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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
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
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
Sequence in context: A019375 A078784 A117196 * A064739 A112860 A152312
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, May 30 2018
STATUS
approved