A326661 Rectangular array in 3 columns that solve the complementary equation c(n) = a(n) + b(3n), where a(1) = 1; see Comments.

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, 79, 21, 22, 88, 23, 24, 97, 26, 27, 107, 28, 29, 115, 30, 31, 124, 32, 33, 133, 34, 36, 142, 37, 38, 151, 39, 40, 160, 41, 42, 169, 44, 45, 179, 46, 47, 187, 48

Offset: 1

Author

Clark Kimberling, Jul 16 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 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).

Examples

			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.
n           a(n)      b(n)     c(n)
---------------------------
1             1        2        7
2             3        4       16
3             5        6       25
4             8        9       35
5            10       11       43
6            12       13       52
7            14       15       61
8            17       18       74
9            19       20       79
10           21       22       88
		

Crossrefs

Cf. A309157, A326662, A326664.

Programs

Mathematica

mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
a = b = c = {}; h = 1; k = 3;
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 *)

Extensions

Replaced a(0)->a(1) in NAME. - R. J. Mathar, Jun 19 2021