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 A095190 Doubled Thue-Morse sequence: the A010060 sequence replacing 0 with 0,0 and 1 with 1,1. 7
 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, 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, 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,1 COMMENTS If b(n)=A010060, then a(2n)=b(n), a(2n+1)=b(n). 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. REFERENCES Mignosi, F.; Restivo, A.; Sciortino, M. Words and forbidden factors. WORDS (Rouen, 1999). Theoret. Comput. Sci. 273 (2002), no. 1-2, 99--117. MR1872445 (2002m:68096) - From N. J. A. Sloane, Jul 10 2012 LINKS FORMULA a(n) = A096273(n) mod 2. - Benoit Cloitre, Jun 29 2004 a(n) = mod(A000120(floor(n/2)), 2) = mod(A010060(floor(n/2)), 2). - Paul Barry, Jan 07 2005 a(n)=mod(-1+sum{k=0..n, mod(C(n, 2k), 2)}, 3). - Paul Barry, Jan 14 2005 a(n)=mod(log_2(sum{k=0..n, mod(C(n,2k),2)}),2). - Paul Barry, Jun 12 2006 EXAMPLE 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 ... PROG (PARI) a(n)=hammingweight(n\2)%2 \\ Charles R Greathouse IV, May 08 2016 CROSSREFS Cf. A010059, A010060, A096288, A096289. Sequence in context: A011657 A072126 A111113 * A131735 A131736 A152228 Adjacent sequences:  A095187 A095188 A095189 * A095191 A095192 A095193 KEYWORD easy,nonn AUTHOR Miklos Kristof and Peter Boros, Jun 21 2004 STATUS approved

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