popular dilutions

Marc LeBrun mlb at well.com
Thu Oct 11 15:41:28 CEST 2001

 >> (note that the 5 that appears as the first dilution of 3 has a different
 >> "successor" than undiluted 5 does).

 > A dumb question: Why? (it has a different successor in that case)

Not dumb, it faked me out too, which is why I made a point to mention it.

The reason is that you don't dilute the diluting 0s, only the original bits.

The first two dilutions of ...xyz are ...0x0y0z then ...00x00y00z, but if 
you start over and redilute the first dilution value it's like diluting 
...abcdef with a=c=e=0, and b=x, d=y, f=z, which dilutes to ...0a0b0c0d0e0f 
= ...0000x000y000z.

The dilutions of ...0x0y0z are a subsequence of those of ...xyz.  As a 
consequence the potency calculation of ...0x0y0z can "skip over" the 
composite that determines the potency of ...xyz.

Also the recursive definition is subtly different than what one might at 
first think (ie program<;-).


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