A295365 - OEIS

A295365

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

1, 2, 3, 20, 41, 82, 147, 256, 433, 722, 1191, 1952, 3185, 5182, 8415, 13648, 22117, 35823, 58002, 93891, 151962, 245925, 397962, 643965, 1042008, 1686057, 2728152, 4414299, 7142544, 11556939, 18699582

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OFFSET

0,2

COMMENTS

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A295357 for a guide to related sequences.

LINKS

Table of n, a(n) for n=0..30.

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

FORMULA

a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622).

EXAMPLE

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

b(2) = 7 (least "new number")

a(3) = a(2) + a(1) + b(2) + b(1) + b(0) = 20

Complement: (b(n)) = (4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, ...)

MATHEMATICA

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