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a(n) = 4*a(n-2) - 4*a(n-4) + 25*a(n-6).
8

%I #23 Sep 08 2022 08:45:18

%S 1,1,1,37,37,169,169,553,553,2461,2461,11857,11857,51409,51409,219733,

%T 219733,969721,969721,4285177,4285177,18755149,18755149,82122913,

%U 82122913,360600481,360600481,1582788997,1582788997,6941826889

%N a(n) = 4*a(n-2) - 4*a(n-4) + 25*a(n-6).

%H Vincenzo Librandi, <a href="/A107248/b107248.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0, 4, 0, -4, 0, 25).

%F From _R. J. Mathar_, Jul 03 2009: (Start)

%F a(n) = 4*a(n-2) - 4*a(n-4) + 25*a(n-6).

%F G.f.: -(1+x-3*x^2+33*x^3+37*x^4+25*x^5)/((5*x^3-2*x^2+1)*(5*x^3+2*x^2-1)).

%F (End)

%t CoefficientList[Series[-(1 + x - 3 x^2 + 33 x^3 + 37 x^4 + 25 x^5)/((5 x^3 - 2 x^2 + 1) (5 x^3 + 2 x^2 - 1)), {x, 0, 29}], x] (* _Michael De Vlieger_, Feb 20 2018 *)

%t LinearRecurrence[{0, 4, 0, -4, 0, 25}, {1, 1, 1, 37, 37, 169}, 30] (* _Vincenzo Librandi_, Feb 21 2018 *)

%o (Magma) I:=[1,1,1,37,37,169]; [n le 6 select I[n] else 4*Self(n-2)-4*Self(n-4)+25*Self(n-6): n in [1..40]]; // _Vincenzo Librandi_, Feb 21 2018

%Y Cf. A007493.

%K nonn,easy,less

%O 0,4

%A _Roger L. Bagula_, Jun 08 2005

%E Edited by _N. J. A. Sloane_, May 13 2006

%E New name using Mathar's recurrence from _Joerg Arndt_, Feb 20 2018