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Search: id:A035928
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| A035928 |
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Numbers n such that BCR(n) = n, where BCR = binary-complement-and-reverse = take one's complement then reverse bit order. |
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+0
7
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| 2, 10, 12, 38, 42, 52, 56, 142, 150, 170, 178, 204, 212, 232, 240, 542, 558, 598, 614, 666, 682, 722, 738, 796, 812, 852, 868, 920, 936, 976, 992, 2110, 2142, 2222, 2254, 2358, 2390, 2470, 2502, 2618, 2650, 2730, 2762, 2866, 2898, 2978, 3010, 3132, 3164
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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Numbers n such that A036044(n) = n.
Also: numbers such that n+BR(n) is in A000225={2^k-1} (with BR = binary reversed). - M. F. Hasler (Maximilian.Hasler(AT)gmail.com), Dec 17 2007
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FORMULA
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If offset were 0, a(2n+1) - a(2n) = 2^[log2(n)+1].
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EXAMPLE
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38 is such a number because 38=100110; complement to get 011001, then reverse bit order to get 100110.
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MAPLE
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[seq(ReflectBinSeq(j, (floor_log_2(j)+1)), j=1..256)];
ReflectBinSeq := (x, n) -> (((2^n)*x)+binrevcompl(x));
binrevcompl := proc(nn) local n, z; n := nn; z := 0; while(n <> 0) do z := 2*z + ((n+1) mod 2); n := floor(n/2); od; RETURN(z); end;
floor_log_2 := proc(n) local nn, i: nn := n; for i from -1 to n do if(0 = nn) then RETURN(i); fi: nn := floor(nn/2); od: end; # Computes essentially the same as floor(log[2](n))
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PROGRAM
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(PARI) for(n=1, 1000, l=length(binary(n)); b=binary(n); if(sum(i=1, l, abs(component(b, i)-component(b, l+1-i)))==l, print1(n, ", ")))
(PARI) for(i=0, 999, if(Set(vecextract(t=binary(i), "-1..1")+t)==["1"], print1(i", "))) - M. F. Hasler (Maximilian.Hasler(AT)gmail.com), Dec 17 2007
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CROSSREFS
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Cf. A061855.
Cf. A000225.
Sequence in context: A080139 A055701 A154391 this_sequence A014486 A166751 A071162
Adjacent sequences: A035925 A035926 A035927 this_sequence A035929 A035930 A035931
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KEYWORD
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nonn,nice,easy
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AUTHOR
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Mike Keith (domnei(AT)aol.com)
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EXTENSIONS
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More terms from Erich Friedman (erich.friedman(AT)stetson.edu).
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