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A095190
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Doubled Thue-Morse sequence: the A010060 sequence replacing 0 by 0,0 and 1 by 1,1.
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5
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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; internal format)
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OFFSET
| 0,1
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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.
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FORMULA
| a(n) = A096273(n) mod 2 - Benoit Cloitre (benoit7848c(AT)orange.fr), Jun 29 2004
a(n)=mod(A000120(floor(n/2)), 2)=mod(A010060(floor(n/2)), 2). - Paul Barry (pbarry(AT)wit.ie), Jan 07 2005
a(n)=mod(-1+sum{k=0..n, mod(C(n, 2k), 2)}, 3); - Paul Barry (pbarry(AT)wit.ie), Jan 14 2005
a(n)=mod(log_2(sum{k=0..n, mod(C(n,2k),2)}),2); - Paul Barry (pbarry(AT)wit.ie), Jun 12 2006
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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 ...
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CROSSREFS
| Cf. A010059, A010060, A096288, A096289.
Sequence in context: A011657 A072126 A111113 * A131735 A131736 A152228
Adjacent sequences: A095187 A095188 A095189 * A095191 A095192 A095193
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KEYWORD
| easy,nonn
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AUTHOR
| Miklos Kristof and Peter Boros (kristmikl(AT)freemail.hu), Jun 21 2004
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