20161201, 14:46  #1 
Aug 2002
20446_{8} Posts 
December 2016

20161202, 03:10  #3  
Romulan Interpreter
"name field"
Jun 2011
Thailand
10011100011101_{2} Posts 
Quote:
Quote:
Programming an algorithm for higher N, however, may turn out to be... interesting... I will go for it this month. edit: OTOH, I agree with Serge, formulation is silly, they could say "weighting coins", or everything else (we just had a similar problem recently), eventually it comes to that: having 27 coins and being allowed 4 weightings with a "non weightmarked" balance scale, tell if all coins have the same weight.. Last fiddled with by LaurV on 20161202 at 03:16 

20170102, 15:18  #4 
Aug 2002
2·4,243 Posts 

20170102, 16:26  #5 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
10011010110001_{2} Posts 
Believe it or not, I've been waiting for the solution, because I expected it to be illuminating  and it certainly was. This is not in OEIS {2,4, ?, 30,114, ...}
I am now playing with this technique to see that this nontrivial (i.e. not a power of 2) set of solutions only arises at n>=4. I spent a few weekends tinkering with n=3, and it would explain why this was a dead end. Is a(3) = 8? 
20170102, 21:33  #6  
"Robert Gerbicz"
Oct 2005
Hungary
1573_{10} Posts 
Quote:
BCDE;FGIJ GIJ;AFH BCDEF;AGHIJ it was found very quickly with my code that used simply random permutations on each side. Not solved this puzzle, but I wasn't that far: for the first contest and for N=30 there are only 15 different cases, up to symmetry, for the 2nd contest there is still not that many cases, but if you want only a few cases to check then you can arrive to the solution's group idea (use very few groups). 

20170102, 21:41  #7 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
5·7·283 Posts 
Brilliant!

20170103, 03:06  #8 
Jan 2017
149 Posts 
Optimal answers to the original problem are a(1) = 2, a(2) = 4, a(3) = 10.
Optimal values for the "linear algebra with contest_count+1 groups" approach should be a(1) = 2, a(2) = 4, a(3) = 10, a(4) = 30, a(5) = 114, a(6) = 454. I haven't given much thought to how plausible it is that there could be a better solution which does not divide into contests+1 groups that way. 
20170103, 07:51  #10 
"Robert Gerbicz"
Oct 2005
Hungary
11^{2}·13 Posts 
a(0)=1 and it is optimal.

20170103, 16:01  #11 
Jan 2017
95_{16} Posts 
If you want the 6contest solutions for the sequence, here are the two ways to get 454:
Code:
group sizes 108, 90, 80, 58, 44, 39, 35 1 1 1 0 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 1 1 0 1 1 1 1 1 1 group sizes 130, 85, 76, 62, 42, 35, 24 1 1 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 0 1 1 1 1 1 1 
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