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a(n+1) = 9*a(n) - 6, a(0) = 2.
1

%I #11 Sep 08 2022 08:45:32

%S 2,12,102,912,8202,73812,664302,5978712,53808402,484275612,4358480502,

%T 39226324512,353036920602,3177332285412,28595990568702,

%U 257363915118312,2316275236064802,20846477124583212,187618294121248902,1688564647091240112,15197081823821161002

%N a(n+1) = 9*a(n) - 6, a(0) = 2.

%H Colin Barker, <a href="/A137483/b137483.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 (10,-9).

%F From _Colin Barker_, Feb 05 2016: (Start)

%F a(n) = (3+5*9^n)/4.

%F a(n) = 10*a(n-1)-9*a(n-2) for n>1.

%F G.f.: 2*(1-4*x) / ((1-x)*(1-9*x)).

%F (End)

%t RecurrenceTable[{a[1] == 2, a[n] == 9 a[n-1] - 6}, a, {n, 30}] (* _Vincenzo Librandi_, Feb 06 2016 *)

%o (PARI) a(n) = (3+5*9^n)/4 \\ _Colin Barker_, Feb 05 2016

%o (PARI) Vec(2*(1-4*x)/((1-x)*(1-9*x)) + O(x^25)) \\ _Colin Barker_, Feb 05 2016

%o (Magma) [n le 1 select 2 else 9*Self(n-1)-6: n in [1..30]]; // _Vincenzo Librandi_, Feb 06 2016

%K nonn,easy

%O 0,1

%A _Paul Curtz_, Apr 22 2008