OFFSET
1,1
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)
EXAMPLE
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,...).
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 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Clark Kimberling, Feb 10 2022
STATUS
approved