[seqfan] Re: ApSimon's Mints counterexample

[Tanya Khovanova]

There are 4 sequences in the database that have a subsequence 1, 2, 4, 8, 15, 28, 51. If the next term is not 92 or 93, then the ApSimon's mint sequence becomes unique.

Also if (12,12,7,7,1,2,0) and (12,0,8,2,7,3,2) is the only solution for 7 mints, then a(8) >= 64. Proof: If there is a solution for 8 mints, then removing the mint that uses the largest number of coins we get a solution for 7 mints.



________________________________
 From: David Applegate <david at research.att.com>
To: seqfan at list.seqfan.eu 
Cc: kostyaknop at gmail.com 
Sent: Friday, June 20, 2014 12:34 PM
Subject: [seqfan] Re: ApSimon's Mints counterexample
 

I can confirm Robert's minimum of 28 coins for 6 mints, and add 51 coins
for 7 mints, using (12,12,7,7,1,2,0) and (12,0,8,2,7,3,2).  I found
these using an exhaustive search written in C++.  In case someone is
interested, I've attached the undocumented (and not written to be
readable) source code.

The search took 2 seconds for 6 mints, and 29 minutes for 7 mints.
I'm running a search for 8 mints, which has established at least 59
coins, and is still running.  However, given the trend of nearly
doubling the number of coins for each additional mint, I expect that
the answer is over 100, and since the search space is (I believe)
exponential in the number of coins, that search is unlikely to get
there for a long time.

-Dave


_______________________________________________

Seqfan Mailing list - http://list.seqfan.eu/