A294399 Solution of the complementary equation a(n) = a(n-1) + b(n-2) + 3, where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4.

1, 3, 8, 15, 23, 32, 42, 54, 67, 81, 96, 112, 129, 148, 168, 189, 211, 234, 258, 283, 310, 338, 367, 397, 428, 460, 493, 527, 563, 600, 638, 677, 717, 758, 800, 843, 887, 933, 980, 1028, 1077, 1127, 1178, 1230, 1283, 1337, 1392, 1448, 1506, 1565, 1625, 1686

Offset: 0

Author

Clark Kimberling, Oct 30 2017

Keywords

• nonn
• easy

Comments

The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A022940 for a guide to related sequences.

Examples

			a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
a(2)  = a(1) + b(0) + 3 = 8
Complement: (b(n)) = (2, 4, 5, 6, 7, 10, 11, 12, 13, 14, 16, ...)
		

Links

• Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.

Crossrefs

Cf. A293076, A293765, A022940.

Programs

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] + b[n - 2] + 3;
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
Table[a[n], {n, 0, 40}]  (* A294399 *)
Table[b[n], {n, 0, 10}]