cp's OEIS Frontend

A189386 a(n) = n+[ns/r]+[nt/r]; r=1, s=sqrt(2), t=1/sqrt(3), []=floor.

%I A189386 #10 Oct 20 2017 14:30:53
%S A189386 2,5,8,11,14,17,20,23,26,29,32,34,38,41,44,47,50,53,55,59,62,65,68,70,
%T A189386 74,77,80,83,86,89,91,95,98,101,104,106,110,112,116,119,121,125,127,
%U A189386 131,133,137,140,142,146,148,152,155,157,161,163,167,169,173,176,178,182,184,188,190,193,197,199,203,205,208,211,214,218,220,224,226,229,233,235,239,241,244,247
%N A189386 a(n) = n+[ns/r]+[nt/r]; r=1, s=sqrt(2), t=1/sqrt(3), []=floor.
%C A189386 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 A189386 a(n)=n+[ns/r]+[nt/r],
%C A189386 b(n)=n+[nr/s]+[nt/s],
%C A189386 c(n)=n+[nr/t]+[ns/t], where []=floor.
%C A189386 Taking r=1, s=sqrt(2), t=1/sqrt(3) gives
%C A189386 a=A189386, b=A189387, c=A189388.
%H A189386 G. C. Greubel, <a href="/A189386/b189386.txt">Table of n, a(n) for n = 1..5000</a>
%t A189386 r=1; s=2^(1/2); t=3^(-1/2);
%t A189386 a[n_] := n + Floor[n*s/r] + Floor[n*t/r];
%t A189386 b[n_] := n + Floor[n*r/s] + Floor[n*t/s];
%t A189386 c[n_] := n + Floor[n*r/t] + Floor[n*s/t]
%t A189386 Table[a[n], {n, 1, 120}]  (*A189386*)
%t A189386 Table[b[n], {n, 1, 120}]  (*A189387*)
%t A189386 Table[c[n], {n, 1, 120}]  (*A189388*)
%Y A189386 Cf. A189387, A189388, A189361, A189383, A189395.
%K A189386 nonn
%O A189386 1,1
%A A189386 _Clark Kimberling_, Apr 21 2011