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From Chai Wah Wu, Nov 01 2018: (Start)
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) for n > 5 (conjectured).
G.f.: (2*x^5 + x^4 + 3*x^3 + 1)/((x - 1)*(x + 1)*(2*x - 1)) (conjectured). (End)
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Conjecture: a(n) = Fibonacci(2*n+1) if n <= 3, for n > 3, a(n) = 2*a(n-1) + 2 if n is even, a(n) = 2*a(n-1) + 5 if n is odd. It would follow that a(n) = 2^(n+1) - 4 + (n mod 2) for n >= 3. - David A. Corneth, Jul 23 2017
Conjecture: a(n) = Fibonacci(2*n+1) if n <= 3, for n > 3, a(n) = 2*a(n-1) + 2 if n is even, a(n) = 2*a(n-1) + 5 if n is odd. - David A. Corneth, Jul 23 2017
proposed
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