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%I #25 Mar 14 2024 12:14:50
%S 0,0,0,1,1,0,2,3,3,2,0,1,5,4,6,7,7,6,4,5,1,0,2,3,11,10,8,9,13,12,14,
%T 15,15,14,12,13,9,8,10,11,3,2,0,1,5,4,6,7,23,22,20,21,17,16,18,19,27,
%U 26,24,25,29,28,30,31,31,30,28,29,25,24,26,27,19,18,16
%N a(n) = not (n XOR (n shift 1)).
%C In other words, a(n) = the binary complement of (n XOR (n shift 1)). - _N. J. A. Sloane_, Mar 14 2024
%C Values of n for which a(n) = 0 are given by A000975(k).
%C Values of n for which a(n) = 1 are given by A097072(k) (for k>1).
%H Paolo P. Lava, <a href="/A279645/b279645.txt">Table of n, a(n) for n = 0..10000</a>
%H Paolo P. Lava, <a href="/A279645/a279645.txt">First occurrence of n in A279645, for n <= 1000</a>
%H Paolo P. Lava, <a href="/A279645/a279645.jpg">Graph of the first 10000 terms</a>
%H Paolo P. Lava, <a href="/A279645/a279645_1.jpg">Graph of the first occurrence of n in A279645, for n <= 1000</a>
%e 5044 converted to base 2 is 1001110110100.
%e Then consider each pair of adjacent bits starting from MSD: if they are equal, 00 or 11, set 1 otherwise 0:
%e (1+0) -> 0
%e (0+0) -> 1
%e (0+1) -> 0
%e (1+1) -> 1
%e (1+1) -> 1
%e (1+0) -> 0
%e (0+1) -> 0
%e (1+1) -> 1
%e (1+0) -> 0
%e (0+1) -> 0
%e (1+0) -> 0
%e (0+0) -> 1
%e We get 10110010001, so a(5044) = 1425.
%p P:=proc(n) local a,b,k; a:=0; b:=convert(n,base,2);
%p for k to nops(b)-1 do a:=a+((((b[k]+b[k+1]) mod 2)+1) mod 2)*2^(k-1); od; RETURN(a); end;
%p [seq(P(i),i=0..100)];
%p # alternative ( _R. J. Mathar_, Jun 22 2020)
%p A279645 := proc(n)
%p local dgs,L ;
%p dgs := convert(n,base,2) ;
%p L := [] ;
%p for i from 2 to nops(dgs) do
%p if op(i,dgs) = op(i-1,dgs) then
%p L := [op(L),1] ;
%p else
%p L := [op(L),0] ;
%p fi ;
%p end do:
%p add( op(i,L)*2^(i-1),i=1..nops(L)) ;
%p end proc:
%t a[n_] := Partition[IntegerDigits[n, 2], 2, 1] /. {b_Integer, c_Integer} :> If[b == c, 1, 0] // FromDigits[#, 2]&;
%t Table[a[n], {n, 0, 100}] (* _Jean-François Alcover_, Aug 08 2023 *)
%Y Cf. A000975, A038554, A097072.
%K nonn,easy,look
%O 0,7
%A _Paolo P. Lava_, Dec 16 2016
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