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%I #4 May 30 2018 13:58:16
%S 2,12,21,29,39,48,57,66,74,84,93,101,111,120,129,138,146,156,165,173,
%T 183,191,201,210,218,228,237,245,255,263,273,282,290,300,309,317,327,
%U 335,345,354,362,372,381,390,399,407,417,426,434,444,453,462,471,479
%N Solution (a(n)) of the complementary equation a(n) = b(3n) + b(5n); see Comments.
%C Define complementary sequences a(n) and b(n) recursively:
%C b(n) = least new,
%C a(n) = b(3n) + 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} = {7,8,9,10,11,12,13,14,15}. See A304799 for a guide to related sequences.
%H Clark Kimberling, <a href="/A304813/b304813.txt">Table of n, a(n) for n = 0..9999</a>
%e b(0) = 1, so that a(0) = 2. Since a(1) = b(3) + b(5), we must have a(1) >= 10, so that b(1) = 3, b(2) = 4, b(3) = 5, b(4) = 6, b(5) = 7, and a(1) = 12.
%t mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
%t h = 3; 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] (* A304813 *)
%t Take[b, 200] (* A304814 *)
%t (* _Peter J. C. Moses_, May 14 2008 *)
%Y Cf. A304799, A304814.
%K nonn,easy
%O 0,1
%A _Clark Kimberling_, May 30 2018
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