|
%I #4 May 13 2018 01:11:08
%S 1,4,16,50,155,468,1410,4234,12709,38132,114404,343218,1029663,
%T 3088997,9267001,27801012,83403047,250209151,750627465,2251882406,
%U 6755647231,20266941705,60800825129,182402475400,547207426215,1641622278659,4924866835993,14774600507994
%N Solution (a(n)) of the complementary equation in Comments.
%C Define sequences a(n) and b(n) recursively, starting with a(0) = 1, b(0) = 2:
%C b(n) = least new;
%C a(n) = 3*a(n-1) + x(0)*b(n) + x(1)*b(n-1) + ... + x(n)b(0),
%C where "least new k" means the least positive integer not yet placed, and x(k) = (-1)^k for k >= 0.
%C ***
%C It appears that a(n)/a(n-1) -> 3 and that {a(n) - 3*a(n-1), n >= 1} is unbounded.
%e b(1) = least not in {a(0),b(0)} = 3;
%e a(1) = 3*a(0) + b(1) - b(0) = 3*1 + 3 - 2 = 4.
%t mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
%t c = 3; a = {1}; b = {2}; x = {-1};
%t Do[AppendTo[b, mex[Flatten[{a, b}], 1]];
%t AppendTo[x, -Last[x]];
%t AppendTo[a, c Last[a] - (Reverse[x] b // Total)], {25}]
%t Grid[{Join[{"n"}, Range[0, # - 1]], Join[{"a(n)"}, Take[a, #]],
%t Join[{"b(n)"}, Take[b, #]], Join[{"x(n)"}, Take[-x, #]]},
%t Alignment -> ".",
%t Dividers -> {{2 -> Red, -1 -> Blue}, {2 -> Red, -1 -> Blue}}] &[10]
%t (* _Peter J. C. Moses_, May 10 2018 *)
%Y Cf. A298741, A298877.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, May 12 2018
|
