



6, 13, 23, 30, 37, 47, 54, 61, 71, 78, 88, 95, 102, 112, 119, 126, 136, 143, 150, 160, 167, 177, 184, 191, 201, 208, 215, 225, 232, 238, 249, 256, 266, 273, 279, 290, 297, 303, 314, 320, 331, 338, 344, 355, 361, 368, 378, 385, 392, 402, 409, 419, 426, 433
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OFFSET

1,1


COMMENTS

This is the fourth of four sequences that partition the positive integers. Suppose that u = (u(n)) and v = (v(n)) are increasing sequences of positive integers. Let u' and v' be their (increasing) complements, and consider these four sequences:
(1) u o v, defined by (u o v)(n) = u(v(n));
(2) u o v';
(3) u' o v;
(4) u' o 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 A356091, u, v, u', v', are the Beatty sequences given by u(n) = floor(n*sqrt(2)) and v(n) = floor(n*sqrt(3), so that r = sqrt(2), s = sqrt(3), r' = 2 + sqrt(2), s' = (3 + sqrt(3))/2.


LINKS



EXAMPLE

(1) u o v = (1, 4, 7, 8, 11, 14, 16, 18, 21, 24, 26, ...) = A356088
(2) u o v' = (2, 5, 9, 12, 15, 19, 22, 25, 29, 32, 36, ...) = A356089
(3) u' o v = (3, 10, 17, 20, 27, 34, 40, 44, 51, 58, 64, ...) = A356090
(4) u' o v' = (6, 13, 23, 30, 37, 47, 54, 61, 71, 78, 88, ...) = A356091


MATHEMATICA

z = 600; zz = 100;
u = Table[Floor[n*Sqrt[2]], {n, 1, z}]; (* A001951 *)
u1 = Complement[Range[Max[u]], u]; (* A001952 *)
v = Table[Floor[n*Sqrt[3]], {n, 1, z}]; (* A022838 *)
v1 = Complement[Range[Max[v]], v]; (* A054406 *)
Table[u[[v[[n]]]], {n, 1, zz}] (* A356088 *)
Table[u[[v1[[n]]]], {n, 1, zz}] (* A356089 *)
Table[u1[[v[[n]]]], {n, 1, zz}] (* A356090 *)
Table[u1[[v1[[n]]]], {n, 1, zz}] (* A356091 *)


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



