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A114483
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s(1)={1}. s(2)={1,0}. If a(n) = 0, s(n+2) = s(n+1) U s(n) U {1}. If a(n) = 1, s(n+2) = s(n+1) U s(n+1) U {1}. (U represents concatenation of finite sequences.) {a(n)} is the limit of {s(n)} as n -> infinity.
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2
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1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1
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OFFSET
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1,1
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COMMENTS
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Number of terms in s(n) is A112361(n).
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LINKS
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EXAMPLE
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s(3) = {1,0,1,0,1}, s(4) = {1,0,1,0,1,1,0,1}, s(5) = {1,0,1,0,1,1,0,1,1,0,1,0,1,1,0,1,1}
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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