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Tanya Khovanova, <a href="http://arxiv.org/abs/1406.3012">Attacking ApSimon's Mints</a>, arXiv:1406.3012, [math.HO], 2014.

R. J. Mathar, <a href="http://arxiv.org/abs/1407.3613">ApSimon's mint problem with three or more weighings</a>, arXiv:1407.3613, [math.CO], 2014.

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Guy and Nowakowski give a(6) <=38 and a(7)<=74. Li improves this to a(6) <=31 and a(7)<=64. a(6)=28 is given by exhaustive search of all variants up to 27 coins and the solution (0,1,2,1,8,10) , , (1,2,2,5,5,0) with 1+2+2+5+8+10=28 coins. David Applegate finds a(7)=51 with (12,12,7,7,1,2,0), (12,0,8,2,7,3,2). - R. J. Mathar, Jun 20 2014

The unique solution for a(8)=90 is (27,1,12,12,6,1,0,4), (3,15,13,3,7,6,6,4) as determined by exhaustive search. There are a total of three solutions for a(7)=51: the one given above, (15,10,6,1,2,1,0), (0,10,9,7,4,4,2), and (15,6,9,1,4,3,1), (0,10,6,7,4,4,2). - David Applegate, Jul 03 2014

A pair of coin vectors gives a solution if every non-empty nonempty subset sum has a different ratio. (1,2,1,0) and (4,0,1,1) is a solution for 4 mints using 4+2+1+1=8 coins because 1:4, 2:0, 1:1, 0:1, (1+2):(4+0)=3:4, (1+1):(4+1)=2:5, (1+0):(4+1)=1:5, (2+1):(0+1)=3:1, (2+0):(0+1)=2:1, (1+0):(1+1)=1:2, (1+2+1):(4+0+1)=4:5, (1+1+0):(4+1+1)=2:6, (2+1+0):(0+1+1)=3:2, (1+2+0):(4+0+1)=3:5, (1+2+1+0):(4+0+1+1)=4:6 are all distinct ratios.

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Tanya Khovanova, <a href="http://blog.tanyakhovanova.com/2014/06/apsimons-mints/">ApSimon’s Mints</a>, Math Blog, June 2014.

Tanya Khovanova, <a href="http://blog.tanyakhovanova.com/2014/12/apsimons-mints-investigation/">ApSimon’s Mints Investigation</a>, Math Blog, December 2014.

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R. J. Mathar, <a href="http://arxiv.org/abs/1407.3613">ApSimon's mint problem with three or more weighings</a>, arXiv:1407.3613, 2014.

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