cp's OEIS Frontend

A189361 a(n) = n + floor(n*s/r) + floor(n*t/r); r=1, s=sqrt(2), t=sqrt(3).

%I A189361 #10 Sep 22 2017 03:00:30
%S A189361 3,7,12,15,20,24,28,32,36,41,45,48,53,57,61,65,70,74,77,82,86,91,94,
%T A189361 98,103,107,111,115,120,123,127,132,136,140,144,148,153,156,161,165,
%U A189361 169,173,177,182,185,190,194,198,202,206,211,215,218,223,227,231,235,240,244,247,252,256,261,264,268,273,277,281,285,289,293,297,302,306,310,314,318,323,326,331,335,339,343
%N A189361 a(n) = n + floor(n*s/r) + floor(n*t/r); r=1, s=sqrt(2), t=sqrt(3).
%C A189361 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 A189361 a(n) = n + [n*s/r] + [n*t/r],
%C A189361 b(n) = n + [n*r/s] + [n*t/s],
%C A189361 c(n) = n + [n*r/t] + [n*s/t], where []=floor.
%C A189361 Taking r=1, s=sqrt(2), t=sqrt(3) gives
%C A189361 a=A189361, b=A189362, c=A189363.
%H A189361 G. C. Greubel, <a href="/A189361/b189361.txt">Table of n, a(n) for n = 1..1000</a>
%t A189361 r = 1; s = 2^(1/2); t = 3^(1/2);
%t A189361 a[n_] := n + Floor[n*s/r] + Floor[n*t/r];
%t A189361 b[n_] := n + Floor[n*r/s] + Floor[n*t/s];
%t A189361 c[n_] := n + Floor[n*r/t] + Floor[n*s/t]
%t A189361 Table[a[n], {n, 1, 120}]  (*A189361*)
%t A189361 Table[b[n], {n, 1, 120}]  (*A189362*)
%t A189361 Table[c[n], {n, 1, 120}]  (*A189363*)
%Y A189361 Cf. A189362, A189363, A184812, A189383, A189386, A189395.
%K A189361 nonn
%O A189361 1,1
%A A189361 _Clark Kimberling_, Apr 20 2011