A347469 For irrational r > 1, let B(r) denote the Beatty sequence for r. Let (s(n)) be the increasing sequence of numbers in both B(sqrt(2)) and B(sqrt(3)). Then this sequence gives the numbers k such that s(k) = floor(k*sqrt(6)).

1, 20, 21, 23, 49, 70, 71, 72, 73, 74, 75, 76, 78, 98, 101, 102, 117, 148, 194, 215, 216, 250, 257, 262, 299, 300, 307, 310, 344, 346, 357, 360, 361, 448, 1071, 1075, 1083, 1114, 1143, 1160, 1203, 1681, 1722, 1725, 1727, 1737, 1740, 1741, 1770, 1771, 1783

Offset: 1

Author

Clark Kimberling, Oct 31 2021

Keywords

• nonn

Comments

It is conjectured that this sequence is infinite.

Examples

			(B(sqrt(2)) and B(sqrt(3))) - B(sqrt(6)) = (0, -1, 1, 1, 3, 3, 5, 5, 5, 3, 5, 5, 3, 5, 4, 3, 2, 2, 1, 0, 0, -1, 0, -1, -1, -1, -1, -1, -1, -2, ...), so that a(4) = 23, the position of the 4th 0.
		

Crossrefs

Cf. A001951, A022838, A022840, A346308, A347467, A347468.

Programs

Mathematica

z = 10000; r = Sqrt[2]; s = Sqrt[3];
u = Table[Floor[n r], {n, 0, z}];  (*A001951*)
v = Table[Floor[n s], {n, 1, z}];  (*A022838*)
w = Intersection[u, v];  (*A346308*)
zz = -1 + Length[w];
t = Table[Floor[n*r*s], {n, 0, zz}]; (* A022840 *)
d = w - t;
Flatten[Position[d, 0]] (* A347469 *)