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

Author

Clark Kimberling, Feb 09 2018

Keywords

• nonn
• easy

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.

Examples

			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, ...)
		

Links

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

Crossrefs

Cf. A297999, A298000, A297830.

Programs

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 *)