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%I #17 Jan 08 2023 11:41:48
%S 2,5,9,12,16,19,22,26,29,33,36,39,43,46,50,53,57,60,63,67,70,74,77,80,
%T 84,87,91,94,98,101,104,108,111,115,118,121,125,128,132,135,138,142,
%U 145,149,152,156,159,162,166,169,173,176,179,183,186,190,193,197,200,203,207,210,214,217,220,224,227,231,234,237,241,244,248,251,255,258,261,265,268,272,275,278,282,285,289,292,296,299,302,306,309,313,316,319,323,326,330,333,337,340,343,347,350,354,357,360,364,367,371,374,377,381,384,388,391,395,398,401,405,408
%N a(n) = n + [rn/t] + [sn/t] + [un/t], where []=floor and r=2^(1/2), s=r+1, t=r+2, u=r+3.
%C From _Clark Kimberling_, Jan 26 2011: (Start)
%C The sequences A184920-A184923 partition the positive integers:
%C A184920: 7, 15, 24, 31, 40, 48, 55, 64, ...
%C A184921: 3, 8, 13, 18, 23, 27, 32, 37, ...
%C A184922: 2, 5, 9, 12, 16, 19, 22, 26, 29, ...
%C A184923: 1, 4, 6, 10, 11, 14, 17, 20, 21, ...
%C Jointly rank the sets {h*r}, {i*s}, {j*t}, {k*u}, where h>=1, i>=1, j>=1, k>=1. The position of n*t in the joint ranking is n + [rn/t] + [sn/t] + [un/t], and likewise for the positions of n*s, n*s, and n*u.
%C (end)
%C Since [rn/t] = sqrt(2) - 1, [sn/t] = sqrt(2)/2, and [un/t] = 2 - sqrt(2)/2, we find using [-x] = -[x] - 1 for noninteger x, that a(n) = floor(n*(2+sqrt(2))) - 1 = A001952(n) - 1. - _Michel Dekking_, Feb 22 2018
%C From _Clark Kimberling_, Dec 27 2022: (Start)
%C 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 (increasing) complements, and consider these four sequences:
%C (1) u o v, defined by (u o v)(n) = u(v(n));
%C (2) u o v';
%C (3) u' o v;
%C (4) v' o u'.
%C Every positive integer is in exactly one of the four sequences.
%C 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
%C 1/(r*r') + 1/(r*s') + 1/(s*s') + 1/(s*r') = 1. For A184922, u, v, u', v', are the Beatty sequences given by u(n) = floor(n*sqrt(2)) and v(n) = floor((1+sqrt(2))/2)*n), so that r = sqrt(2), s = (1+sqrt(2))/2, r' = (2+sqrt(2))/2, s' = 1 + 1/sqrt(2).
%C (1) u o v = (2, 5, 9, 12, 16, 19, 22, 26, 29, 33, 36, 39, 43, ...) = A184922
%C (2) u o v' = (1, 4, 7, 8, 11, 14, 15, 18, 21, 24, 25, 28, 31, ...) = A188376
%C (3) u' o v = (6, 13, 23, 30, 40, 47, 54, 64, 71, 81, 88, 95, ...) = A359351
%C (4) u' o v' = (3, 10, 17, 20, 27, 34, 37, 44, 51, 58, 61, 68, ...) = A188396
%C For results of intersections instead of intersections, see A003151. For the reverse composites, v o u, v' o u, v o u', v' o u', see A341239.
%C (end)
%F a(n) = floor(n*(2+sqrt(2))) - 1. - _Michel Dekking_, Feb 22 2018
%t z = 100; zz = 10;
%t u = Table[Floor[n Sqrt[2]], {n, 1, z}]
%t u1 = Complement[Range[Max[u]], u]
%t v = Table[Floor[n (1 + Sqrt[2])], {n, 1, z}]
%t v1 = Complement[Range[Max[v]], v]
%t Table[u[[v[[n]]]], {n, 1, zz}]; (* A184922 *)
%t Table[u[[v1[[n]]]], {n, 1, zz}]; (* A188376 *)
%t Table[u1[[v[[n]]]], {n, 1, zz}]; (* A359351 *)
%t Table[u1[[v1[[n]]]], {n, 1, zz}]; (* A188396 *)
%Y Cf. A184920, A184921, A184922, A184923, A188376, A188396, A359351.
%K nonn
%O 1,1
%A _Clark Kimberling_, Jan 26 2011
%E Name corrected by _Michel Dekking_, Feb 22 2018
%E Edited by _Clark Kimberling_, Dec 27 2022
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