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 A304811 Solution (a(n)) of the complementary equation a(n) = b(2n) + b(5n) ; see Comments. 3
 2, 11, 19, 26, 34, 43, 51, 58, 67, 75, 83, 91, 99, 107, 114, 123, 131, 138, 146, 155, 163, 170, 179, 187, 195, 203, 211, 219, 226, 235, 243, 250, 258, 267, 275, 282, 291, 299, 306, 314, 323, 331, 338, 347, 355, 363, 370, 379, 387, 394, 403, 411, 418, 426 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,1 COMMENTS Define complementary sequences a(n) and b(n) recursively: b(n) = least new, a(n) = b(2n) + b(5n), where "least new" means the least positive integer not yet placed.  Empirically, {a(n) - 8*n: n >= 0} = {2,3} and {7*b(n) - 8*n: n >= 0} = {6,7,8,9,10,11,12,13}.  See A304799 for a guide to related sequences. LINKS Clark Kimberling, Table of n, a(n) for n = 0..9999 EXAMPLE b(0) = 1, so that a(0) = 2.  Since a(1) = b(2) + b(5), we must have a(1) >= 9, so that b(1) = 3, b(2) = 4, b(3) = 5, b(4) = 6, b(5) = 7, b(6) = 8, and a(1) = 11. MATHEMATICA mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]); h = 2; 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]  (* A304811 *) Take[b, 200]  (* A304812 *) (* Peter J. C. Moses, May 14 2008 *) CROSSREFS Cf. A304799, A304812. Sequence in context: A019375 A078784 A117196 * A064739 A112860 A152312 Adjacent sequences:  A304808 A304809 A304810 * A304812 A304813 A304814 KEYWORD nonn,easy AUTHOR Clark Kimberling, May 30 2018 STATUS approved

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Last modified May 12 07:28 EDT 2021. Contains 343821 sequences. (Running on oeis4.)