cp's OEIS Frontend

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)).

%I A347469 #10 Nov 20 2021 21:25:45
%S A347469 1,20,21,23,49,70,71,72,73,74,75,76,78,98,101,102,117,148,194,215,216,
%T A347469 250,257,262,299,300,307,310,344,346,357,360,361,448,1071,1075,1083,
%U A347469 1114,1143,1160,1203,1681,1722,1725,1727,1737,1740,1741,1770,1771,1783
%N 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)).
%C A347469 It is conjectured that this sequence is infinite.
%e A347469 (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.
%t A347469 z = 10000; r = Sqrt[2]; s = Sqrt[3];
%t A347469 u = Table[Floor[n r], {n, 0, z}];  (*A001951*)
%t A347469 v = Table[Floor[n s], {n, 1, z}];  (*A022838*)
%t A347469 w = Intersection[u, v];  (*A346308*)
%t A347469 zz = -1 + Length[w];
%t A347469 t = Table[Floor[n*r*s], {n, 0, zz}]; (* A022840 *)
%t A347469 d = w - t;
%t A347469 Flatten[Position[d, 0]] (* A347469 *)
%Y A347469 Cf. A001951, A022838, A022840, A346308, A347467, A347468.
%K A347469 nonn
%O A347469 1,2
%A A347469 _Clark Kimberling_, Oct 31 2021