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a(n) = 2*F(n-1) + 2*F(n-3) + 10*F(n-5) + 9*F(n-8) where n >= 8 and F = A000045.
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%I #33 Sep 08 2022 08:46:18

%S 56,97,153,250,403,653,1056,1709,2765,4474,7239,11713,18952,30665,

%T 49617,80282,129899,210181,340080,550261,890341,1440602,2330943,

%U 3771545,6102488,9874033,15976521,25850554,41827075,67677629,109504704,177182333,286687037

%N a(n) = 2*F(n-1) + 2*F(n-3) + 10*F(n-5) + 9*F(n-8) where n >= 8 and F = A000045.

%H Vincenzo Librandi, <a href="/A280932/b280932.txt">Table of n, a(n) for n = 8..1100</a>

%H H. Zhao and X. Li, <a href="http://www.fq.math.ca/44-1.html">On the Fibonacci numbers of trees</a>, The Fibonacci Quarterly, Vol. 44, Number 1 (2006), page 37.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (1,1).

%F G.f.: x^8*(56 + 41*x)/(1 - x - x^2).

%F a(n) = a(n-1) + a(n-2).

%F From the g.f.: a(n) = 56*F(n-7) + 41*F(n-8) = 41*F(n-6) + 15*F(n-7) = 15*F(n-5) + 26*F(n-6) = 26*F(n-4) - 11*F(n-5) = -11*F(n-3) + 37*F(n-4) = 37*F(n-2) - 48*F(n-3) = -48*F(n-1) + 85*F(n-2) = 85*F(n) - 133*F(n-1), and so on.

%t LinearRecurrence[{1, 1}, {56, 97}, 35]

%o (Magma) [2*Fibonacci(n-1)+2*Fibonacci(n-3)+10*Fibonacci(n-5)+9*Fibonacci(n-8): n in [8..40]];

%o (Magma) a0:=56; a1:=97; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]];

%Y Cf. A000045, A022130, A101156, A280931.

%K nonn,easy

%O 8,1

%A _Vincenzo Librandi_, Jan 24 2017

%E Corrected and extended by _Bruno Berselli_, Jan 24 2017