OFFSET
0,1
COMMENTS
Conjecture: a(n) is the number of letters (0's and 1's) in the n-th iteration of the mapping 00->0101, 1->1001, starting with 00; see A288600.
LINKS
Clark Kimberling, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (2,0,-1).
FORMULA
a(n) = 2*a(n-1) - a(n-3) for n >= 3, where a(0) = 2, a(1) = 4, a(2) = 6, a(3) = 8.
G.f.: 2*(1 - x^2 - x^3)/(1 - 2*x + x^3).
a(n) = Fibonacci(n) + Lucas(n) + 2 for n >= 1. - Greg Dresden, Oct 10 2020
MATHEMATICA
Join[{2}, LinearRecurrence[{2, 0, -1}, {4, 6, 8}, 40]]
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jun 14 2017
STATUS
approved