Timeline for partitioning a number into two sets based on sum of digits
Current License: CC BY-SA 2.5
6 events
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| Feb 13, 2011 at 1:32 | comment | added | ARupinski | That is a nice short proof of the result I cited. I didn't feel like running through hard estimates on the maximal size of minimal partitions, so I decided just to give a sketch of the proof. Another method of proof I considered talking about was via the theory of minimal covering sets; see for example: M. Lothaire,"Combinatorics on Words" in Encylopedia of mathematics and its applications. Vol, 17, Addison-Wesley, 1983. pp. xix+238, ISBN 0-201-13516-7. MR 84g:05002 From Lothaire's result, one can deduce the finiteness of minimal partitions as a corollary, but this seemed too advanced. | |
| Feb 12, 2011 at 20:33 | comment | added | Gerhard Paseman | Also, arguing modulo i, any appropriately large multiset of i digits has a nonempty (consecutive) subset summing to 0 mod i, so I think one can bring the number of occurrences of i down to 9. I am still glad of your original analysis though. Gerhard "Ask Me About System Design" Paseman, 2011.02.12 | |
| Feb 12, 2011 at 20:25 | history | edited | Tony Huynh | CC BY-SA 2.5 |
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| Feb 12, 2011 at 20:24 | comment | added | Gerhard Paseman | You can also use divisibility conditions to improve the estimate for certain numbers. E. G. there are fewer than 36 occurences of 6. Gerhard "Ask Me About System Design" Paseman, 2011.02.12 | |
| Feb 12, 2011 at 20:12 | comment | added | Gerhard Paseman | You should be able to tighten the estimate to 43, as there are at most 8 nonzero digits to sum over in S2. Gerhard "Ask Me About System Design" Paseman, 2011.02.12 | |
| Feb 12, 2011 at 20:05 | history | answered | Tony Huynh | CC BY-SA 2.5 |