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A309157
Rectangular array in 3 columns that solve the complementary equation c(n) = a(n) + b(2n), where a(1) = 1; see Comments.
3
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
1,2
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).
LINKS
Clark Kimberling and Peter J. C. Moses, Complementary Equations with Advanced Subscripts, J. Int. Seq. 24 (2021) Article 21.3.3.
EXAMPLE
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
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 *)
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).
Sequence in context: A026183 A026199 A026207 * A261645 A193724 A266407
KEYWORD
nonn,tabf,easy
AUTHOR
Clark Kimberling, Jul 15 2019
STATUS
approved