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%I #31 Jul 28 2023 10:36:48
%S 0,0,1,1,1,1,0,0,1,1,0,0,0,0,1,1,1,1,0,0,0,0,1,1,0,0,1,1,1,1,0,0,1,1,
%T 0,0,0,0,1,1,0,0,1,1,1,1,0,0,0,0,1,1,1,1,0,0,1,1,0,0,0,0,1,1,1,1,0,0,
%U 0,0,1,1,0,0,1,1,1,1,0,0,0,0,1,1,1,1,0,0,1,1,0,0,0,0,1,1,0,0,1,1,1,1,0,0,1
%N Doubled Thue-Morse sequence: a(2n) = A010060(n), a(2n+1) = A010060(n).
%C The A010060 sequence replacing 0 with 0,0 and 1 with 1,1.
%C Let n = Sum(c(k)*2^k), c(k) = 0,1, be the binary form of n, n = Sum(d(k)*3^k), d(k) = 0,1,2, the ternary form, n = Sum(e(k)*5^k), e(k) = 0,1,2,3,4, the base 5 form. Then a(n) = Sum(c(k)+d(k)) mod 2 = Sum(c(k)+e(k)) mod 2.
%H Claude Lenormand, <a href="/A318921/a318921.pdf">Deux transformations sur les mots</a>, Preprint, 5 pages, Nov 17 2003. Apparently unpublished. This is a scanned copy of the version that the author sent to me in 2003. - _N. J. A. Sloane_, Sep 09 2018. See page 2 for a different construction of this same sequence.
%H F. Mignosi, A. Restivo, and M. Sciortino, <a href="http://dx.doi.org/10.1016/S0304-3975(00)00436-9">Words and forbidden factors</a>, WORDS (Rouen, 1999). Theoret. Comput. Sci. 273 (2002), no. 1-2, 99--117. MR1872445 (2002m:68096). [_N. J. A. Sloane_, Jul 10 2012]
%F a(n) = A096273(n) mod 2. - _Benoit Cloitre_, Jun 29 2004
%F a(n) = mod(A000120(floor(n/2)), 2) = mod(A010060(floor(n/2)), 2). - _Paul Barry_, Jan 07 2005
%F a(n) = mod(-1 + Sum_{k=0..n} mod(C(n, 2k), 2), 3). - _Paul Barry_, Jan 14 2005
%F a(n) = mod(log_2(Sum_{k=0..n} mod(C(n, 2k),2)),2). - _Paul Barry_, Jun 12 2006
%e The Thue-Morse sequence is: 0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 ... so a(n) = 0 0 1 1 1 1 0 0 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 0 0 1 1 1 1 0 0 ...
%t a[n_] := Mod[DigitCount[Floor[n/2], 2, 1], 2]; Array[a, 100, 0] (* _Amiram Eldar_, Jul 28 2023 *)
%o (PARI) a(n)=hammingweight(n\2)%2 \\ _Charles R Greathouse IV_, May 08 2016
%Y Cf. A000120, A010059, A010060, A096273, A096288, A096289.
%K easy,nonn
%O 0,1
%A _Miklos Kristof_ and Peter Boros, Jun 21 2004
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