17 has 10 such permutations, and 18 has exactly 1. Computational challenge: find the unique permutation of 0..17 which has the property that the product of two consecutive elements when written in base 17 (and is not 1 digit in base 17) appears somewhere in this permutation. Victor On Mon, Mar 5, 2012 at 5:00 PM, Victor Miller <victorsmiller@gmail.com> wrote:
Here are a few more members of the sequences (here's the whole sequence starting at 1):
1,2,6,14,20,27,68,41,29,58,20,0,18,25,0,0
Hmm. Lot's of 0's.
Victor
On Mon, Mar 5, 2012 at 9:25 AM, Hans Havermann <gladhobo@teksavvy.com> wrote:
On Monday 05 March 2012 05:23:17 Hans Havermann wrote:
Victor Miller:
I wrote a backtracking program to calculate the number of such sequences for base b=2 through 13. Here are the numbers 2,6,14,20,27,68,41,29,58,20,0,18.
I believe that you may have included in your count numbers that begin with a zero.
Gareth McCaughan:
Quite right, too. Don't you think that's more natural?
If by 'natural' you mean 'complete', yes. I was trying to ascertain why the counts were different for bases 2-5 & 7-8 from Jean-Paul Davalan's solution sets at the bottom of this page:
http://www.cetteadressecomportecinquantesignes.com/DixChiffres.htm
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