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n+[ns/r]+[nt/r]; r=1, s=sin(1), t=cos(1).
3

%I #4 Mar 30 2012 18:57:26

%S 1,4,6,9,11,14,15,18,20,23,25,28,30,32,35,37,40,42,44,46,49,51,54,56,

%T 59,61,63,66,68,71,73,75,77,80,82,85,87,89,92,94,97,99,102,104,106,

%U 108,111,113,116,119,120,123,125,128,130,133,134,137,139,142,144,147,150,151,154,156,159,161,164,165,168,170,173,175,178,180,182

%N n+[ns/r]+[nt/r]; r=1, s=sin(1), t=cos(1).

%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+[ns/r]+[nt/r],

%C b(n)=n+[nr/s]+[nt/s],

%C c(n)=n+[nr/t]+[ns/t], where []=floor.

%C Taking r=1, s=sin(1), t=cos(1) gives

%C a=A189756, b=A189757, c=A189758.

%F a(n)=n+[n*sin(1)]+[n*cos(1)].

%t r=1; s=Sin[1]; t=Cos[1];

%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}] (*A189756*)

%t Table[b[n], {n, 1, 120}] (*A189757*)

%t Table[c[n], {n, 1, 120}] (*A189758*)

%Y Cf. A189757, A189758.

%K nonn

%O 1,2

%A _Clark Kimberling_, Apr 26 2011