A297997 Solution (b(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.

3, 4, 5, 6, 8, 10, 12, 13, 14, 16, 17, 19, 20, 23, 25, 26, 27, 30, 31, 32, 35, 36, 37, 38, 39, 41, 44, 46, 47, 48, 49, 50, 52, 54, 55, 56, 57, 58, 60, 62, 64, 66, 67, 68, 70, 71, 72, 73, 75, 78, 80, 82, 84, 85, 86, 88, 89, 92, 94, 96, 98, 99, 100, 102, 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 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: A092253 A073849 A288597 * A136497 A055563 A090161

Adjacent sequences: A297994 A297995 A297996 * A297998 A297999 A298000

Keyword

nonn,easy

Author

Clark Kimberling, Feb 04 2018

Status

approved