[seqfan] Re: ApSimon's Mints counterexample

[israel at math.ubc.ca]

For 6 mints I can get 29 coins:

P = [ 0 3 3 3 4 4], Q = [ 10 0 1 5 1 3]

Trying for 28...

Cheers,
Robert

On Jun 18 2014, Tanya Khovanova wrote:

>Dear SeqFans,
>
> Konstantin Knop posted an example for 6 mints using 30 coins as a comment 
> for my blog post:
>
>P=(0,1,1,2,4,8) and Q=(9,6,1,1,1,1) ->  sum = 9+6+1+2+4+8=30.
>
>I checked it. It gives different ratios.
>http://blog.tanyakhovanova.com/?p=501
>
>
>That means Konstantin proved that the sequence http://oeis.org/A007673
>is wrong.
>
> BTW, someone should checked 4 and 5 mints. ApSimon gives two solutions 
> for 4 mints with 8 coins: P=(0,1,2,3), Q=(1,2,2,0) and P=(0,1,1,4), 
> Q=(2,0,1,1). And one solution for 5 coins with 15 mints: P=(1,0,1,4,5), 
> Q=(1,2,2,5,0). He claims that he proved optimality, but it is not 
> included in the book.
>
>Tanya
>
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