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a(n) = n + [n*s/r] + [n*t/r]; r=1, s=1/sqrt(2), t=sqrt(3).
8

%I #12 Sep 22 2017 03:00:22

%S 2,6,10,12,16,20,23,26,30,34,37,40,44,47,50,54,58,61,64,68,71,75,78,

%T 81,85,89,92,95,99,102,105,109,113,116,119,123,127,129,133,137,140,

%U 143,147,151,153,157,161,164,167,171,175,178,181,185,188,191,195,199,202,205,209,212,216,219,222,226,230,233,236,240,243,246,250,254,257,260,264,268,270,274,278,281,284

%N a(n) = n + [n*s/r] + [n*t/r]; r=1, s=1/sqrt(2), t=sqrt(3).

%C 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

%C a(n) = n + [n*s/r] + [n*t/r],

%C b(n) = n + [n*r/s] + [n*t/s],

%C c(n) = n + [n*r/t] + [n*s/t], where []=floor.

%C Taking r=1, s=1/sqrt(2), t=sqrt(3) gives

%C a=A189395, b=A189396, c=A189397.

%H G. C. Greubel, <a href="/A189395/b189395.txt">Table of n, a(n) for n = 1..1000</a>

%t r=1; s=2^(-1/2); t=3^(1/2);

%t a[n_] := n + Floor[n*s/r] + Floor[n*t/r];

%t b[n_] := n + Floor[n*r/s] + Floor[n*t/s];

%t c[n_] := n + Floor[n*r/t] + Floor[n*s/t]

%t Table[a[n], {n, 1, 120}] (*A189395*)

%t Table[b[n], {n, 1, 120}] (*A189396*)

%t Table[c[n], {n, 1, 120}] (*A189397*)

%Y Cf. A189396, A189397, A189361, A189383, A189386.

%K nonn

%O 1,1

%A _Clark Kimberling_, Apr 21 2011