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%I #9 Aug 06 2018 05:02:48
%S 1,3,4,5,6,7,8,9,10,12,13,14,15,16,17,18,20,21,22,23,24,25,27,28,29,
%T 30,31,32,33,35,36,37,38,39,40,41,42,44,45,46,47,48,49,50,52,53,54,55,
%U 56,57,59,60,61,62,63,64,65,66,68,69,70,71,72,73,74,76,77
%N Solution (b(n)) of the complementary equation a(n) = b(2n) + b(5n); see Comments.
%C Define complementary sequences a(n) and b(n) recursively:
%C b(n) = least new,
%C a(n) = b(2n) + 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} = {6,7,8,9,10,11,12,13}. See A304799 for a guide to related sequences.
%H Clark Kimberling, <a href="/A304812/b304812.txt">Table of n, a(n) for n = 0..10000</a>
%e b(0) = 1, so that a(0) = 2. Since a(1) = b(2) + b(5), we must have a(1) >= 9, so that b(1) = 3, b(2) = 4, b(3) = 5, b(4) = 6, b(5) = 7, b(6) = 8, and a(1) = 11.
%t mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
%t h = 2; 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]]], {501}];
%t Take[a, 200] (* A304811 *)
%t Take[b, 200] (* A304812 *)
%t (* _Peter J. C. Moses_, May 14 2008 *)
%Y Cf. A304799, A304811.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, May 30 2018
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