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A082410
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a(1)=0 and a(n) is built with the rule: for any k>=0, if a(1),a(2),......,a(2^k+1) are known next 2^k terms are given as follows: a(2k^+1+i)=1-a(2^k+1-i) for 1<=i<=2^k
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4
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0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| It seems that a(n) is A014577 shifted right twice.
Complement of characteristic function of A060833.
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FORMULA
| n>=2 sum(k=1, n, a(k))=(n+A037834(n-1))/2
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EXAMPLE
| 3 first terms are 0,1,1 therefore a(4)=a(3+1)=1-a(3-1)=1-a(2)=0, a(5)=a(3+2)=1-a(3-2)=1-a(1)=1 and sequence begins 0,1,1,0,1,...
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CROSSREFS
| Sequence in context: A080886 A083924 A171587 * A094217 A174784 A092220
Adjacent sequences: A082407 A082408 A082409 * A082411 A082412 A082413
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KEYWORD
| nonn
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AUTHOR
| Benoit Cloitre (benoit7848c(AT)orange.fr), Apr 24 2003
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