A356101 - OEIS

%I #22 Jun 05 2023 08:56:17

%S 1,3,9,12,14,16,19,21,25,27,30,32,37,43,45,48,50,56,59,61,63,66,72,74,

%T 77,79,85,88,90,92,95,97,101,103,106,108,110,113,119,121,124,126,132,

%U 135,137,139,142,144,148,150,153,155,161,166,168,171,173,177,179

%N Intersection of A000201 and A022839.

%C This is the second of four sequences that partition the positive integers. See A351415.

%e 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:

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

%e (2)  u ^ v';

%e (3)  u' ^ v;

%e (4)  u' ^ v'.

%e 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 r = (1+sqrt(5))/2, s = sqrt(5), r' = (3+sqrt(5))/2, s' = (5 + sqrt(5))/4.

%e (1)  u ^ v = (4, 6, 8, 11, 17, 22, 24, 29, 33, 35, 38, 40, 42, ...) =  A351415

%e (2)  u ^ v' = (1, 3, 9, 12, 14, 16, 19, 21, 25, 27, 30, 32, 37, ...) =  A356101

%e (3)  u' ^ v = (2, 13, 15, 20, 26, 31, 44, 49, 60, 62, 73, 78, ...) = A356102

%e (4)  u' ^ v' = (5, 7, 10, 18, 23, 28, 34, 36, 39, 41, 47, 52, 54, ...) = A356103

%t z = 200;

%t r = (1 + Sqrt[5])/2; u = Table[Floor[n*r], {n, 1, z}]  (* A000201 *)

%t u1 = Take[Complement[Range[1000], u], z]  (* A001950 *)

%t r1 = Sqrt[5]; v = Table[Floor[n*r1], {n, 1, z}]  (* A022839 *)

%t v1 = Take[Complement[Range[1000], v], z]  (* A108598 *)

%t Intersection[u, v]   (* A351415 *)

%t Intersection[u, v1]  (* A356101 *)

%t Intersection[u1, v]  (* A356102 *)

%t Intersection[u1, v1] (* A356103 *)

%Y Cf. u = A000201, u' = A001950, v = A022839, v' = A108598, A351415, A356102, A356103, A356104 (results of composition instead of intersections), A190509 (composites, reversed order).

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Sep 04 2022