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Nov 30, 2018 at 12:11 comment added Claude Chaunier I agree, this gives some structure to the problem :)
Nov 30, 2018 at 4:22 history bounty ended Harry Altman
Nov 30, 2018 at 4:21 vote accept Harry Altman
Nov 30, 2018 at 4:21 comment added Harry Altman OK, Claude was first, but this is easily the most helpful answer, so I'm going to award this one the bounty!
Nov 28, 2018 at 9:20 history edited RaphaelB4 CC BY-SA 4.0
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Nov 28, 2018 at 9:03 history edited RaphaelB4 CC BY-SA 4.0
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Nov 28, 2018 at 8:45 comment added RaphaelB4 @Claude Chaunier yes that it (sorry I made the calculation on paper for weight = 1,2,...,n and I miss the 0)! Ok let me correct it.
Nov 28, 2018 at 5:25 comment added Harry Altman Oh, wonderful! With this, it's easy to compute; no need to program anything complicated. Can check possibilities much more easily. :) You may want to go back and clean this up, you seem to have gotten some things mixed up in the middle, but your statement in the comment I can certainly verify.
Nov 27, 2018 at 20:46 comment added Claude Chaunier I keep getting this wrong. Here is your great result in another form : $v_1(1 b_1b_2\dots b_n) = 1 + v_1(b_1b_2\dots b_n)$ and $v_1(0b_1b_2\dots b_n) = \frac{1}{n+2} + \frac{n+3}{n+2} v_1(b_1b_2\dots b_n)$.
Nov 27, 2018 at 17:06 history answered RaphaelB4 CC BY-SA 4.0