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A329925
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Beatty sequence for (1+sqrt(41))/5.
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3
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1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20, 22, 23, 25, 26, 28, 29, 31, 32, 34, 35, 37, 38, 39, 41, 42, 44, 45, 47, 48, 50, 51, 53, 54, 56, 57, 59, 60, 62, 63, 65, 66, 68, 69, 71, 72, 74, 75, 76, 78, 79, 81, 82, 84, 85, 87, 88, 90, 91, 93, 94, 96, 97
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OFFSET
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1,2
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COMMENTS
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Let r = (1+sqrt(41))/5. Then (floor(n*r)) and (floor(n*r + 8r/5)) are a pair of Beatty sequences; i.e., every positive integer is in exactly one of the sequences. 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*r), where r = (1+sqrt(41))/5.
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MATHEMATICA
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t = 8/5; r = Simplify[(2 - t + Sqrt[t^2 + 4])/2]; s = Simplify[r/(r - 1)];
Table[Floor[r*n], {n, 1, 200}] (* A329925 *)
Table[Floor[s*n], {n, 1, 200}] (* A329926 *)
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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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