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A329994
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Beatty sequence for 2^x, where 1/x^2 + 1/2^x = 1.
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3
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2, 4, 7, 9, 12, 14, 17, 19, 22, 24, 27, 29, 31, 34, 36, 39, 41, 44, 46, 49, 51, 54, 56, 59, 61, 63, 66, 68, 71, 73, 76, 78, 81, 83, 86, 88, 90, 93, 95, 98, 100, 103, 105, 108, 110, 113, 115, 118, 120, 122, 125, 127, 130, 132, 135, 137, 140, 142, 145, 147
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OFFSET
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1,1
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COMMENTS
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Let x be the solution of 1/x^2 + 1/2^x = 1. Then (floor(n x^2)) and (floor(n 2^x)) 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*2^x), where x = 1.298192... is the constant in A329992.
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MATHEMATICA
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r = x /. FindRoot[1/x^2 + 1/2^x == 1, {x, 1, 10}, WorkingPrecision -> 120]
Table[Floor[n*r^2], {n, 1, 250}] (* A329993 *)
Table[Floor[n*2^r], {n, 1, 250}] (* A329994 *)
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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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EXTENSIONS
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STATUS
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approved
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