Search: id:A035928
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%I A035928
%S A035928 2,10,12,38,42,52,56,142,150,170,178,204,212,232,240,542,558,598,614,
%T A035928 666,682,722,738,796,812,852,868,920,936,976,992,2110,2142,2222,2254,
%U A035928 2358,2390,2470,2502,2618,2650,2730,2762,2866,2898,2978,3010,3132,3164
%N A035928 Numbers n such that BCR(n) = n, where BCR = binary-complement-and-reverse 
               = take one's complement then reverse bit order.
%C A035928 Numbers n such that A036044(n) = n.
%C A035928 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
%F A035928 If offset were 0, a(2n+1) - a(2n) = 2^[log2(n)+1].
%e A035928 38 is such a number because 38=100110; complement to get 011001, then 
               reverse bit order to get 100110.
%p A035928 [seq(ReflectBinSeq(j,(floor_log_2(j)+1)),j=1..256)];
%p A035928 ReflectBinSeq := (x,n) -> (((2^n)*x)+binrevcompl(x));
%p A035928 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;
%p A035928 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))
%o A035928 (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,",")))
%o A035928 (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
%Y A035928 Cf. A061855.
%Y A035928 Cf. A000225.
%Y A035928 Sequence in context: A080139 A055701 A154391 this_sequence A014486 A166751 
               A071162
%Y A035928 Adjacent sequences: A035925 A035926 A035927 this_sequence A035929 A035930 
               A035931
%K A035928 nonn,nice,easy
%O A035928 1,1
%A A035928 Mike Keith (domnei(AT)aol.com)
%E A035928 More terms from Erich Friedman (erich.friedman(AT)stetson.edu).

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