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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. 6
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 (list; graph; refs; listen; history; text; internal format)
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

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Last modified February 21 02:57 EST 2019. Contains 320364 sequences. (Running on oeis4.)