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 A305129 Solution (a(n)) of the complementary equation a(n) = 2*a(n-1) - a(n-2) + b(n); see Comments. 3
 1, 2, 8, 20, 39, 67, 105, 154, 215, 289, 377, 480, 599, 735, 889, 1062, 1256, 1472, 1711, 1974, 2262, 2576, 2917, 3286, 3684, 4112, 4571, 5062, 5586, 6144, 6737, 7366, 8032, 8736, 9480, 10265, 11092, 11962, 12876, 13835, 14840, 15892, 16992, 18141, 19340 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS 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), 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 1's and 2's. LINKS Clark Kimberling, Table of n, a(n) for n = 0..10000 EXAMPLE b(0) = least not in {a(0), a(1)} = 3; a(2) = 2*a(1) - a(0) +  b(2) must exceed  = 2*2 -1 + 5 = 8, so that b(0) = 3, b(1) = 4, b(2) = 5, b(3) = 6, b(4) =7, and a(2) = 8. MATHEMATICA a = {1, 2}; b = {3, 4, 5}; mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]); Do[AppendTo[a, 2 Last[a] - a[[-2]] + Last[b]];   AppendTo[b, mex[Flatten[{a, b}], Last[b]]], {200}]; a (* Peter J. C. Moses, May 30 2018 *) CROSSREFS Cf. A305329, A305330. Sequence in context: A025219 A032767 A294869 * A032633 A294437 A007290 Adjacent sequences:  A305126 A305127 A305128 * A305130 A305131 A305132 KEYWORD nonn,easy AUTHOR Clark Kimberling, May 30 2018 STATUS approved

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Last modified June 23 21:56 EDT 2021. Contains 345402 sequences. (Running on oeis4.)