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A304799
Solution (a(n)) of the complementary equation a(n) = b(n) + b(2n); see Comments.
19
2, 7, 10, 14, 18, 23, 26, 31, 34, 38, 43, 46, 50, 55, 59, 62, 66, 71, 74, 79, 82, 86, 90, 95, 98, 103, 106, 110, 115, 118, 122, 126, 131, 134, 139, 142, 146, 151, 154, 158, 162, 167, 170, 174, 179, 182, 187, 191, 194, 199, 203, 206, 210, 215, 218, 223, 226
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
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