

A346308


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


17



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
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OFFSET

1,2


COMMENTS

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



FORMULA

In general, if r and s are irrational numbers greater than 1, and a(n) is the nth 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)).


EXAMPLE

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.


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 *)


PROG

(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)))


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



