A295998 Solution of the complementary equation a(n) = 2*a(n-2) + b(n-2), where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.

1, 2, 5, 8, 16, 23, 41, 56, 93, 124, 199, 262, 413, 541, 844, 1101, 1708, 2223, 3438, 4470, 6901, 8966, 13829, 17960, 27687, 35950, 55405, 71932, 110843, 143898, 221721, 287832, 443479, 575702, 886997, 1151444, 1774036, 2302931, 3548116, 4605907, 7096278

Offset: 0

Author

Clark Kimberling, Dec 02 2017

Keywords

• nonn
• easy

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. a(n)/a(n-1) -> 1.298123759410105...

See A295860 for a guide to related sequences.

Links

• Clark Kimberling, Table of n, a(n) for n = 0..999
• Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.

Crossrefs

Cf. A001622, A000045, A294860.

Programs

Mathematica

mex[t_] := NestWhile[# + 1 &, 1, MemberQ[t, #] &];
a[0] = 1; a[1] = 2; b[0] = 3;
a[n_] := a[n] = 2 a[n - 2] + b[n - 2];  (* A295998 *)
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
Table[a[n], {n, 0, 100}];
Table[b[n], {n, 0, 30}]

Formula

a(0) = 1, a(1) = 2, b(0) = 3, so that a(2) = 5, b(1) = 4.

Complement: (b(n)) = (3, 4, 6, 7, 9, 10, 11, 12, 13, 14, 15, 17, 18, ...)