OFFSET
0,2
COMMENTS
The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. The initial values of each sequence in the following guide are a(0) = 1, a(2) = 3, b(0) = 2, b(1) = 4:
A294413: a(n) = a(n-1) + a(n-2) - b(n-1) + 6
A294414: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2)
A294415: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + 1
A294416: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) + n
A294417: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - n
A294418: a(n) = a(n-1) + a(n-2) + b(n-1) + 2*b(n-2)
A294419: a(n) = a(n-1) + a(n-2) + 2*b(n-1) + 2*b(n-2)
A294420: a(n) = a(n-1) + a(n-2) + 2*b(n-1) + b(n-2)
A294421: a(n) = a(n-1) + a(n-2) + 2*b(n-1) - b(n-2)
A294422: a(n) = a(n-1) + a(n-2) + b(n-1) - b(n-2) + 1
A294423: a(n) = a(n-1) + a(n-2) + b(n-1) - b(n-2) + n
A294424: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - 1
A294425: a(n) = a(n-1) + a(n-2) + b(n-1) + b(n-2) - 2
A294426: a(n) = a(n-1) + 2*a(n-2) + b(n-1) + b(n-2) - 3
Conjecture: a(n)/a(n-1) -> (1 + sqrt(5))/2, the golden ratio.
LINKS
Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.
EXAMPLE
a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
a(2) = a(1) + a(0) + b(1) + b(0) = 6
Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 9, 11, 12, 13, ...)
MATHEMATICA
mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 3; b[0] = 2; b[1] = 4;
a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n - 1] + b[n - 2];
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
Table[a[n], {n, 0, 40}] (* A294414 *)
Table[b[n], {n, 0, 10}]
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Oct 31 2017
EXTENSIONS
Definition corrected by Georg Fischer, Sep 27 2020
STATUS
approved