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A326661
Rectangular array in 3 columns that solve the complementary equation c(n) = a(n) + b(3n), where a(1) = 1; see Comments.
3
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
1,2
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).
EXAMPLE
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
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 *)
CROSSREFS
KEYWORD
nonn,tabf,easy
AUTHOR
Clark Kimberling, Jul 16 2019
EXTENSIONS
Replaced a(0)->a(1) in NAME. - R. J. Mathar, Jun 19 2021
STATUS
approved