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a(n) = 5*a(n-1) + 5*a(n-2) with a(0) = 0, a(1) = 5.
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%I #31 Apr 21 2023 06:15:37

%S 0,5,25,150,875,5125,30000,175625,1028125,6018750,35234375,206265625,

%T 1207500000,7068828125,41381640625,242252343750,1418169921875,

%U 8302111328125,48601406250000,284517587890625,1665594970703125

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

%H G. C. Greubel, <a href="/A106565/b106565.txt">Table of n, a(n) for n = 0..1000</a>

%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>

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

%F Equals 5*A057088(n). - _T. D. Noe_, Feb 17 2006

%F From _Philippe Deléham_, Nov 19 2008: (Start)

%F a(n) = 5*a(n-1) + 5*a(n-2), n > 1; a(0)=0, a(1)=5.

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

%F a(n) = (1/6)*5^((n+1)/2)*((1-(-1)^n)*Lucas(2*n) + (1+(-1)^n)*sqrt(5)*Fibonacci(2*n)). - _G. C. Greubel_, Sep 06 2021

%t LinearRecurrence[{5,5}, {0,5}, 40] (* _G. C. Greubel_, Sep 06 2021 *)

%o (Magma) I:=[0,5]; [n le 2 select I[n] else 5*(Self(n-1) +Self(n-2)): n in [1..41]]; // _G. C. Greubel_, Sep 06 2021

%o (Sage) [5*lucas_number1(n, 5, -5) for n in (0..40)] # _G. C. Greubel_, Sep 06 2021

%Y Cf. A057088.

%K nonn,easy

%O 0,2

%A _Roger L. Bagula_, May 30 2005

%E Name changed by _G. C. Greubel_, Sep 06 2021