A297826 Solution (a(n)) of the near-complementary equation a(n) = a(1)*b(n-1) - a(0)*b(n-2) + n, where a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, and (b(n)) is the increasing sequence of positive integers not in (a(n)). See Comments.

1, 2, 7, 9, 11, 15, 18, 21, 22, 24, 28, 29, 33, 34, 40, 42, 43, 45, 51, 51, 53, 59, 59, 61, 63, 65, 69, 74, 76, 77, 79, 81, 83, 87, 90, 91, 93, 95, 97, 101, 104, 107, 110, 111, 113, 117, 118, 120, 122, 126, 131, 133, 136, 139, 140, 142, 146, 147, 153, 155

Offset

0,2

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) 0 <= a(k) - a(k-1) <= 6 for k>=1;

(2) if d is in {0,1,2,3,4,5,6}, then a(k) = a(k-1) + d for infinitely many k.

***

See A297830 for a guide to related sequences.

Links

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

Example

a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, so that a(2) = 7.
Complement: (b(n)) = (3, 4, 5, 6, 8,10,12,13,14,16, ...)

Mathematica

mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
tbl = {}; a[0] = 1; a[1] = 2; b[0] = 3; b[1] = 4;
a[n_] := a[n] = a[1]*b[n - 1] - a[0]*b[n - 2] + n;
b[n_] := b[n] = mex[tbl = Join[{a[n], a[n - 1], b[n - 1]}, tbl], b[n - 1]];
Table[a[n], {n, 0, 300}]  (* A297826 *)
Table[b[n], {n, 0, 300}]  (* A297997 *)
(* Peter J. C. Moses, Jan 03 2017 *)

Crossrefs

Cf. A297997, A297830.

Sequence in context: A022424 A360944 A136498 * A288598 A277737 A082371

Adjacent sequences: A297823 A297824 A297825 * A297827 A297828 A297829

Keyword

nonn,easy

Author

Clark Kimberling, Feb 04 2018

Status

approved