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 1-4 of 4 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)).

A351415 Intersection of Beatty sequences for (1+sqrt(5))/2 and sqrt(5).

Original entry on oeis.org

4, 6, 8, 11, 17, 22, 24, 29, 33, 35, 38, 40, 42, 46, 51, 53, 55, 58, 64, 67, 69, 71, 76, 80, 82, 84, 87, 93, 98, 100, 105, 111, 114, 116, 118, 122, 127, 129, 131, 134, 140, 145, 147, 152, 156, 158, 160, 163, 165, 169, 174, 176, 181, 187, 190, 192, 194, 199
Offset: 1

Views

Author

Clark Kimberling, Feb 10 2022

Keywords

Comments

Conjecture: every term of the difference sequence is in {2,3,4,5,6}, and each occurs infinitely many times.
From Clark Kimberling, Jul 29 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 A351415, 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.
(1) u ^ v = (4, 6, 8, 11, 17, 22, 24, 29, 33, 35, 38, 40, 42, ...) = A351415
(2) u ^ v' = (1, 3, 9, 12, 14, 16, 19, 21, 25, 27, 30, 32, ...) = A356101
(3) u' ^ v = (2, 13, 15, 20, 26, 31, 44, 49, 60, 62, 73, 78, ...) = A356102
(4) u' ^ v' = (5, 7, 10, 18, 23, 28, 34, 36, 39, 41, 47, 52, ...) = A356103
(End)

Examples

			The two Beatty sequences are (1,3,4,6,8,9,11,12,14,...) and (2,4,6,8,11,13,15,17,...), with common terms forming the sequence (4,6,8,11,...).

Crossrefs

Cf. A000201, A022839, A296184, A346308, A347469, A347793.
Cf. A001950, A108598, A356101, A356102, A356103, A356104 (results of composition instead of intersections), A190509 (composites, reversed order).

Programs

  • Mathematica
    z = 200;
r = (1 + Sqrt[5])/2; u = Table[Floor[n*r], {n, 1, z}]  (* A000201 *)
u1 = Take[Complement[Range[1000], u], z]  (* A001950 *)
r1 = Sqrt[5]; v = Table[Floor[n*r1], {n, 1, z}]  (* A022839 *)
v1 = Take[Complement[Range[1000], v], z]  (* A108598 *)
Intersection[u, v]    (* A351415 *)
Intersection[u, v1]   (* A356101 *)
Intersection[u1, v]   (* A356102 *)
Intersection[u1, v1]  (* A356103 *)

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 *)

A347468 Numbers k such that floor(k*sqrt(3)) = floor(h*sqrt(2)) for some h.

Original entry on oeis.org

1, 3, 5, 7, 9, 11, 13, 14, 15, 17, 18, 19, 21, 22, 23, 24, 25, 26, 27, 28, 29, 31, 32, 33, 35, 36, 38, 39, 40, 42, 43, 44, 45, 46, 48, 49, 50, 52, 53, 54, 56, 57, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 81, 82, 84, 85, 86, 88, 89, 90, 92, 93, 94, 95
Offset: 1

Views

Author

Clark Kimberling, Oct 28 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) = 3 because floor(3*sqrt(3)) = floor(4*sqrt(2)).  (For each such k, there is only one such h.)

Crossrefs

Cf. A001951, A022838, A346308, A347467, 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 1-4 of 4 results.

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