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A347469
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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)).
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5
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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
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OFFSET
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1,2
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COMMENTS
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It is conjectured that this sequence is infinite.
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LINKS
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EXAMPLE
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(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.
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MATHEMATICA
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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 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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