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A138954
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Number of complement symmetries in the rotations of the binary expansion of a number.
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2
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0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
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OFFSET
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0,11
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COMMENTS
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It seems that the number of complement rotational symmetries is nonzero iff #0 = #1 in the binary expansion of a number.
The above statement is true in only one direction. It is clearly necessary for the number of 1 bits to equal the number of 0 bits. However, this is not sufficient. The first counterexample is n = 37 with binary expansion 100101 and complement 011010. Values of n for which a(n) is nonzero are therefore a proper subset of A031443. - Andrew Howroyd, Jan 12 2020
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LINKS
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EXAMPLE
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a(2) = 1 because 2 has binary expansion 10 and the complement shows up once in rotations;
a(10) = 2 because 10 has binary expansion 1010 and its complement shows up twice in rotations.
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PROG
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(PARI) a(n)={my(s=0); if(n, my(b=logint(n, 2)+1); if(2*hammingweight(n)==b, my(w=2^b-1-n); for(i=2, b, w=if(w%2, w+2^b, w)\2; if(w==n, s++)))); s} \\ Andrew Howroyd, Jan 12 2020
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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Missing a(8) inserted and terms a(21) and beyond from Andrew Howroyd, Jan 12 2020
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STATUS
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approved
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