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A329826
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Beatty sequence for (5+sqrt(17))/4.
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2
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2, 4, 6, 9, 11, 13, 15, 18, 20, 22, 25, 27, 29, 31, 34, 36, 38, 41, 43, 45, 47, 50, 52, 54, 57, 59, 61, 63, 66, 68, 70, 72, 75, 77, 79, 82, 84, 86, 88, 91, 93, 95, 98, 100, 102, 104, 107, 109, 111, 114, 116, 118, 120, 123, 125, 127, 130, 132, 134, 136, 139
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OFFSET
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1,1
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COMMENTS
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Let r = (3+sqrt(17))/4. Then (floor(n*r)) and (floor(n*r + r/2)) are a pair of Beatty sequences; i.e., every positive integer is in exactly one of the sequences. The sequence (a(n) mod 2)) of 0's and 1's has only two run-lengths: 3 and 4. See the Guide to related sequences at A329825.
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LINKS
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FORMULA
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a(n) = floor(n*s), where s = (5+sqrt(17))/4.
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MATHEMATICA
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t = 1/2; r = Simplify[(2 - t + Sqrt[t^2 + 4])/2]; s = Simplify[r/(r - 1)];
Table[Floor[r*n], {n, 1, 200}] (* A329825 *)
Table[Floor[s*n], {n, 1, 200}] (* A329826 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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