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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
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
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
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Feb 09 2018
STATUS
approved