OFFSET
1,2
COMMENTS
This is the first 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 A356088, 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.
EXAMPLE
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 *)
PROG
(Python)
from math import isqrt
def A356088(n): return isqrt(isqrt(3*n*n)**2<<1) # Chai Wah Wu, Aug 06 2022
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Aug 04 2022
STATUS
approved