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%I #9 Jun 01 2022 12:16:27
%S 2,10,17,24,31,37,44,51,59,66,73,79,86,93,101,108,115,121,128,135,143,
%T 150,157,164,170,177,185,192,199,206,212,220,227,234,241,247,254,262,
%U 269,276,283,289,296,304,311,318,325,331,338,345,353,360,367,373,380
%N Solution (a(n)) of the complementary equation a(n) = b(n) + b(5n) ; see Comments.
%C Define complementary sequences a(n) and b(n) recursively:
%C b(n) = least new,
%C a(n) = b(n) + b(5n),
%C where "least new" means the least positive integer not yet placed. Empirically, {a(n) - 6*n: n >= 0} = {2,3} and {5*b(n) - 6*n: n >= 0} = {5,6,7,8,9,10,11}. See A304799 for a guide to related sequences.
%H Clark Kimberling, <a href="/A304805/b304805.txt">Table of n, a(n) for n = 0..10000</a>
%e b(0) = 1, so that a(0) = 2. Since a(1) = b(1) + b(5), we must have a(1) >= 10, so that b(1) = 3, b(2) = 4, b(3) = 5, ..., b(7) = 9, and a(1) = 10.
%t mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
%t h = 1; 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] (* A304805 *)
%t Take[b, 200] (* A304806 *)
%t (* _Peter J. C. Moses_, May 14 2008 *)
%Y Cf. A304799, A304806.
%K nonn,easy
%O 0,1
%A _Clark Kimberling_, May 28 2018
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