A305330 Solution (a(n)) of the complementary equation a(n) = 2*a(n-1) - a(n-2) + b(n-1) + b(n); see Comments.

1, 2, 12, 33, 67, 116, 182, 267, 373, 503, 660, 846, 1063, 1313, 1598, 1920, 2281, 2683, 3128, 3618, 4155, 4741, 5378, 6068, 6813, 7615, 8476, 9398, 10383, 11434, 12554, 13745, 15009, 16348, 17764, 19259, 20835, 22494, 24238, 26069, 27989, 30000, 32104

Offset

0,2

Comments

Define sequences a(n) and b(n) recursively, starting with a(0) = 1, a(1) = 2:

b(n) = least new;

a(n) = 2*a(n-1) - a(n-2) + b(n-1),

where "least new" means the least positive integer not yet placed. It appears that a(n)/a(n-1) -> 1, that {a(n) - a(n-1), n >=1} is unbounded, and that the 3rd difference sequence of (a(n)) consists entirely of 2's and 3's.

Links

Clark Kimberling, Table of n, a(n) for n = 0..10000

Example

b(0) = least not in {a(0), a(1)} = 3;
a(2) = 2*a(1) - a(0) +  b(1) + b(2) must be great enough that b(0) = 3, b(1) = 4, and b(2) = 5, so that a(2) = and b  exceed  = 2*2 -1 + 4 = 7, so that b(0) = 3, b(1) = 4, b(2) = 5, b(3) = 6, and a(2) = 12.

Mathematica

a = {1, 2}; b = {3, 4, 5};
mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
Do[AppendTo[a, 2 Last[a] - a[[-2]] + (b[[-1]] + b[[-2]])];
  AppendTo[b, mex[Flatten[{a, b}], Last[b]]], {120}]; a
 (* Peter J. C. Moses, May 30 2018 *)

Crossrefs

Cf. A305129, A305329.

Sequence in context: A133577 A106185 A098203 * A320542 A242709 A055707

Adjacent sequences: A305327 A305328 A305329 * A305331 A305332 A305333

Keyword

nonn,easy

Author

Clark Kimberling, Jun 02 2018

Status

approved