[seqfan] Re: ApSimon's Mints counterexample

[israel at math.ubc.ca]

Yes, 28: P = [0,1,2,2,5,5], Q = [10,0,1,2,8,1]. And my program proves that 
this is optimal for 6 mints. Also that 8 is optimal for 4 mints, and 15 for 
5 mints. I'll see what it can do for 7 (probably will run overnight).

What I've been doing, by the way, is using  z3
(see http://z3.codeplex.com/).  I guess I'll have to put up a web page
with my programs.

Cheers,
Robert



On Jun 18 2014, israel at math.ubc.ca wrote:

>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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