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A304813
Solution (a(n)) of the complementary equation a(n) = b(3n) + b(5n); see Comments.
3
2, 12, 21, 29, 39, 48, 57, 66, 74, 84, 93, 101, 111, 120, 129, 138, 146, 156, 165, 173, 183, 191, 201, 210, 218, 228, 237, 245, 255, 263, 273, 282, 290, 300, 309, 317, 327, 335, 345, 354, 362, 372, 381, 390, 399, 407, 417, 426, 434, 444, 453, 462, 471, 479
OFFSET
0,1
COMMENTS
Define complementary sequences a(n) and b(n) recursively:
b(n) = least new,
a(n) = b(3n) + 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} = {7,8,9,10,11,12,13,14,15}. See A304799 for a guide to related sequences.
LINKS
EXAMPLE
b(0) = 1, so that a(0) = 2. Since a(1) = b(3) + b(5), we must have a(1) >= 10, so that b(1) = 3, b(2) = 4, b(3) = 5, b(4) = 6, b(5) = 7, and a(1) = 12.
MATHEMATICA
mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
h = 3; 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] (* A304813 *)
Take[b, 200] (* A304814 *)
(* Peter J. C. Moses, May 14 2008 *)
CROSSREFS
Sequence in context: A048955 A034049 A271361 * A279424 A073041 A284595
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, May 30 2018
STATUS
approved