A304803 - OEIS

%I #9 May 26 2018 22:47:39

%S 2,9,15,21,26,32,38,45,51,56,62,68,75,81,87,92,98,105,111,117,122,129,

%T 135,141,146,152,159,165,171,176,182,189,195,201,206,212,218,225,231,

%U 236,242,248,255,261,267,272,279,285,291,297,302,309,315,321,326

%N Solution (a(n)) of the complementary equation a(n) = b(n) + b(4n); see Comments.

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

%C b(n) = least new,

%C a(n) = b(n) + b(4n),

%C where "least new" means the least positive integer not yet placed.  Empirically, {a(n) - 5*n: n >= 0} = {2,3} and {4*b(n) - 5*n: n >= 0} = {4,5,6,7,8,9}.  See A304799 for a guide to related sequences.

%H Clark Kimberling, <a href="/A304803/b304803.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(4), 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) = 9.

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

%t h = 1; k = 4; 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]  (* A304803 *)

%t Take[b, 200]  (* A304804 *)

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

%Y Cf. A304799, A304804.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, May 19 2018