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| NAME
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allocatedSolution (a(n)) of the complementary forequation Clarka(n) = 2*a(n-1) - a(n-2) + b(n-1) + b(n); see KimberlingComments.
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| DATA
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1, 2, 12, 33, 67, 116, 182, 267, 373, 503, 660, 846, 1063, 1313, 1598, 1920, 2281, 2683, 3128, 3618, 4155, 4741, 5378, 6068, 6813, 7615, 8476, 9398, 10383, 11434, 12554, 13745, 15009, 16348, 17764, 19259, 20835, 22494, 24238, 26069, 27989, 30000, 32104
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| OFFSET
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0,2
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| COMMENTS
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Define sequences a(n) and b(n) recursively, starting with a(0) = 1, a(1) = 2:
b(n) = least new;
a(n) = 2*a(n-1) - a(n-2) + b(n-1),
where "least new" means the least positive integer not yet placed. It appears that a(n)/a(n-1) -> 1, that {a(n) - a(n-1), n >=1} is unbounded, and that the 3rd difference sequence of (a(n)) consists entirely of 2's and 3's.
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| LINKS
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Clark Kimberling, <a href="/A305330/b305330.txt">Table of n, a(n) for n = 0..10000</a>
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| EXAMPLE
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b(0) = least not in {a(0), a(1)} = 3;
a(2) = 2*a(1) - a(0) + b(1) + b(2) must be great enough that b(0) = 3, b(1) = 4, and b(2) = 5, so that a(2) = and b exceed = 2*2 -1 + 4 = 7, so that b(0) = 3, b(1) = 4, b(2) = 5, b(3) = 6, and a(2) = 12.
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| MATHEMATICA
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a = {1, 2}; b = {3, 4, 5};
mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
Do[AppendTo[a, 2 Last[a] - a[[-2]] + (b[[-1]] + b[[-2]])];
AppendTo[b, mex[Flatten[{a, b}], Last[b]]], {120}]; a
(* Peter J. C. Moses, May 30 2018 *)
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| CROSSREFS
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Cf. A305129, A305329.
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| KEYWORD
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allocated
nonn,easy
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| AUTHOR
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Clark Kimberling, Jun 02 2018
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| STATUS
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approved
editing
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