Timeline for A prime sequence can be partitioned into two sets of equal or consecutive sum
Current License: CC BY-SA 3.0
16 events
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| Jan 6, 2015 at 19:40 | answer | added | The Masked Avenger | timeline score: 0 | |
| Jan 6, 2015 at 19:23 | comment | added | The Masked Avenger | This will also follow from the assertion that for n large enough, all but 6 positive integers between 0 and S_n, the sum of the first n primes, are realized as a subset sum, with the exceptions smaller than 7 or bigger than S_n - 7. | |
| S Jan 6, 2015 at 18:58 | history | suggested | alexwlchan | CC BY-SA 3.0 |
added mathjax and formatting to make the post easier to read
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| Jan 6, 2015 at 18:46 | review | Suggested edits | |||
| S Jan 6, 2015 at 18:58 | |||||
| Aug 15, 2010 at 11:52 | vote | accept | user8140 | ||
| Aug 15, 2010 at 11:30 | vote | accept | user8140 | ||
| Aug 15, 2010 at 11:50 | |||||
| Aug 13, 2010 at 9:22 | answer | added | dvitek | timeline score: 5 | |
| Aug 10, 2010 at 17:18 | comment | added | S. Carnahan♦ | Sorry, by "small case analysis at the end" I meant that the smallest primes needed to be arranged by hand, not that the greedy algorithm could be proved to work by case analysis. | |
| Aug 5, 2010 at 6:57 | comment | added | user8140 | Thank you, Scoot,Mariano. @Mariano,Yes! According to the greedy algorithm: (31+19+17+7+5) -(29+23+13+11+3)=79-79=0, then how to put 2? While (31+29+17+3)=80, 23+13+19+11+7+5+2=80, to be more beautiful, 31+29-23-19+17-13-11-7-5+3-2=0 | |
| Aug 5, 2010 at 5:57 | vote | accept | user8140 | ||
| Aug 5, 2010 at 6:58 | |||||
| Aug 5, 2010 at 5:48 | comment | added | Mariano Suárez-Álvarez | @Scott, I think that when you start with the first 11 primes that greedy algorithm does not work. | |
| Aug 5, 2010 at 5:11 | comment | added | Mariano Suárez-Álvarez | Google should find various algorithms to solve the so-called partition problem en.wikipedia.org/wiki/Partition_problem, which can be solved efficiently in lots of cases, it seems. | |
| Aug 5, 2010 at 5:11 | comment | added | S. Carnahan♦ | There is no need to write a program, because the statement is true. Fix n, and starting from the largest prime in P[n] and following decreasing order, put each prime into the pile with the smaller sum. Standard estimates like Bertrand's postulate (and improvements - see Wikipedia) imply the difference between the two piles will be less than, say, 3/2 the next prime on the list. This reduces to a small case analysis at the end. | |
| Aug 5, 2010 at 5:02 | answer | added | Darsh Ranjan | timeline score: 1 | |
| Aug 5, 2010 at 4:15 | history | edited | user8140 | CC BY-SA 2.5 |
added 3 characters in body
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| Aug 5, 2010 at 4:04 | history | asked | user8140 | CC BY-SA 2.5 |