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A304805 Solution (a(n)) of the complementary equation a(n) = b(n) + b(5n) ; see Comments. 3
2, 10, 17, 24, 31, 37, 44, 51, 59, 66, 73, 79, 86, 93, 101, 108, 115, 121, 128, 135, 143, 150, 157, 164, 170, 177, 185, 192, 199, 206, 212, 220, 227, 234, 241, 247, 254, 262, 269, 276, 283, 289, 296, 304, 311, 318, 325, 331, 338, 345, 353, 360, 367, 373, 380 (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(n) + b(5n),

where "least new" means the least positive integer not yet placed.  Empirically, {a(n) - 6*n: n >= 0} = {2,3} and {5*b(n) - 6*n: n >= 0} = {5,6,7,8,9,10,11}.  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(1) + b(5), we must have a(1) >= 10, so that b(1) = 3, b(2) = 4, b(3) = 5, ..., b(7) = 9, and a(1) = 10.

MATHEMATICA

mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);

h = 1; 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]  (* A304805 *)

Take[b, 200]  (* A304806 *)

(* _Peter J. C.  Moses_, May 14 2008 *)

CROSSREFS

Cf. A304799, A304806.

Sequence in context: A305093 A316753 A304809 * A214086 A127492 A258974

Adjacent sequences:  A304802 A304803 A304804 * A304806 A304807 A304808

KEYWORD

nonn,easy

AUTHOR

Clark Kimberling, May 28 2018

STATUS

approved

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Last modified May 12 06:54 EDT 2021. Contains 343820 sequences. (Running on oeis4.)