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a(n) = 5*a(n-1) + 4*a(n-2), with a(0) = 4, a(1) = 4.
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%I #26 Sep 08 2022 08:45:18

%S 0,4,20,116,660,3764,21460,122356,697620,3977524,22678100,129300596,

%T 737215380,4203279284,23965257940,136639406836,779058065940,

%U 4441847957044,25325472048980,144394752073076,823275648561300,4693957251098804,26762888849739220

%N a(n) = 5*a(n-1) + 4*a(n-2), with a(0) = 4, a(1) = 4.

%H Vincenzo Librandi, <a href="/A106567/b106567.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (5,4).

%F a(n) = 4*A015537(n).

%F From _Chai Wah Wu_, Mar 21 2018: (Start)

%F a(n) = 5*a(n-1) + 4*a(n-2) for n > 1.

%F G.f.: 4*x/(1 - 5*x - 4*x^2). (End)

%F a(n) = 4*(p^n - q^n)/(p - q), where 2*p = 5 + sqrt(41), 2*q = 5 - sqrt(41). - _G. C. Greubel_, Sep 06 2021

%t CoefficientList[Series[4*x/(1-5*x-4*x^2), {x, 0, 30}], x] (* _Vincenzo Librandi_, Mar 22 2018 *)

%o (Magma) I:=[0,4]; [n le 2 select I[n] else 5*Self(n-1) +4*Self(n-2): n in [1..30]]; // _Vincenzo Librandi_, Mar 22 2018

%o (PARI) a(n) = (([0,4; 1,5]^n)*[0,1]~)[1]; \\ _Michel Marcus_, Mar 22 2018

%o (Sage)

%o def A106567_list(prec):

%o P.<x> = PowerSeriesRing(ZZ, prec)

%o return P( 4*x/(1-5*x-4*x^2) ).list()

%o A106567_list(30) # _G. C. Greubel_, Sep 06 2021

%Y Cf. A015537.

%K nonn,easy,less

%O 0,2

%A _Roger L. Bagula_, May 30 2005

%E Edited by _N. J. A. Sloane_, Apr 30 2006

%E New name after Chai Wah Wu, by _Bruno Berselli_, Mar 22 2018