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

%I #10 Sep 23 2017 13:05:52

%S 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 77,82,84,89,91,93,94,96,101,108,111,115,118,119,122,123,125,130,134,

%U 137,140,141,144,147,148,149,151,152,154,159,164,171,175,178,181

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

%C 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.

%C Conjecture: the limits g = lim G(n)/n, e = lim E(n)/n, el = lim L(n) exist; if so, then 1/g + 1/e + 1/el = 1.) In particular, A259724, A259725 and A259726 partition the positive integers.

%H G. C. Greubel, <a href="/A259726/b259726.txt">Table of n, a(n) for n = 1..1000</a>

%t z = 1000; r = Sqrt[2]; s = Sqrt[3];

%t u = Table[Floor[r*Floor[s*n]], {n, 1, z}];

%t v = Table[Floor[s*Floor[r*n]], {n, 1, z}];

%t Select[Range[400], u[[#]] < v[[#]] &] (* A259724 *)

%t Select[Range[200], u[[#]] == v[[#]] &] (* A259725 *)

%t Select[Range[200], u[[#]] > v[[#]] &] (* A259726 *)

%Y Cf. A259724, A259745.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Jul 15 2015

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