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 A304802 Solution (b(n)) of the complementary equation a(n) = b(n) + b(3n); see Comments. 3
 1, 3, 4, 5, 6, 7, 9, 10, 11, 12, 14, 15, 16, 18, 19, 20, 21, 23, 24, 25, 26, 28, 29, 30, 31, 32, 34, 35, 36, 37, 39, 40, 41, 43, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 59, 60, 61, 63, 64, 65, 66, 67, 69, 70, 71, 72, 74, 75, 76, 78, 79, 80, 81, 83 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS 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. LINKS Clark Kimberling, Table of n, a(n) for n = 0..10000 EXAMPLE 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. MATHEMATICA 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}]; Take[a, 200] (* A304801 *) Take[b, 200] (* A304802 *) (* Peter J. C. Moses, May 14 2008 *) CROSSREFS Cf. A304799, A304801. Sequence in context: A039179 A247914 A184518 * A288228 A002328 A039133 Adjacent sequences: A304799 A304800 A304801 * A304803 A304804 A304805 KEYWORD nonn,easy AUTHOR Clark Kimberling, May 19 2018 STATUS approved

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Last modified December 2 09:50 EST 2023. Contains 367517 sequences. (Running on oeis4.)