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%I #28 Feb 12 2021 16:03:17
%S 1,2,5,3,4,12,6,7,20,8,9,26,10,11,33,13,14,41,15,16,47,17,18,54,19,21,
%T 61,22,23,68,24,25,75,27,28,83,29,30,89,31,32,96,34,35,104,36,37,110,
%U 38,39,117,40,42,124,43,44,131,45,46,138,48,49,146,50,51
%N Rectangular array in 3 columns that solve the complementary equation c(n) = a(n) + b(2n), 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 these 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).
%H Clark Kimberling and Peter J. C. Moses, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL24/Kimberling/kimb23.html">Complementary Equations with Advanced Subscripts</a>, J. Int. Seq. 24 (2021) Article 21.3.3.
%e c(1) = a(1) + b(2) > = 1 + 3, so that
%e a(2) = mex{1,2} = 3;
%e b(2) = mex{1,2,3} = 4;
%e c(1) = 5.
%e Then c(2) = a(2) + b(4) >= 3 + 8, so that
%e a(3) = 6, b(3) = 7;
%e a(4) = 8, b(4) = 9;
%e c(2) = a(2) + b(4) = 3 + 9 = 12.
%e n a(n) b(n) c(n)
%e --------------------
%e 1 1 2 5
%e 2 3 4 12
%e 3 6 7 20
%e 4 8 9 26
%e 5 10 11 33
%e 6 13 14 41
%e 7 15 16 47
%e 8 17 18 54
%e 9 19 21 61
%e 10 22 23 68
%t mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
%t a = b = c = {}; h = 1; k = 2;
%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. A326663 (3rd column),
%Y A101544 solves c(n) = a(n) + b(n),
%Y A326661 solves c(n) = a(n) + b(3n),
%Y A326662 solves c(n) = a(2n) + b(2n).
%K nonn,tabf,easy
%O 1,2
%A _Clark Kimberling_, Jul 15 2019
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