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A309157
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Rectangular array in 3 columns that solve the complementary equation c(n) = a(n) + b(2n), where a(1) = 1; see Comments.
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3
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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
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OFFSET
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1,2
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COMMENTS
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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).
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LINKS
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EXAMPLE
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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
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MATHEMATICA
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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]];
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CROSSREFS
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A326661 solves c(n) = a(n) + b(3n),
A326662 solves c(n) = a(2n) + b(2n).
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KEYWORD
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nonn,tabf,easy
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AUTHOR
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STATUS
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approved
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