OFFSET
0,2
COMMENTS
The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. Guide to related sequencres:
A295053: a(n) = a(n-1) + a(n-2) + b(0) + b(1) + ... + b(n-1), a(0) = 1, a(1) = 2, b(0) = 3
A295054: a(n) = a(n-1) + a(n-2) + b(1) + b(2) + ... + b(n-1), a(0) = 1, a(1) = 2, b(0) = 3
A295055: a(n) = a(n-2) + b(1) + b(2) + ... + b(n-1), a(0) = 1, a(1) = 2, b(0) = 3
A295056: a(n) = 2*a(n-1) + b(n-1), a(0) = 1, a(1) = 4, b(0) = 2, b(1) = 3
A295057: a(n) = 2*a(n-1) + b(n-1), a(0) = 2, a(1) = 5, b(0) = 1
A295058: a(n) = 2*a(n-1) - b(n-1), a(0) = 3, a(1) = 5, b(0) = 1
A295059: a(n) = 2*a(n-1) + b(n-2), a(0) = 1, a(1) = 4, b(0) = 2, b(1) = 3
A295060: a(n) = 2*a(n-1) - b(n-2), a(0) = 3, a(1) = 5, b(0) = 1, b(1) = 2
A295061: a(n) = 4*a(n-1) + b(n-1), a(0) = 1, a(1) = 3, b(0) = 2
A295062: a(n) = 4*a(n-2) + b(n-2), a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4
A295063: a(n) = 4*a(n-2) + b(n-1) + b(n-2), a(0) = 1, a(1) = 3, b(0) = 2
A295064: a(n) = 8*a(n-3) + b(n-1), a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2
A295065: a(n) = 8*a(n-3) + b(n-2), a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2
A295066: a(n) = 2*a(n-2) + b(n-1), a(0) = 1, a(1) = 3, b(0) = 2
A295067: a(n) = 2*a(n-2) + b(n-2), a(0) = 1, a(1) = 3, b(0) = 2
A295068: a(n) = 2*a(n-2) - b(n-1) + n, a(0) = 3, a(1) = 4, b(0) = 1
A295069: a(n) = 2*a(n-2) - b(n-2) + n, a(0) = 3, a(1) = 4, b(0) = 1
A295070: a(n) = a(n-2) + b(n-1) + b(n-2), a(0) = 3, a(1) = 2, b(0) = 3
A295133: a(n) = 3*a(n-1) + b(n-1), a(0) = 1, a(1) = 2, b(0) = 3
A295134: a(n) = 3*a(n-1) + b(n-1) - 1, a(0) = 1, a(1) = 2, b(0) = 3
A295135: a(n) = 3*a(n-1) + b(n-1) - 2, a(0) = 1, a(1) = 2, b(0) = 3
A295136: a(n) = 3*a(n-1) + b(n-1) - 3, a(0) = 1, a(1) = 2, b(0) = 3
A295137: a(n) = 3*a(n-1) + b(n-1) - n, a(0) = 1, a(1) = 2, b(0) = 3
A295138: a(n) = 3*a(n-2) + b(n-1), a(0) = 1, a(1) = 2, b(0) = 3
A295139: a(n) = 3*a(n-1) + b(n-2), a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4
A295140: a(n) = 3*a(n-1) - b(n-2) + 4, a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4
A295141: a(n) = 2*a(n-1) + a(n-2) + b(n-2), a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4
A295142: a(n) = 2*a(n-1) + a(n-2) + b(n-2), a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4
A295143: a(n) = 2*a(n-1) + a(n-1) + b(n-1), a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4
A295144: a(n) = 2*a(n-1) + a(n-2) + b(n-1), a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4
A295145: a(n) = a(n-1) + 2*a(n-2) + b(n-2), a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4
A295146: a(n) = a(n-1) + 2*a(n-2) + b(n-2), a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4
A295147: a(n) = a(n-1) + 2*a(n-2) + b(n-1), a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4
A295148: a(n) = a(n-1) + 2*a(n-2) + b(n-1), a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4
LINKS
Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.
EXAMPLE
a(0) = 1, a(1) = 2, b(0) = 3
b(1) = 4 (least "new number")
a(2) = a(1) + a(0) + b(0) + b(1) = 10
Complement: (b(n)) = (3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, ...)
MATHEMATICA
mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 2; b[0] = 3;
a[n_] := a[n] = a[n - 1] + a[n - 2] + Sum[b[k], {k, 0, 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, 18}] (* A295053 *)
Table[b[n], {n, 0, 10}]
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Nov 18 2017
STATUS
approved