A346308 Intersection of Beatty sequences for sqrt(2) and sqrt(3).

1, 5, 8, 12, 15, 19, 22, 24, 25, 29, 31, 32, 36, 38, 39, 41, 43, 45, 46, 48, 50, 53, 55, 57, 60, 62, 65, 67, 69, 72, 74, 76, 77, 79, 83, 84, 86, 90, 91, 93, 96, 98, 100, 103, 107, 110, 114, 117, 121, 124, 128, 131, 135, 138, 140, 142, 145, 147, 148, 152, 154

Offset: 1

Author

Clark Kimberling, Sep 11 2021

Keywords

• nonn

Comments

Let d(n) = a(n) - A022840(n). Conjecture: (d(n)) is unbounded below and above, and d(n) = 0 for infinitely many n.

From Clark Kimberling, Jul 26 2022: (Start)

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 complements, and assume that the following four sequences are infinite:

(1) u ^ v = intersection of u and v (in increasing order);

(2) u ^ v';

(3) u' ^ v;

(4) u' ^ v'.

Every positive integer is in exactly one of the four sequences. For A346308, 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. (See A356052.)

(End)

Examples

			Beatty sequence for sqrt(2): (1,2,4,5,7,8,9,11,12,14,...).
Beatty sequence for sqrt(3): (1,3,5,6,8,10,12,13,15,...).
a(n) = (1,5,8,12,...).
In the notation in Comments:
(1)  u ^ v = (1, 5, 8, 12, 15, 19, 22, 24, 25, 29, 31, 32, ...) =  A346308.
(2)  u ^ v' = (2, 4, 7, 9, 11, 14, 16, 18, 21, 26, 28, 33, 35, ...) =  A356085.
(3)  u' ^ v = (3, 6, 10, 13, 17, 20, 27, 34, 51, 58, 64, 71, 81, ...) = A356086.
(4)  u' ^ v' = (23, 30, 37, 40, 44, 47, 54, 61, 68, 75, 78, 85, ...) = A356087.
		

Crossrefs

Intersection of A001951 and A022838.

Cf. A346308, A347467, A347468, A347469.

Cf. A001952, A022838, A054406, A356085, A356086, A356087, A356088 (composites instead of intersections).

Programs

Mathematica

z = 200;
r = Sqrt[2]; u = Table[Floor[n*r], {n, 1, z}]  (* A001951 *)
u1 = Take[Complement[Range[1000], u], z]  (* A001952 *)
r1 = Sqrt[3]; v = Table[Floor[n*r1], {n, 1, z}]  (* A022838 *)
v1 = Take[Complement[Range[1000], v], z]  (* A054406 *)
t1 = Intersection[u, v]    (* A346308 *)
t2 = Intersection[u, v1]   (* A356085 *)
t3 = Intersection[u1, v]   (* A356086 *)
t4 = Intersection[u1, v1]  (* A356087 *)

Python

from math import isqrt
from itertools import count, islice
def A346308_gen(): # generator of terms
    return filter(lambda n:n == isqrt(3*(isqrt(n**2//3)+1)**2),(isqrt(n*n<<1) for n in count(1)))
A346308_list = list(islice(A346308_gen(),30)) # Chai Wah Wu, Aug 06 2022

Formula

In general, if r and s are irrational numbers greater than 1, and a(n) is the n-th term of the intersection (assumed nonempty) of the Beatty sequences for r and s, then a(n) = floor(r*ceiling(a(n)/r)) = floor(s*ceiling(a(n)/s)).