20161201, 14:46  #1 
"Mike"
Aug 2002
1111100101100_{2} Posts 
December 2016

20161202, 03:10  #3  
Romulan Interpreter
Jun 2011
Thailand
2^{3}·19·61 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 
"Mike"
Aug 2002
2^{2}·3·5·7·19 Posts 

20170102, 16:26  #5 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3×3,109 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
2×7×103 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
246F_{16} Posts 
Brilliant!

20170103, 03:06  #8 
Jan 2017
3·29 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
2×7×103 Posts 
a(0)=1 and it is optimal.

20170103, 16:01  #11 
Jan 2017
87_{10} 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 
Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
December 2017  Batalov  Puzzles  4  20180104 04:33 
December 2015  Xyzzy  Puzzles  15  20160106 10:23 
December 2014  Xyzzy  Puzzles  13  20150102 19:41 
Conference in Amsterdam 12 December  fivemack  Information & Answers  6  20111212 13:13 
Server update in December  ltd  Prime Sierpinski Project  4  20101217 13:14 