%I #4 May 30 2018 13:58:35
%S 1,3,4,5,6,7,8,9,10,11,12,14,15,16,17,18,19,20,21,23,24,25,26,27,28,
%T 29,30,31,32,34,35,36,37,38,39,40,41,42,44,45,46,47,48,49,50,51,52,54,
%U 55,56,57,58,59,60,61,62,64,65,66,67,68,69,70,71,73,74,75
%N Solution (b(n)) of the complementary equation a(n) = b(4n) + b(5n); see Comments.
%C Define complementary sequences a(n) and b(n) recursively:
%C b(n) = least new,
%C a(n) = b(4n) + b(5n),
%C where "least new" means the least positive integer not yet placed. Empirically, {a(n) - 8*n: n >= 0} = {2,3} and {7*b(n) - 8*n: n >= 0} = {8,9,10,11,12,13,14,15,16,17}. See A304799 for a guide to related sequences.
%H Clark Kimberling, <a href="/A304816/b304816.txt">Table of n, a(n) for n = 0..10000</a>
%e b(0) = 1, so that a(0) = 2. Since a(1) = b(4) + b(5), we must have a(1) >= 11, so that b(1) = 3, b(2) = 4, b(3) = 5, b(4) = 6, b(5) = 7, and a(1) = 13.
%t mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
%t h = 4; k = 5; a = {}; b = {1};
%t AppendTo[a, mex[Flatten[{a, b}], 1]];
%t Do[Do[AppendTo[b, mex[Flatten[{a, b}], Last[b]]], {k}];
%t AppendTo[a, Last[b] + b[[1 + (Length[b] - 1)/k h]]], {500}];
%t Take[a, 200] (* A304815 *)
%t Take[b, 200] (* A304816 *)
%t (* _Peter J. C. Moses_, May 14 2008 *)
%Y Cf. A304799, A304815.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, May 30 2018