A356182 - OEIS

%I #8 Sep 05 2022 10:04:34

%S 5,10,17,22,29,34,39,46,51,58,64,69,76,81,88,93,100,105,110,117,122,

%T 129,135,140,147,152,159,164,171,176,181,188,193,200,206,211,218,223,

%U 230,235,240,247,252,259,265,271,277,282,289,294,301,306,311,318,323

%N a(n) = A022838(A001952(n)).

%C This is the third 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:

%C (1) v o u, defined by (v o u)(n)  = v(u(n));

%C (2) v' o u;

%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 A356088 to A356091.

%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 A356182, 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.

%e (1)  v o u = (1, 3, 6, 8, 12, 13, 15, 19, 20, 24, 25, 27, 31, 32, ...) = A356180

%e (2)  v' o u = (2, 4, 9, 11, 16, 18, 21, 26, 28, 33, 35, 37, 42, 44, ...) = A356181

%e (3)  v o u' = (5, 10, 17, 22, 29, 34, 39, 46, 51, 58, 64, 69, 76, ...) = A356182

%e (4)  v' o u' = (7, 14, 23, 30, 40, 47, 54, 63, 70, 80, 87, 94, 104, ...) = A356183

%t z = 800; zz = 100;

%t u = Table[Floor[n*Sqrt[2]], {n, 1, z}];  (* A001951 *)

%t u1 = Complement[Range[Max[u]], u];       (* A001952 *)

%t v = Table[Floor[n*Sqrt[3]], {n, 1, z}];  (* A022838 *)

%t v1 = Complement[Range[Max[v]], v];  (* A054406 *)

%t Table[v[[u[[n]]]], {n, 1, zz}]      (* A356180 *)

%t Table[v1[[u[[n]]]], {n, 1, zz}]     (* A356181) *)

%t Table[v[[u1[[n]]]], {n, 1, zz}]     (* A356182 *)

%t Table[v1[[u1[[n]]]], {n, 1, zz}]    (* A356183 *)

%o (Python)

%o from math import isqrt

%o def A356182(n): return isqrt(3*((k:=n<<1)+isqrt(k*n))**2) # _Chai Wah Wu_, Sep 05 2022

%Y Cf. A001951, A001952, A022838, A054406, A346308 (intersections), A356088 (reverse composites), A356180, A356181, A356183.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Aug 24 2022