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A304800 Solution (b(n)) of the complementary equation a(n) = b(n) + b(2n); see Comments. 3

%I

%S 1,3,4,5,6,8,9,11,12,13,15,16,17,19,20,21,22,24,25,27,28,29,30,32,33,

%T 35,36,37,39,40,41,42,44,45,47,48,49,51,52,53,54,56,57,58,60,61,63,64,

%U 65,67,68,69,70,72,73,75,76,77,78,80,81,83,84,85,87,88

%N Solution (b(n)) of the complementary equation a(n) = b(n) + b(2n); see Comments.

%C Define complementary sequences a(n) and b(n) recursively:

%C b(n) = least new,

%C a(n) = b(n) + b(2n),

%C where "least new" means the least positive integer not yet placed. See A304799 for a guide to related sequences. Empirically, {a(n) - 3*n: n >= 0} = {2,3} and {2*b(n) - 3*n : n >= 0} = {2,3,4,5}.

%H Clark Kimberling, <a href="/A304800/b304800.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(2), we must have a(1) >= 7, so that b(1) = 3, b(2) = 4, b(3) = 5, b(4) = 6, and a(1) = 7.

%t mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);

%t h = 1; k = 2; 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] (* A304799 *)

%t Take[b, 200] (* A304800 *)

%t (* _Peter J. C. Moses_, May 14 2008 *)

%Y Cf. A304799.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, May 19 2018

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Last modified June 21 19:59 EDT 2021. Contains 345365 sequences. (Running on oeis4.)