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A259726 Numbers k such that [r[s*k]] > [s[r*k]], where r = sqrt(2), s=sqrt(3), and [ ] = floor.

%I A259726 #14 Nov 15 2024 03:06:52
%S A259726 2,3,6,7,9,11,12,14,26,33,36,40,41,43,48,52,55,59,62,65,67,70,72,74,
%T A259726 77,82,84,89,91,93,94,96,101,108,111,115,118,119,122,123,125,130,134,
%U A259726 137,140,141,144,147,148,149,151,152,154,159,164,171,175,178,181
%N A259726 Numbers k such that [r[s*k]] > [s[r*k]], where r = sqrt(2), s=sqrt(3), and [ ] = floor.
%C A259726 Suppose that r and s are distinct real numbers, and let f(r,s,k) = [s[r*k]] - [r[s*k]]. Let (G(n)) be the sequence of those k for which f(r,s,k) > 0, let (E(n)) be those for which f(r,s,k) = 0, and (L(n)), those for which f(r,s,k) < 0. Clearly (G(n), E(n), L(n)) partition the positive integers. In particular, A259724, A259725, A259726 partition the positive integers.
%C A259726 Conjecture: the limits g = lim G(n)/n, e = lim E(n)/n, el = lim L(n)/n exist; if so, then 1/g + 1/e + 1/el = 1.
%H A259726 G. C. Greubel, <a href="/A259726/b259726.txt">Table of n, a(n) for n = 1..1000</a>
%t A259726 z = 1000; r = Sqrt[2]; s = Sqrt[3];
%t A259726 u = Table[Floor[r*Floor[s*n]], {n, 1, z}];
%t A259726 v = Table[Floor[s*Floor[r*n]], {n, 1, z}];
%t A259726 Select[Range[400], u[[#]] < v[[#]] &]  (* A259724 *)
%t A259726 Select[Range[200], u[[#]] == v[[#]] &] (* A259725 *)
%t A259726 Select[Range[200], u[[#]] > v[[#]] &]  (* A259726 *)
%Y A259726 Cf. A259724, A259745.
%K A259726 nonn,easy
%O A259726 1,1
%A A259726 _Clark Kimberling_, Jul 15 2015