OFFSET
1,1
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) v o u, defined by (v o u)(n) = v(u(n));
(2) u o v';
(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 A356104 to A356107.
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 A356217 u, v, u', v', are the Beatty sequences given by u(n) = floor(n*(1+sqrt(5))/2) and v(n) = floor(n*sqrt(5)), so that r = (1+sqrt(5))/2, s = sqrt(5), r' = (3+sqrt(5))/2, s' = (5 + sqrt(5))/4.
EXAMPLE
MATHEMATICA
z = 1000;
u = Table[Floor[n*(1 + Sqrt[5])/2], {n, 1, z}]; (* A000201 *)
u1 = Complement[Range[Max[u]], u]; (* A001950 *)
v = Table[Floor[n*Sqrt[5]], {n, 1, z}]; (* A022839 *)
v1 = Complement[Range[Max[v]], v]; (* A108598 *)
Table[v[[u[[n]]]], {n, 1, z/4}] (* A356217 *)
Table[v1[[u[[n]]]], {n, 1, z/4}] (* A356218 *)
Table[v[[u1[[n]]]], {n, 1, z/4}] (* A190509 *)
Table[v1[[u1[[n]]]], {n, 1, z/4}] (* A356220 *)
PROG
(Python)
from math import isqrt
def A356217(n): return isqrt(5*(n+isqrt(5*n**2)>>1)**2) # Chai Wah Wu, Oct 14 2022
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Oct 02 2022
STATUS
approved