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A304801
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Solution (a(n)) of the complementary equation a(n) = b(n) + b(3n); see Comments.
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3
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2, 8, 13, 17, 22, 27, 33, 38, 42, 47, 53, 58, 62, 68, 73, 77, 82, 88, 93, 97, 102, 108, 113, 117, 122, 127, 133, 138, 142, 147, 153, 158, 162, 168, 173, 177, 182, 188, 193, 198, 202, 207, 213, 218, 222, 227, 233, 238, 242, 248, 253, 257, 262, 267, 273, 278
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OFFSET
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0,1
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COMMENTS
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Define complementary sequences a(n) and b(n) recursively:
b(n) = least new,
a(n) = b(n) + b(3n),
where "least new" means the least positive integer not yet placed. Empirically, {a(n) - 4*n: n >= 0} = {2,3} and {3*b(n) - 4*n: n >= 0} = {3,4,5,6,7}. See A304799 for a guide to related sequences.
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LINKS
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EXAMPLE
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b(0) = 1, so that a(0) = 2. Since a(1) = b(1) + b(3), we must have a(1) >= 8, so that b(1) = 3, b(2) = 4, b(3) = 5, b(4) = 6, b(5) = 7, and a(1) = 8.
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MATHEMATICA
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mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
h = 1; k = 3; a = {}; b = {1};
AppendTo[a, mex[Flatten[{a, b}], 1]];
Do[Do[AppendTo[b, mex[Flatten[{a, b}], Last[b]]], {k}];
AppendTo[a, Last[b] + b[[1 + (Length[b] - 1)/k h]]], {500}];
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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