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August 24th, 2015, 04:08 PM   #21
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 Originally Posted by al-mahed In that case, I find it unlikely since with only 11 numbers Mobel provided 11 5-tuples. Seems that adding 13 more numbers to choose from would pass the 13 mark.
Adding options does not necessarily increase the number of certain kinds of possible tuples.
ex: if you have many options (>=44) you can go 1-2-3-4-5, 1-2-6-7-8, 1-2-9-10-11, ... , 1-2- 3k - 3k+1 - 3k+2 , ...
but for a group of tuples that not everyone share the same pair, i.e. not the case above, think about why a pair cannot appear more than 4 times in all tuples. What kind of bad thing will happen?
Similarly, think about why a pair cannot appear exactly 3 times while also have tuples not containing that pair.

If you see why a pair cannot appear for more than 4 times, then imagine choosing the first tuple, the choice of which is immaterial. Suppose it's 1-2-3-4-5. Can you have more than 31 tuples in total, if each of the new tuple have to share a pair with 1-2-3-4-5, yet each pair in 1-2-3-4-5 cannot appear for more than 4 times? Having 14 options and having 14000000 options yield the same answer.

Using this, you can shrink the case with each pair appearing at most twice to 11 and shrink the case where each pair can appear four times to 13. The case when everyone share the same pair is relatively simple.

August 25th, 2015, 08:49 AM   #22
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Quote:
 Originally Posted by tanjz12 Adding options does not necessarily increase the number of certain kinds of possible tuples. ex: if you have many options (>=44) you can go 1-2-3-4-5, 1-2-6-7-8, 1-2-9-10-11, ... , 1-2- 3k - 3k+1 - 3k+2 , ... but for a group of tuples that not everyone share the same pair, i.e. not the case above, think about why a pair cannot appear more than 4 times in all tuples. What kind of bad thing will happen? Similarly, think about why a pair cannot appear exactly 3 times while also have tuples not containing that pair. If you see why a pair cannot appear for more than 4 times, then imagine choosing the first tuple, the choice of which is immaterial. Suppose it's 1-2-3-4-5. Can you have more than 31 tuples in total, if each of the new tuple have to share a pair with 1-2-3-4-5, yet each pair in 1-2-3-4-5 cannot appear for more than 4 times? Having 14 options and having 14000000 options yield the same answer. Using this, you can shrink the case with each pair appearing at most twice to 11 and shrink the case where each pair can appear four times to 13. The case when everyone share the same pair is relatively simple.
i do not understand what you mean when you talk about options.
Please can you explain it giving examples?

August 25th, 2015, 07:41 PM   #23
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 Originally Posted by mobel i do not understand what you mean when you talk about options. Please can you explain it giving examples?
Sorry, when I'm talking about "options", I just mean the symbols available for forming tuples.
Like if you changed the number of possible symbols from 11 to 12, the maximum number of tuples you can form is still 11.

 August 27th, 2015, 08:30 AM #24 Banned Camp   Joined: Dec 2013 Posts: 1,117 Thanks: 41 I abandon this thread. I came to the conclusion that combinatorics as long as it is hard few will be interested in. Thank you all for trying!
August 27th, 2015, 10:13 AM   #25
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 Originally Posted by mobel I abandon this thread. I came to the conclusion that combinatorics as long as it is hard few will be interested in. Thank you all for trying!
You got answers, it is not that hard, it is perhaps not so interesting as CRG said, especially if it is not related to any other nice result or open problem.

 August 27th, 2015, 10:23 AM #26 Banned Camp   Joined: Dec 2013 Posts: 1,117 Thanks: 41 For sure I got answers but not what I really needed. I thanked you for this. I did not publish more algorithms since no one tried those published. Im not programmer. Starting from today I will not post any subject related to combinatorics. All the results I have from posters are not new for me. It is not only here but in other forums too. Number theory and combinatorics finished! no more posts about those 2 subjects. Anyway thanks lot.
August 27th, 2015, 10:38 AM   #27
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Quote:
 Originally Posted by mobel Number theory and combinatorics finished! no more posts about those 2 subjects. Anyway thanks lot.
Number Theory? Oh no, that's a shame, don't do that . I mean, combinatorics ok (who cares...) but NT is the queen of all sciences.

August 27th, 2015, 10:54 AM   #28
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Quote:
 Originally Posted by al-mahed You got answers, it is not that hard, it is perhaps not so interesting as CRG said, especially if it is not related to any other nice result or open problem.
Quite the contrary: I think that the problem itself is interesting, though it's not my cup of tea (more combinatorial than number-theoretic). I'd like to see a solution for the problem, and I don't think anyone has proven that they have one. There have been ideas presented in this thread, and some of them may work, and even be efficient, but as far as I can tell no one understands all their details well enough to implement them.

If someone does, though, it would probably merit a new sequence at the OEIS.

 August 27th, 2015, 11:23 AM #29 Banned Camp   Joined: Dec 2013 Posts: 1,117 Thanks: 41 I have designed until now 6 algorithms to solve this problem. I have published on internet 3 of them. I did not abandon the problem. Im still thinking to solve it. But because I`m not programmer I will solve it with pen and paper. I have hard time to focus because of health problems. Anyway thank you all.

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