A304806 - OEIS

%I #4 May 29 2018 20:41:51

%S 1,3,4,5,6,7,8,9,11,12,13,14,15,16,18,19,20,21,22,23,25,26,27,28,29,

%T 30,32,33,34,35,36,38,39,40,41,42,43,45,46,47,48,49,50,52,53,54,55,56,

%U 57,58,60,61,62,63,64,65,67,68,69,70,71,72,74,75,76,77,78

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

%C Define complementary sequences a(n) and b(n) recursively:

%C b(n) = least new,

%C a(n) = b(n) + b(5n),

%C 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.

%H Clark Kimberling, <a href="/A304806/b304806.txt">Table of n, a(n) for n = 0..10000</a>

%e 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.

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

%t h = 1; k = 5; a = {}; b = {1};

%t AppendTo[a, mex[Flatten[{a, b}], 1]];

%t Do[Do[AppendTo[b, mex[Flatten[{a, b}], Last[b]]], {k}];

%t   AppendTo[a, Last[b] + b[[1 + (Length[b] - 1)/k h]]], {500}];

%t Take[a, 200]  (* A304805 *)

%t Take[b, 200]  (* A304806 *)

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

%Y Cf. A304799, A304805.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, May 28 2018