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A186447
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a(n)=a(floor(n/3)+a(n-1)*floor(n/4)) XOR a(floor(n/2))
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0
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1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1
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OFFSET
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0
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COMMENTS
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A simple unpredictable binary sequence.
Conjecture: All finite binary words appear in the sequence infinitely many times.
The sequence appears to have a slight bias towards 0. From n=0 through n=999, there are 510 1's. But after 10000 terms, the sequence has produced only 4900 1's. And after 10000000 terms, the sequence has produced 4910267 1's.
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LINKS
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EXAMPLE
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For n=20, a(n)=a(10) XOR a(floor(20/3)+a(19)*5)
=0 XOR a(11)=0 XOR 0 =0.
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MATHEMATICA
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f[0] = f[1] = 1; f[n_] := f[n] =
Mod[f[Floor[n/3] + f[n - 1] Floor[n/4]] + f[Floor[n/2]], 2]; Table[f[n], {n, 0, 100}]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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