A094688 Convolution of Fibonacci(n) and 3^n.
0, 1, 4, 14, 45, 140, 428, 1297, 3912, 11770, 35365, 106184, 318696, 956321, 2869340, 8608630, 25826877, 77482228, 232449268, 697351985, 2092062720, 6276199106, 18828615029, 56485873744, 169457667600, 508373077825, 1525119354868
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (4,-2,-3).
Programs
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Magma
I:=[0,1,4]; [n le 3 select I[n] else 4*Self(n-1)-2*Self(n-2) -3*Self(n-3): n in [1..41]]; // Vincenzo Librandi, Jun 24 2012
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Mathematica
LinearRecurrence[{4,-2,-3},{0,1,4},40] (* Vincenzo Librandi, Jun 24 2012 *) Table[(3^(n+1) -LucasL[n+2])/5, {n,0,40}] (* Vladimir Reshetnikov, Sep 27 2016 *) -
PARI
a(n)=(3^(n+1)-fibonacci(n+1)-fibonacci(n+3))/5 \\ Charles R Greathouse IV, Jun 28 2011
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SageMath
[(3^(n+1) -lucas_number2(n+2,1,-1))/5 for n in range(41)] # G. C. Greubel, Feb 09 2023
Formula
G.f.: x/((1-3*x)*(1-x-x^2)).
a(n) = (1/5)*(3^(n+1) - Lucas(n+2)).
a(n) = 4*a(n-1) - 2*a(n-2) - 3*a(n-3).
a(n) = A101220(3, 3, n). - Ross La Haye, Jan 28 2005
a(n) = a(n-1) + a(n-2) + 3^(n-1) for n > 1, with a(0) = 0, a(1) = 1. - Ross La Haye, Aug 20 2005
a(n) = 3*a(n-1) + Fibonacci(n), where a(0) = 0. - Taras Goy, Mar 24 2019