A309157 Rectangular array in 3 columns that solve the complementary equation c(n) = a(n) + b(2n), where a(1) = 1; see Comments.

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, 61, 22, 23, 68, 24, 25, 75, 27, 28, 83, 29, 30, 89, 31, 32, 96, 34, 35, 104, 36, 37, 110, 38, 39, 117, 40, 42, 124, 43, 44, 131, 45, 46, 138, 48, 49, 146, 50, 51

Offset: 1

Author

Clark Kimberling, Jul 15 2019

Keywords

• nonn
• tabf
• easy

Comments

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).

Examples

			c(1) = a(1) + b(2) > = 1 + 3, so that
a(2) = mex{1,2} = 3;
b(2) = mex{1,2,3} = 4;
c(1) = 5.
Then c(2) = a(2) + b(4) >= 3 + 8, so that
a(3) = 6, b(3) = 7;
a(4) = 8, b(4) = 9;
c(2) = a(2) + b(4) = 3 + 9 = 12.
   n    a(n) b(n) c(n)
  --------------------
   1      1    2    5
   2      3    4   12
   3      6    7   20
   4      8    9   26
   5     10   11   33
   6     13   14   41
   7     15   16   47
   8     17   18   54
   9     19   21   61
  10     22   23   68
		

Links

• Clark Kimberling and Peter J. C. Moses, Complementary Equations with Advanced Subscripts, J. Int. Seq. 24 (2021) Article 21.3.3.

Crossrefs

Cf. A326663 (3rd column),

A101544 solves c(n) = a(n) + b(n),

A326661 solves c(n) = a(n) + b(3n),

A326662 solves c(n) = a(2n) + b(2n).

Programs

Mathematica

mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
a = b = c = {}; h = 1; k = 2;
Do[Do[AppendTo[a,
  mex[Flatten[{a, b, c}], Max[Last[a /. {} -> {0}], 1]]];
  AppendTo[b, mex[Flatten[{a, b, c}], Max[Last[b /. {} -> {0}], 1]]], {k}];
  AppendTo[c, a[[h Length[a]/k]] + Last[b]], {150}];
{a, b, c} // ColumnForm
a = Take[a, Length[c]]; b = Take[b, Length[c]];
Flatten[Transpose[{a, b, c}]] (* Peter J. C. Moses, Jul 04 2019 *)