Gareth, One of us miscalculated. I get 0000->(Empty): 0 0001->01: 2 0010->10: 2 0011->(Empty): 0 0100->01: 2 0101->0101: 4 0110->(Empty): 0 0111->01: 2 1000->10: 2 1001->(Empty): 0 1010->1010: 4 1011->10: 2 1100->(Empty): 0 1101->01: 2 1110->10: 2 1111->(Empty): 0 0+2+2+0+2+4+0+2+2+0+4+2+0+2+2+0=24 And 24/16=3/2, not 7/4. Can you find my mistake (or yours)? Jim On Thursday, May 26, 2016, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
On 26/05/2016 16:14, James Propp wrote:
One question that comes to mind is, if you randomly pick an integer between
2^(n-1) and 2^n-1, what's the expected number of uneliminated bits that remain, or equivalently, what is the expected number of bit-runs of odd length? (If you prefer, you can look at all bit strings of length n, whether they begin with a 1 or a 0; the answer is the same.)
I get the sequence of answers 1, 1, 3/2, 3/2, ...; I'm guessing the pattern continues as 2, 2, 5/2, 5/2, etc.
I get a different answer. Using the alternative formulation (all length-n bit-strings) and starting at length 0, I get 0, 1, 1, 3/2, 7/4, 17/8, 39/16, 89/32, ... which doesn't seem especially beautiful.
(The corresponding integer sequence 0, 2, 4, 12, 28, 68, ... doesn't appear to be in OEIS.)
Have I miscalculated?
-- g
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