A351415
Intersection of Beatty sequences for (1+sqrt(5))/2 and sqrt(5).
Original entry on oeis.org
4, 6, 8, 11, 17, 22, 24, 29, 33, 35, 38, 40, 42, 46, 51, 53, 55, 58, 64, 67, 69, 71, 76, 80, 82, 84, 87, 93, 98, 100, 105, 111, 114, 116, 118, 122, 127, 129, 131, 134, 140, 145, 147, 152, 156, 158, 160, 163, 165, 169, 174, 176, 181, 187, 190, 192, 194, 199
Offset: 1
The two Beatty sequences are (1,3,4,6,8,9,11,12,14,...) and (2,4,6,8,11,13,15,17,...), with common terms forming the sequence (4,6,8,11,...).
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z = 200;
r = (1 + Sqrt[5])/2; u = Table[Floor[n*r], {n, 1, z}] (* A000201 *)
u1 = Take[Complement[Range[1000], u], z] (* A001950 *)
r1 = Sqrt[5]; v = Table[Floor[n*r1], {n, 1, z}] (* A022839 *)
v1 = Take[Complement[Range[1000], v], z] (* A108598 *)
Intersection[u, v] (* A351415 *)
Intersection[u, v1] (* A356101 *)
Intersection[u1, v] (* A356102 *)
Intersection[u1, v1] (* A356103 *)
Original entry on oeis.org
1, 3, 9, 12, 14, 16, 19, 21, 25, 27, 30, 32, 37, 43, 45, 48, 50, 56, 59, 61, 63, 66, 72, 74, 77, 79, 85, 88, 90, 92, 95, 97, 101, 103, 106, 108, 110, 113, 119, 121, 124, 126, 132, 135, 137, 139, 142, 144, 148, 150, 153, 155, 161, 166, 168, 171, 173, 177, 179
Offset: 1
Starting with a general overview, suppose that u = (u(n)) and v = (v(n)) are increasing sequences of positive integers. Let u' and v' be their complements, and assume that the following four sequences are infinite:
(1) u ^ v = intersection of u and v (in increasing order);
(2) u ^ v';
(3) u' ^ v;
(4) u' ^ v'.
Every positive integer is in exactly one of the four sequences. For A351415, u, v, u', v', are the Beatty sequences given by u(n) = floor(n*(1+sqrt(5))/2) and v(n) = floor(n*sqrt(5)), so r = (1+sqrt(5))/2, s = sqrt(5), r' = (3+sqrt(5))/2, s' = (5 + sqrt(5))/4.
(1) u ^ v = (4, 6, 8, 11, 17, 22, 24, 29, 33, 35, 38, 40, 42, ...) = A351415
(2) u ^ v' = (1, 3, 9, 12, 14, 16, 19, 21, 25, 27, 30, 32, 37, ...) = A356101
(3) u' ^ v = (2, 13, 15, 20, 26, 31, 44, 49, 60, 62, 73, 78, ...) = A356102
(4) u' ^ v' = (5, 7, 10, 18, 23, 28, 34, 36, 39, 41, 47, 52, 54, ...) = A356103
-
z = 200;
r = (1 + Sqrt[5])/2; u = Table[Floor[n*r], {n, 1, z}] (* A000201 *)
u1 = Take[Complement[Range[1000], u], z] (* A001950 *)
r1 = Sqrt[5]; v = Table[Floor[n*r1], {n, 1, z}] (* A022839 *)
v1 = Take[Complement[Range[1000], v], z] (* A108598 *)
Intersection[u, v] (* A351415 *)
Intersection[u, v1] (* A356101 *)
Intersection[u1, v] (* A356102 *)
Intersection[u1, v1] (* A356103 *)
Original entry on oeis.org
5, 7, 10, 18, 23, 28, 34, 36, 39, 41, 47, 52, 54, 57, 65, 68, 70, 75, 81, 83, 86, 94, 99, 104, 112, 115, 117, 123, 128, 130, 133, 141, 146, 151, 157, 159, 162, 164, 170, 175, 180, 188, 191, 193, 198, 204, 206, 209, 217, 222, 227, 233, 235, 238, 240, 246, 251
Offset: 1
Starting with a general overview, suppose that u = (u(n)) and v = (v(n)) are increasing sequences of positive integers. Let u' and v' be their complements, and assume that the following four sequences are infinite:
(1) u ^ v = intersection of u and v (in increasing order);
(2) u ^ v';
(3) u' ^ v;
(4) u' ^ v'.
Every positive integer is in exactly one of the four sequences. For A351415, u, v, u', v', are the Beatty sequences given by u(n) = floor(n*(1+sqrt(5))/2) and v(n) = floor(n*sqrt(5)), so that r = (1+sqrt(5))/2, s = sqrt(5), r' = (3+sqrt(5))/2, s' = (5 + sqrt(5))/4.
(1) u ^ v = (4, 6, 8, 11, 17, 22, 24, 29, 33, 35, 38, 40, 42, ...) = A351415
(2) u ^ v' = (1, 3, 9, 12, 14, 16, 19, 21, 25, 27, 30, 32, 37, ...) = A356101
(3) u' ^ v = (2, 13, 15, 20, 26, 31, 44, 49, 60, 62, 73, 78, ...) = A356102
(4) u' ^ v' = (5, 7, 10, 18, 23, 28, 34, 36, 39, 41, 47, 52, 54, ...) = A356103
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z = 200;
r = (1 + Sqrt[5])/2; u = Table[Floor[n*r], {n, 1, z}] (* A000201 *)
u1 = Take[Complement[Range[1000], u], z] (* A001950 *)
r1 = Sqrt[5]; v = Table[Floor[n*r1], {n, 1, z}] (* A022839 *)
v1 = Take[Complement[Range[1000], v], z] (* A108598 *)
Intersection[u, v] (* A351415 *)
Intersection[u, v1] (* A356101 *)
Intersection[u1, v] (* A356102 *)
Intersection[u1, v1] (* A356103 *)
Showing 1-3 of 3 results.
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