OFFSET
0,2
COMMENTS
The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. Guide to related sequences:
A295613: a(n) = 2*a(n-1) - a(n-3) + b(n-1); a(0) = 1, a(1) = 2, a(2) = 3, b(0) = 4, b(1) = 5, b(2) = 6.
A295614: a(n) = 2*a(n-1) - a(n-3) + b(n-1); a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6.
A295615: a(n) = 2*a(n-1) - a(n-3) + b(n-1); a(0) = 2, a(1) = 4, a(2) = 6, b(0) = 1, b(1) = 3, b(2) = 5.
A295616: a(n) = 2*a(n-1) - a(n-3) + b(n-2); a(0) = 1, a(1) = 2, a(2) = 3, b(0) = 4, b(1) = 5, b(2) = 6.
A295617: a(n) = 2*a(n-1) - a(n-3) + b(n-2); a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2, b(1) = 4, b(2) = 6.
A295618: a(n) = 2*a(n-1) - a(n-3) + b(n-2); a(0) = 2, a(1) = 4, a(2) = 6, b(0) = 1, b(1) = 3, b(2) = 5.
LINKS
Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.
EXAMPLE
a(0) = 1, a(1) = 2, a(2) = 3, b(0) = 4, b(1) = 5, b(2) = 6, so that
b(3) = 7 (least "new number")
a(3) = 2*a(2) - a(0) + b(2) = 11
Complement: (b(n)) = (4, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15, ...)
MATHEMATICA
mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 2; a[2] = 3; b[0] = 4; b[1] = 5; b[2] = 6;
a[n_] := a[n] = 2 a[n - 1] - a[n - 3] + b[n - 1];
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
Table[a[n], {n, 0, 30}] (* A295613 *)
Table[b[n], {n, 0, 20}] (* complement *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Nov 25 2017
STATUS
approved