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A304815 Solution (a(n)) of the complementary equation a(n) = b(4n) + b(5n); see Comments. 3
2, 13, 22, 33, 43, 53, 63, 72, 83, 92, 103, 112, 123, 133, 143, 153, 163, 173, 182, 193, 203, 213, 223, 233, 243, 253, 263, 272, 283, 292, 303, 313, 323, 333, 342, 353, 362, 373, 382, 393, 403, 413, 423, 432, 443, 452, 463, 472, 483, 493, 503, 513, 522, 533 (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(4n) + 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} = {8,9,10,11,12,13,14,15,16,17}. See A304799 for a guide to related sequences.
LINKS
EXAMPLE
b(0) = 1, so that a(0) = 2. Since a(1) = b(4) + b(5), we must have a(1) >= 11, so that b(1) = 3, b(2) = 4, b(3) = 5, b(4) = 6, b(5) = 7, and a(1) = 13.
MATHEMATICA
mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
h = 4; 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] (* A304815 *)
Take[b, 200] (* A304816 *)
(* Peter J. C. Moses, May 14 2008 *)
CROSSREFS
Sequence in context: A127485 A061385 A366208 * A156179 A090519 A018540
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, May 30 2018
STATUS
approved

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Last modified March 28 03:28 EDT 2024. Contains 371235 sequences. (Running on oeis4.)