paseman | If there is an example for each integer k, then it is also an example for each (k-1), so there are either infinitely many for each k, or there is a k for which there is none. Gerhard "Ask Me About System Design" Paseman, 2010.12.28 | |
| Dec 29, 2010 at 6:53 | comment | added | sleepless in beantown | In deference to @Frictionless-Jellyfish 's answer, I was thinking of deleting my comment-like answer, except for the fact that $M$ divisble by $k!$ does not make as much sense as having $M$ divisible by the product of the first $k$ prime numbers, or of $k$ distinct prime numbers. $2010$ factors into 2x3x5x67, and is not divisible by $4!=24$. The first example with 4 factors has to be divisible by 210, 5 factors has to be divisible by 2310. | |
| Dec 29, 2010 at 4:57 | comment | added | sleepless in beantown | Though in a cultural sense, the New Year starts (started) in Israel and India a bit earlier, as their years are phase shifted ahead of the US/EU calendars, and in China the New Year starts a bit later as their year's end is phase shifted after the calendar used to demarcate the passage of time on MO. And numerically they're using different starting points for their "zero" year along with being phase shifted as a difference. Do the Australians hang their calendars upside down? ;>) | |
| Dec 29, 2010 at 4:31 | answer | added | sleepless in beantown | timeline score: 8 | |
| Dec 29, 2010 at 4:26 | comment | added | Wadim Zudilin | Not yet but it will come much earlier than to other places. :-) | |
| Dec 29, 2010 at 4:22 | comment | added | Chandan Singh Dalawat | Is it already the new year in Australia ? | |
| Dec 29, 2010 at 4:17 | history | asked | Wadim Zudilin | CC BY-SA 2.5 |