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 A304806 Solution (b(n)) of the complementary equation a(n) = b(n) + b(5n) ; see Comments. 3
 1, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 18, 19, 20, 21, 22, 23, 25, 26, 27, 28, 29, 30, 32, 33, 34, 35, 36, 38, 39, 40, 41, 42, 43, 45, 46, 47, 48, 49, 50, 52, 53, 54, 55, 56, 57, 58, 60, 61, 62, 63, 64, 65, 67, 68, 69, 70, 71, 72, 74, 75, 76, 77, 78 (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(5n), where "least new" means the least positive integer not yet placed.  Empirically, {a(n) - 6*n: n >= 0} = {2,3} and {5*b(n) - 6*n: n >= 0} = {5,6,7,8,9,10,11}.  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(5), we must have a(1) >= 10, so that b(1) = 3, b(2) = 4, b(3) = 5, ..., b(7) = 9, and a(1) = 10. MATHEMATICA mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]); h = 1; k = 5; 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]  (* A304805 *) Take[b, 200]  (* A304806 *) (* Peter J. C. Moses, May 14 2008 *) CROSSREFS Cf. A304799, A304805. Sequence in context: A297468 A047565 A026466 * A304810 A026469 A258187 Adjacent sequences:  A304803 A304804 A304805 * A304807 A304808 A304809 KEYWORD nonn,easy AUTHOR Clark Kimberling, May 28 2018 STATUS approved

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Last modified May 12 06:54 EDT 2021. Contains 343820 sequences. (Running on oeis4.)