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Numbers h such that floor(k*sqrt(3)) = floor(h*sqrt(2)) for some k.
4

%I #5 Nov 20 2021 21:25:07

%S 1,4,6,9,11,14,16,17,18,21,22,23,26,27,28,29,31,32,33,34,36,38,39,41,

%T 43,44,46,48,49,51,53,54,55,56,59,60,61,64,65,66,68,70,71,73,76,78,81,

%U 83,86,88,91,93,96,98,99,101,103,104,105,108,109,110,113

%N Numbers h such that floor(k*sqrt(3)) = floor(h*sqrt(2)) for some k.

%e Beatty sequence for sqrt(2): (1,2,4,5,7,8,9,11,12,14,...)

%e Beatty sequence for sqrt(3): (1,3,5,6,8,10,12,13,15,...)

%e Intersection: (1,5,8,12,...), as in A346308.

%e a(2) = 4 because floor(3*sqrt(3)) = floor(4*sqrt(2)). (For each such h, there is only one such k.)

%t z = 200; r = Sqrt[2]; s = Sqrt[3];

%t u = Table[Floor[n r], {n, 0, z}]; (*A001951*)

%t v = Table[Floor[n s], {n, 1, z}]; (*A022838*)

%t w = Intersection[u, v] (*A346308*)

%t zz = -1 + Length[w];

%t Table[Ceiling[w[[n]]/r], {n, 1, zz}] (* A347467 *)

%t Table[Ceiling[w[[n]]/s], {n, 1, zz}] (* A347468 *)

%Y Cf. A001951, A022838, A346308, A347468, A347469.

%K nonn

%O 1,2

%A _Clark Kimberling_, Oct 16 2021