

A304812


Solution (b(n)) of the complementary equation a(n) = b(2n) + b(5n); see Comments.


3



1, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, 17, 18, 20, 21, 22, 23, 24, 25, 27, 28, 29, 30, 31, 32, 33, 35, 36, 37, 38, 39, 40, 41, 42, 44, 45, 46, 47, 48, 49, 50, 52, 53, 54, 55, 56, 57, 59, 60, 61, 62, 63, 64, 65, 66, 68, 69, 70, 71, 72, 73, 74, 76, 77
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OFFSET

0,2


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..10000


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]]], {501}];
Take[a, 200] (* A304811 *)
Take[b, 200] (* A304812 *)
(* Peter J. C. Moses, May 14 2008 *)


CROSSREFS

Cf. A304799, A304811.
Sequence in context: A114904 A039220 A048135 * A026500 A091213 A211347
Adjacent sequences: A304809 A304810 A304811 * A304813 A304814 A304815


KEYWORD

nonn,easy


AUTHOR

Clark Kimberling, May 30 2018


STATUS

approved



