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A189459
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a(n) = n + [n*s/r] + [n*t/r]; r=1, s=1+sqrt(2), t=1/2+sqrt(2).
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3
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4, 9, 15, 20, 26, 31, 36, 42, 47, 53, 58, 62, 68, 73, 79, 84, 90, 95, 100, 106, 111, 117, 122, 126, 132, 137, 143, 148, 154, 159, 164, 170, 175, 181, 185, 190, 196, 201, 207, 212, 217, 223, 228, 234, 239, 245, 249, 254, 260, 265, 271, 276, 281, 287, 292, 298, 303, 309, 313, 318, 324, 329, 335, 340, 345, 351, 356, 362, 367, 371, 377, 382, 388, 393, 399, 404, 409, 415, 420, 426, 431, 435, 441
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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+2^(1/2); t=1/2+2^(1/2);
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}] (* A189459 *)
Table[g[n], {n, 1, 120}] (* A189467 *)
Table[h[n], {n, 1, 120}] (* A189468 *)
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PROG
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(PARI) for(n=1, 100, print1(n + floor(n*(1+sqrt(2))) + floor(n*(1 + 2*sqrt(2))/2), ", ")) \\ G. C. Greubel, Apr 13 2018
(Magma) [n + Floor(n*(1 + Sqrt(2))) + Floor(n*(1 + 2*Sqrt(2))/2): 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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