A298110 Solution (b(n)) of the near-complementary equation a(n) = a(1)*b(n) - a(0)*b(n-1) + n, where a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, b(2) = 5, and (b(n)) is the increasing sequence of positive integers not in (a(n)). See Comments.

3, 4, 5, 6, 7, 9, 11, 13, 14, 15, 17, 18, 20, 21, 24, 26, 27, 28, 31, 32, 33, 36, 37, 38, 39, 40, 42, 45, 47, 48, 49, 50, 51, 53, 55, 56, 57, 58, 59, 61, 63, 65, 67, 68, 69, 71, 72, 73, 74, 76, 79, 81, 83, 85, 86, 87, 89, 90, 93, 95, 97, 99, 100, 101, 103

Offset

0,1

Comments

The sequence (a(n)) generated by the equation a(n) = a(1)*b(n-1) - a(0)*b(n-2) + n, with initial values as shown, includes duplicates; e.g. a(18) = a(19) = 51. If the duplicates are removed from (a(n)), the resulting sequence and (b(n)) are complementary. Conjectures:

(1) 1 <= b(k) - b(k-1) <= 3 for k>=1;

(2) if d is in {1,2,3}, then b(k) = b(k-1) + d for infinitely many k.

***

See A298000 and A297830 for guides to related sequences.

Links

Clark Kimberling, Table of n, a(n) for n = 0..2000

Example

a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, b(2) = 5, so that a(2) = 8.
Complement: A298110 = (3,4,5,6,7,9,11,13,14,15,17, ...)

Mathematica

mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
a[0] = 1; a[1] = 2; b[0] = 3; b[1] = 4; b[2] = 5;
a[n_] := a[1]*b[n] - a[0]*b[n - 1] + n;
Table[{a[n], b[n + 1] = mex[Flatten[Map[{a[#], b[#]} &, Range[0, n]]], b[n - 0]]}, {n, 2, 3000}];
Table[a[n], {n, 0, 150}]  (* A297999 *)
Table[b[n], {n, 0, 150}]  (* A298110 *)
(* Peter J. C. Moses, Jan 16 2018 *)

Crossrefs

Cf. A297999, A298000, A297830.

Sequence in context: A039051 A047564 A154536 * A091815 A081692 A161346

Adjacent sequences: A298107 A298108 A298109 * A298111 A298112 A298113

Keyword

nonn,easy

Author

Clark Kimberling, Feb 09 2018

Status

approved