A295615 - OEIS

A295615

Solution of the complementary equation a(n) = 2*a(n-1) - a(n-3) + b(n-1), where a(0) = 2, a(1) = 4, a(2) = 6, b(0) = 1, b(1) = 3, b(2) = 5, and (a(n)) and (b(n)) are increasing complementary sequences.

2, 4, 6, 15, 33, 68, 130, 237, 417, 716, 1208, 2013, 3326, 5461, 8927, 14547, 23653, 38400, 62275, 100920, 163464, 264678, 428462, 693487, 1122324, 1816215, 2938973, 4755653, 7695123, 12451307, 20146996, 32598905, 52746540, 85346122, 138093378, 223440256

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,1

COMMENTS

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295613 for a guide to related sequences.

a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622), so that a( ) has the growth rate of the Fibonacci numbers (A000045).

LINKS

Table of n, a(n) for n=0..35.

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

EXAMPLE

a(0) = 2, a(1) = 4, a(2) = 6, b(0) = 1, b(1) = 3, b(2) = 5, so that

b(3) = 7 (least "new number")

a(3) = 2*a(2) - a(0) + b(2) = 15

Complement: (b(n)) = (1, 3, 5, 7, 8, 9, 10, 11, 12, 13, 14, 16, ...)

MATHEMATICA

mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;

a[0] = 2; a[1] = 4; a[2] = 6; b[0] = 1; b[1] = 3; b[2] = 5;