[seqfan] Re: Variant Binary Representation and A068156
Frank Adams-Watters
franktaw at netscape.net
Sat Apr 18 01:17:19 CEST 2015
I am pretty sure that the smallest number so representable is represented by alternating 1's and 0's.[1] So, at step n, we are adding 2^(n-1) + 2^n = 3*2^(n-1), which agrees with A068156 (see the last formula there).
[1] This representation is not unique; as long as each pair has a 0 and a 1 you will get the same result. So, e.g., 1001 is as good as 1010.
Franklin T. Adams-Watters
-----Original Message-----
From: Dale Gerdemann <dale.gerdemann at gmail.com>
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
Sent: Fri, Apr 17, 2015 3:27 pm
Subject: [seqfan] Variant Binary Representation and A068156
In normal binary, the 1's represent powers of 2 and the 0's are just
place
holders. Suppose, in the interest of fairness and political correctness,
it
has been decided that the 1's and 0's should play equal roles. So a 1 with
n
0's to the left represents 2^n and likewise, a 0 with n 1's to the left
also
represents 2^n.
Now the question: given an equal number of 1's and 0's, what is
the
smallest number that you can construct? I get the answer A068156.
Any
thoughts on why this might be
true?
Dale
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