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Solution (a(n)) of the system of complementary equations defined in Comments.
6

%I #13 Jul 30 2018 09:40:52

%S 1,3,6,8,10,13,15,17,19,22,24,27,29,31,33,36,38,40,43,45,48,50,52,55,

%T 57,59,62,64,66,69,71,73,76,78,80,82,85,87,90,92,94,97,99,101,104,106,

%U 108,111,113,115,118,120,122,125,127,129,132,134,136,139,141

%N Solution (a(n)) of the system of complementary equations defined in Comments.

%C Define sequences a(n), b(n), c(n) recursively, starting with a(0) = 1:

%C a(n) = least new,

%C b(n) = least new,

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

%C where "least new k" means the least positive integer not yet placed. The three sequences partition the positive integers. Empirically, for all n >= 0:

%C 1 <= 3*a(n) - 7*n <= 4,

%C 5 <= 3*b(n) - 7*n <= 8,

%C 4 <= c(n) - 7*n <= 6,

%H Robert Israel, <a href="/A304500/b304500.txt">Table of n, a(n) for n = 0..10000</a>

%e a(0) = 1, b(0) = 2; c(0) = 1 + 2*2 = 5, so that a(1) = 3, so that b(1) = 4, so that c(1) = 11.

%p S:= {$1..200}:

%p for n from 0 do

%p if nops(S) < 2 then break fi;

%p a[n]:= min(S);

%p S:= S minus {a[n]};

%p b[n]:= min(S);

%p c[n]:= a[n]+2*b[n];

%p S:= S minus {b[n],c[n]}

%p od:

%p seq(A[i]i=0..n-1); # _Robert Israel_, Jul 30 2018

%t z = 301;

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

%t a = {}; b = {}; c = {};

%t Do[AppendTo[a,

%t mex[Flatten[{a, b, c}], If[Length[a] == 0, 1, Last[a]]]];

%t AppendTo[b, mex[Flatten[{a, b, c}], Last[a]]];

%t AppendTo[c, Last[a] + 2*Last[b]], {z}];

%t Take[a, 100] (* A304500 *)

%t Take[b, 100] (* A304501 *)

%t Take[c, 100] (* A304502 *)

%t Grid[{Join[{"n"}, Range[0, 20]], Join[{"a(n)"}, Take[a, 21]],

%t Join[{"b(n)"}, Take[b, 21]], Join[{"c(n)"}, Take[c, 21]]},

%t Alignment -> ".", Dividers -> {{2 -> Red, -1 -> Blue}, {2 -> Red, -1 -> Blue}}]

%t (* _Peter J. C. Moses_, Apr 26 2018 *)

%Y Cf. A304497, A304501, A304502.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, May 19 2018