OFFSET
0,1
COMMENTS
Define complementary sequences a(n) and b(n) recursively: both are strictly increasing, b(0) = 1, and a(n) = b(n) + b(2n) for n >= 0. Empirically,
(1) {a(n) - 4*n: n >= 0} = {2,3} and {3*b(n) - 4*n: n >= 0} = {2,3,4,5}.
(2) If the equation for a(n) is generalized to a(n) = b(h*n) + b(k*n), where 1 <= h < k, then {a(n) - (h + k + 1)*n: n >= 0} = {2,3} and {(h + k)*b(n) - (h + k + 1)*n : n >= 0} = {k + h - 1, k + h, ..., 2*k + 2*h - 1}.
(3) {a(n) - a(n-1): n >= 1) = {h+k, h+k+1, h+k+2}.
(4) {k*b(n)-b(k*n): n >= 0} = {k-2, k-1, ..., 2*k-2}
***
Guide to related sequences:
h k (a(n)) (b(n))
LINKS
Clark Kimberling, Table of n, a(n) for n = 0..10000
C. Kimberling and P. J. C. Moses, Complementary Equations with Advanced Subscripts, J. Int. Seq. 24 (2021) Article 21.3.3.
EXAMPLE
b(0) = 1, so that a(0) = 2. Since a(1) = b(1) + b(2), we must have a(1) >= 7, so that b(1) = 3, b(2) = 4, b(3) = 5, b(4) = 6, and a(1) = 7.
MATHEMATICA
mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
h = 1; k = 2; a = {}; b = {1};
AppendTo[a, mex[Flatten[{a, b}], 1]];
Do[Do[AppendTo[b, mex[Flatten[{a, b}], Last[b]]], {k}];
AppendTo[a, Last[b] + b[[1 + (Length[b] - 1)/k h]]], {500}];
Take[a, 200] (* A304799 *)
Take[b, 200] (* A304800 *)
(* Peter J. C. Moses, May 14 2008 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, May 19 2018
EXTENSIONS
Comments edited by Clark Kimberling, Jul 07 2019
STATUS
approved