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1, 5, 7, 9, 11, 15, 19, 21, 22, 24, 26, 28, 32, 36, 38, 42, 45, 49, 53, 55, 57, 59, 63, 65, 66, 70, 72, 74, 76, 80, 82, 84, 86, 89, 91, 93, 97, 101, 103, 107, 111, 114, 118, 120, 124, 128, 130, 132, 135, 137, 141, 145, 147, 149, 151, 155, 156, 158, 162, 164
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OFFSET
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1,2
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COMMENTS
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This is the first of four sequences that partition the positive integers. 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.
Assume that if w is any of the sequences u, v, u', v', then lim_{n->oo) w(n)/n exists and defines the (limiting) density of w. For w = u,v,u',v', denote the densities by r,s,r',s'. Then the densities of sequences (1)-(4) exist, and
1/(r*r') + 1/(r*s') + 1/(s*s') + 1/(s*r') = 1.
For A356052, u, v, u', v', are the Beatty sequences given by u(n) = floor(n*sqrt(2)) and v(n) = floor(n*(1/2 + sqrt(2))), so that r = sqrt(2), s = 1/2 + sqrt(2), r' = 2 + sqrt(2), s' = (9 + 4*sqrt(2))/7.
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LINKS
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EXAMPLE
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(1) u ^ v = (1, 5, 7, 9, 11, 15, 19, 21, 22, 24, 26, 28, ...) = A356052
(2) u ^ v' = (2, 4, 8, 12, 14, 16, 18, 25, 29, 31, 33, 35, ...) = A356053
(3) u' ^ v = (3, 13, 17, 30, 34, 40, 44, 47, 51, 61, 68, ...) = A356054
(4) u' ^ v' = (6, 10, 20, 23, 27, 37, 54, 58, 64, 71, 75, ...) = A356055
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MATHEMATICA
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z = 250;
u = Table[Floor[n (Sqrt[2])], {n, 1, z}] (* A001951 *)
u1 = Complement[Range[Max[u]], u] (* A001952 *)
v = Table[Floor[n (1/2 + Sqrt[2])], {n, 1, z}] (* A137803 *)
v1 = Complement[Range[Max[v]], v] (* A137804 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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