A356141 - OEIS

A356141

a(n) = A137804(A001952(n)).

6, 12, 20, 27, 35, 41, 48, 56, 62, 71, 77, 83, 92, 98, 106, 113, 121, 127, 134, 142, 148, 157, 163, 169, 177, 184, 192, 198, 207, 213, 219, 228, 234, 242, 249, 255, 263, 270, 278, 284, 291, 299, 305, 314, 320, 328, 335, 341, 349, 355, 364, 370, 376, 385, 391

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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) v o u, defined by (v o u)(n) = v(u(n));

(2) v' o u;

(3) v o u';

(4) v' o u'.

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 A356056 to A356059.

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 A356141, 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.

LINKS

Table of n, a(n) for n=1..55.

EXAMPLE

(1) v o u = (1, 3, 7, 9, 13, 15, 17, 21, 22, 26, 28, 30, 34, ...) = A356138

(2) v' o u = (2, 4, 8, 10, 14, 16, 18, 23, 25, 29, 31, 33, 37, ...) = A356139

(3) v o u' = (5, 11, 19, 24, 32, 38, 44, 51, 57, 65, 70, 76, 84, ...) = A356140

(4) v' o u' = (6, 12, 20, 27, 35, 41, 48, 56, 62, 71, 77, 83, 92, ...) = A356141

MATHEMATICA

z = 800;

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*)