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A189469
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a(n) = n + [n*s/r] + [n*t/r]; r=1, s=1+sqrt(2), t=1+sqrt(3).
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3
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5, 11, 18, 23, 30, 36, 42, 48, 54, 61, 67, 72, 79, 85, 91, 97, 104, 110, 115, 122, 128, 135, 140, 146, 153, 159, 165, 171, 178, 183, 189, 196, 202, 208, 214, 220, 227, 232, 239, 245, 251, 257, 263, 270, 275, 282, 288, 294, 300, 306, 313, 319, 324, 331, 337, 343, 349, 356, 362, 367, 374, 380, 387, 392, 398, 405, 411, 417, 423, 429, 435, 441, 448, 454, 460, 466, 472, 479, 484, 491, 497, 503
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OFFSET
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1,1
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COMMENTS
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This is one of three sequences that partition the positive integers. In general, suppose that r, s, t are positive real numbers for which the sets {i/r: i>=1}, {j/s: j>=1}, {k/t: k>=1} are pairwise disjoint. Let a(n) be the rank of n/r when all the numbers in the three sets are jointly ranked. Define b(n) and c(n) as the ranks of n/s and n/t. It is easy to prove that
f(n) = n + [n*s/r] + [n*t/r],
g(n) = n + [n*r/s] + [n*t/s],
h(n) = n + [n*r/t] + [n*s/t], where []=floor.
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LINKS
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MATHEMATICA
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r=1; s=1+Sqrt[2]; t=1+Sqrt[3];
f[n_]:= n + Floor[n*s/r] + Floor[n*t/r];
g[n_]:= n + Floor[n*r/s] + Floor[n*t/s];
h[n_]:= n + Floor[n*r/t] + Floor[n*s/t];
Table[f[n], {n, 1, 120}] (* A189469 *)
Table[g[n], {n, 1, 120}] (* A189470 *)
Table[h[n], {n, 1, 120}] (* A189471 *)
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PROG
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(PARI) for(n=1, 100, print1(n + floor(n*(1+sqrt(2))) + floor(n*(1+sqrt(3))), ", ")) \\ G. C. Greubel, Apr 13 2018
(Magma) [n + Floor(n*(1+Sqrt(2))) + Floor(n*(1+Sqrt(3))): n in [1..100]]; // G. C. Greubel, Apr 13 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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