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A187976
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a(n) = [nr+kr]-[nr]-[kr], where r=sqrt(2), k=6, [ ]=floor.
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4
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0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0
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OFFSET
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1
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COMMENTS
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The frequency of 1 in (a(n)) is equal to 6*sqrt(2)-8. This can be determined from the 6-block substitution sigma-hat of sigma given by sigma(1)=12, sigma(2)=121, which generates A006337. - Michel Dekking, Jan 24 2017
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LINKS
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FORMULA
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a(n) = [(n+6)*r] - [n*r] - [6*r], where r=sqrt(2).
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MATHEMATICA
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r=2^(1/2);
seqA=Table[Floor[(n+6)r]-Floor[n*r]-Floor[6r], {n, 1, 220}] (* A187976 *)
Flatten[Position[seqA, 0] ] (* A187977 *)
Flatten[Position[seqA, 1] ] (* A187978 *)
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PROG
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(PARI) for(n=1, 30, print1(floor((n+6)*sqrt(2)) - floor(n*sqrt(2)) - floor(6*sqrt(2)), ", ")) \\ G. C. Greubel, Jan 31 2018
(Magma) [Floor((n+6)*Sqrt(2)) - Floor(n*Sqrt(2)) - Floor(6*Sqrt(2)): n in [1..30]]; // G. C. Greubel, Jan 31 2018
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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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