%I #9 Jan 08 2023 11:42:37
%S 2,4,9,12,16,19,21,26,28,33,36,38,43,45,50,53,57,60,62,67,70,74,77,79,
%T 84,86,91,94,98,101,103,108,111,115,118,120,125,127,132,135,137,142,
%U 144,149,152,156,159,161,166,168,173,176,178,183,185,190,193,197
%N a(n) = floor(r*floor(s*n)), where r = 1 + sqrt(2) and s = sqrt(2).
%C Conjecture: 1 < r*s*n - a(n) < 3 for n >= 1.
%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) v o u, defined by (v o u)(n) = v(u(n));
%C (2) u o v';
%C (3) v o u';
%C (4) v' o u'.
%C Every positive integer is in exactly one of the four sequences. For the reverse composites, u o v, u o v', u' o v, u' o v', see A184922.
%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.
%C For A341239, 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) v o u = (2, 4, 9, 12, 16, 19, 21, 26, 28, 33, 36, 38, ...) = A341239
%C (2) v' o u = (1, 3, 6, 8, 11, 13, 15, 18, 20, 23, 25, 27, ...) = A286666
%C (3) v o u' = (7, 14, 24, 31, 41, 48, 55, 65, 72, 82, 89, ...) = A188383
%C (4) v' o u' = (5, 10, 17, 22, 29, 34, 39, 46, 51, 58, 63, ...) = A098021
%C (end)
%F a(n) = floor(r*floor(s*n)), where r = 1 + sqrt(2) and s = sqrt(2).
%t z = 140; r = 1 + Sqrt[2]; s = Sqrt[2]; f[x_] := Floor[r*Floor[s*x]];
%t Table[f[n], {n, 1, z}]
%Y Cf. A184922, A341240; A098021, A188383, A341239, A286666.
%K nonn
%O 1,1
%A _Clark Kimberling_, Feb 07 2021