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%I #19 Mar 12 2022 07:56:34
%S 2,8,14,22,26,32,38,42,50,56,62,70,74,82,88,94,98,104,110,118,122,128,
%T 134,138,146,152,158,162,168,174,182,186,194,200,206,214,218,224,230,
%U 234,242,248,254,262,266,274,280,286,290,296,302,310,314,322,328
%N Positions of 1,1 in the Thue-Morse sequence (A010060).
%C Every positive integer lies in exactly one of these four sequences: A248056, A091855, A091855, A248057.
%H Clark Kimberling, <a href="/A248057/b248057.txt">Table of n, a(n) for n = 1..1000</a>
%F a(n) = 2*A091855(n) for n >= 1.
%e Thue-Morse sequence: 0,1,1,0,1,0,0,1,1,0,0,1,0,1,1,..., so that a(1) = 2 and a(2) = 8.
%t z = 400; u = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {1, 0}}] &, {0}, 9] (* A010060 *)
%t v = Rest[u]
%t t1 = Table[If[u[[n]] == 0 && v[[n]] == 0, 1, 0], {n, 1, z}];
%t t2 = Table[If[u[[n]] == 0 && v[[n]] == 1, 1, 0], {n, 1, z}];
%t t3 = Table[If[u[[n]] == 1 && v[[n]] == 0, 1, 0], {n, 1, z}];
%t t4 = Table[If[u[[n]] == 1 && v[[n]] == 1, 1, 0], {n, 1, z}];
%t Flatten[Position[t1, 1]] (* A248056 *)
%t Flatten[Position[t2, 1]] (* A091855 *)
%t Flatten[Position[t3, 1]] (* A091785 *)
%t Flatten[Position[t4, 1]] (* A248057 *)
%t SequencePosition[ThueMorse[Range[400]],{1,1}][[All,2]] (* Requires Mathematica version 10 or later *) (* _Harvey P. Dale_, May 16 2017 *)
%o (PARI) t(n)=hammingweight(n)%2;
%o for(n=1,500,if(t(n)==1&&t(n-1)==1,print1(n,", "))); \\ _Joerg Arndt_, Mar 12 2022
%Y Cf. A010060, A091855, A091785, A248056.
%K nonn,easy
%O 1,1
%A _Clark Kimberling_, Sep 30 2014
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