%I
%S 1,2,7,3,4,16,5,6,25,8,9,35,10,11,43,12,13,52,14,15,61,17,18,71,19,20,
%T 79,21,22,88,23,24,97,26,27,107,28,29,115,30,31,124,32,33,133,34,36,
%U 142,37,38,151,39,40,160,41,42,169,44,45,179,46,47,187,48
%N Rectangular array in 3 columns that solve the complementary equation c(n) = a(n) + b(3n), where a(1) = 1; see Comments.
%C Let A = (a(n)), B = (b(n)), and C = (c(n)). A unique solution (A,B,C) exists for the following conditions: (1) A,B,C must partition the positive integers, and (2) A and B are defined by mex (minimal excludant, as in A067017); that is, a(n) is the least "new" positive integer, and likewise for b(n).
%e c(1) = a(1) + b(3) >= 1 + 6, so that b(1) = mex{1} = 2; a(2) = mex{1,2} = 3; b(2) = mex{1,2,3} = 4; a(3)= mex{1,2,3,4} = 5, a(4) = mex{1,2,3,4,5} = 6, c(1) = 7.
%e n a(n) b(n) c(n)
%e 
%e 1 1 2 7
%e 2 3 4 16
%e 3 5 6 25
%e 4 8 9 35
%e 5 10 11 43
%e 6 12 13 52
%e 7 14 15 61
%e 8 17 18 74
%e 9 19 20 79
%e 10 21 22 88
%t mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
%t a = b = c = {}; h = 1; k = 3;
%t Do[Do[AppendTo[a,
%t mex[Flatten[{a, b, c}], Max[Last[a /. {} > {0}], 1]]];
%t AppendTo[b, mex[Flatten[{a, b, c}], Max[Last[b /. {} > {0}], 1]]], {k}];
%t AppendTo[c, a[[h Length[a]/k]] + Last[b]], {150}];
%t {a, b, c} // ColumnForm
%t a = Take[a, Length[c]]; b = Take[b, Length[c]];
%t Flatten[Transpose[{a, b, c}]](* _Peter J. C. Moses_, Jul 04 2019 *)
%Y Cf. A309157, A326662, A326664.
%K nonn,tabl,easy,changed
%O 1,2
%A _Clark Kimberling_, Jul 16 2019
%E Replaced a(0)>a(1) in NAME.  _R. J. Mathar_, Jun 19 2021
