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 A190329 a(n) = n + [n*s/r] + [n*t/r]; r=1, s=sqrt(2), t=1/s. 4
 2, 5, 9, 11, 15, 18, 20, 24, 27, 31, 33, 36, 40, 42, 46, 49, 53, 55, 58, 62, 64, 68, 71, 73, 77, 80, 84, 86, 90, 93, 95, 99, 102, 106, 108, 111, 115, 117, 121, 124, 126, 130, 133, 137, 139, 143, 146, 148, 152, 155, 159, 161, 164, 168, 170, 174, 177, 181, 183, 186, 190, 192, 196, 199, 201, 205, 208, 212, 214, 217, 221, 223, 227, 230 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS 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 a(n) = n + [n*s/r] + [n*t/r], b(n) = n + [n*r/s] + [n*t/s], c(n) = n + [n*r/t] + [n*s/t], where []=floor. Taking r=1, s=sqrt(2), t=1/s gives a=A190326, b=A190327, c=A190328. LINKS G. C. Greubel, Table of n, a(n) for n = 1..10000 FORMULA A190329:  a(n) = n + [n*sqrt(2)] + [n/sqrt(2)]. A190330:  b(n) = n + [n/sqrt(2)] + [n/2)]. A187338:  c(n) = 3*n + [n*sqrt(2)]. MATHEMATICA r=1; s=2^(1/2); t=1/s; a[n_] := n + Floor[n*s/r] + Floor[n*t/r]; b[n_] := n + Floor[n*r/s] + Floor[n*t/s]; c[n_] := n + Floor[n*r/t] + Floor[n*s/t]; Table[a[n], {n, 1, 120}]  (*A190329*) Table[b[n], {n, 1, 120}]  (*A190330*) Table[c[n], {n, 1, 120}]  (*A187338*) PROG (PARI) for(n=1, 50, print1(n + floor(n*sqrt(2)) + floor(n/sqrt(2)), ", ")) \\ G. C. Greubel, Jan 29 2018 (MAGMA) [n + Floor(n*Sqrt(2)) + Floor(n/Sqrt(2)): n in [1..50]]; // G. C. Greubel, Jan 29 2018 CROSSREFS Cf. A190330, A187338. Sequence in context: A189472 A190369 A067568 * A329791 A189522 A210510 Adjacent sequences:  A190326 A190327 A190328 * A190330 A190331 A190332 KEYWORD nonn AUTHOR Clark Kimberling, May 08 2011 STATUS approved

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Last modified October 27 19:33 EDT 2021. Contains 348287 sequences. (Running on oeis4.)