cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

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

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Sep 11 2021

Keywords

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)).

A347469 For irrational r > 1, let B(r) denote the Beatty sequence for r. Let (s(n)) be the increasing sequence of numbers in both B(sqrt(2)) and B(sqrt(3)). Then this sequence gives the numbers k such that s(k) = floor(k*sqrt(6)).

Original entry on oeis.org

1, 20, 21, 23, 49, 70, 71, 72, 73, 74, 75, 76, 78, 98, 101, 102, 117, 148, 194, 215, 216, 250, 257, 262, 299, 300, 307, 310, 344, 346, 357, 360, 361, 448, 1071, 1075, 1083, 1114, 1143, 1160, 1203, 1681, 1722, 1725, 1727, 1737, 1740, 1741, 1770, 1771, 1783
Offset: 1

Views

Author

Clark Kimberling, Oct 31 2021

Keywords

Comments

It is conjectured that this sequence is infinite.

Examples

			(B(sqrt(2)) and B(sqrt(3))) - B(sqrt(6)) = (0, -1, 1, 1, 3, 3, 5, 5, 5, 3, 5, 5, 3, 5, 4, 3, 2, 2, 1, 0, 0, -1, 0, -1, -1, -1, -1, -1, -1, -2, ...), so that a(4) = 23, the position of the 4th 0.

Crossrefs

Cf. A001951, A022838, A022840, A346308, A347467, A347468.

Programs

  • Mathematica
    z = 10000; r = Sqrt[2]; s = Sqrt[3];
u = Table[Floor[n r], {n, 0, z}];  (*A001951*)
v = Table[Floor[n s], {n, 1, z}];  (*A022838*)
w = Intersection[u, v];  (*A346308*)
zz = -1 + Length[w];
t = Table[Floor[n*r*s], {n, 0, zz}]; (* A022840 *)
d = w - t;
Flatten[Position[d, 0]] (* A347469 *)

A347467 Numbers h such that floor(k*sqrt(3)) = floor(h*sqrt(2)) for some k.

Original entry on oeis.org

1, 4, 6, 9, 11, 14, 16, 17, 18, 21, 22, 23, 26, 27, 28, 29, 31, 32, 33, 34, 36, 38, 39, 41, 43, 44, 46, 48, 49, 51, 53, 54, 55, 56, 59, 60, 61, 64, 65, 66, 68, 70, 71, 73, 76, 78, 81, 83, 86, 88, 91, 93, 96, 98, 99, 101, 103, 104, 105, 108, 109, 110, 113
Offset: 1

Views

Author

Clark Kimberling, Oct 16 2021

Keywords

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,...)
Intersection: (1,5,8,12,...), as in A346308.
a(2) = 4 because floor(3*sqrt(3)) = floor(4*sqrt(2)).  (For each such h, there is only one such k.)

Crossrefs

Cf. A001951, A022838, A346308, A347468, A347469.

Programs

  • Mathematica
    z = 200; r = Sqrt[2]; s = Sqrt[3];
u = Table[Floor[n r], {n, 0, z}]; (*A001951*)
v = Table[Floor[n s], {n, 1, z}]; (*A022838*)
w = Intersection[u, v]  (*A346308*)
zz = -1 + Length[w];
Table[Ceiling[w[[n]]/r], {n, 1, zz}] (* A347467 *)
Table[Ceiling[w[[n]]/s], {n, 1, zz}] (* A347468 *)
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.