Biquanimous numbers

[David W. Wilson]

Bill Thurston:  Thanks for your detailed discussion of this problem I
presented.  My viscera are vindicated.  I think I understand the crux
of the argument, but there are some details that I don't quite understand.
Right now, I have other things to do (they expect me to do work here),
but I will take this subject up with you some other time in order to
fully understand the FA construction.

Others: Thanks for your replies.  The biquanimous numbers (base 2 thru 10)
should be EIS sequences, I think (I don't have time to generate them right
now).  Also, I wanted to count the biquanimous numbers of various lengths,
out of curiosity.  I suspect that in the limit, 1/2 of the base-b numbers
are biquanimous, which could be proved or disproved from the FA construction.

My viscera are acting up again... I suspect that in the limit, half of
the base-b numbers are biquanimous.  I can prove this true of bases 2
and 3.  The quick statistacal argument:  Half the numbers have odd digit
sum, hence are non-biquanimous.  The remaining half have even digit sum,
a long random one of these would very likely split into two same-sum
pieces, hence almost all of these are biquanimous.

- Dave Wilson