A022940 a(n) = a(n-1) + b(n-2) for n >= 3, a( ) increasing, given a(1) = 1, a(2) = 3; where b( ) is complement of a( ).

1, 3, 5, 9, 15, 22, 30, 40, 51, 63, 76, 90, 106, 123, 141, 160, 180, 201, 224, 248, 273, 299, 326, 354, 383, 414, 446, 479, 513, 548, 584, 621, 659, 698, 739, 781, 824, 868, 913, 959, 1006, 1054, 1103, 1153, 1205, 1258, 1312, 1367, 1423, 1480

Offset

1,2

Comments

From Clark Kimberling, Oct 30 2017: (Start)

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:

here: a(n) = a(n-1) + b(n-2) [with a different offset]

A294397: a(n) = a(n-1) + b(n-2) + 1;

A294398: a(n) = a(n-1) + b(n-2) + 2;

A294399: a(n) = a(n-1) + b(n-2) + 3;

A294400: a(n) = a(n-1) + b(n-2) + n;

A294401: a(n) = a(n-1) + b(n-2) + 2*n.

(End)

Links

Ivan Neretin, Table of n, a(n) for n = 1..10000

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

Example

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

Mathematica

Fold[Append[#1, #1[[-1]] + Complement[Range[Max@#1 + 1], #1][[#2]]] &, {1, 3}, Range[50]] (* Ivan Neretin, Apr 04 2016 *)

Crossrefs

Cf. A005228 and references therein.

Cf. A293076, A293765, A294381.

Sequence in context: A029518 A061954 A095039 * A025207 A027688 A118403

Adjacent sequences: A022937 A022938 A022939 * A022941 A022942 A022943

Keyword

nonn,easy

Author

Clark Kimberling

Status

approved