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%I #4 May 30 2018 13:58:30
%S 2,13,22,33,43,53,63,72,83,92,103,112,123,133,143,153,163,173,182,193,
%T 203,213,223,233,243,253,263,272,283,292,303,313,323,333,342,353,362,
%U 373,382,393,403,413,423,432,443,452,463,472,483,493,503,513,522,533
%N Solution (a(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="/A304815/b304815.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, A304816.
%K nonn,easy
%O 0,1
%A _Clark Kimberling_, May 30 2018
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