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a(n) = (3*9^n + 1)/2.
4

%I #16 Sep 14 2024 15:51:08

%S 2,14,122,1094,9842,88574,797162,7174454,64570082,581130734,

%T 5230176602,47071589414,423644304722,3812798742494,34315188682442,

%U 308836698141974,2779530283277762,25015772549499854,225141952945498682,2026277576509488134,18236498188585393202

%N a(n) = (3*9^n + 1)/2.

%H Vincenzo Librandi, <a href="/A199560/b199560.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 a(n) = 2*A066443(n).

%F a(n) = 9*a(n-1) - 4.

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

%F G.f.: 2*(1-3*x)/((1-x)*(1-9*x)).

%F From _Elmo R. Oliveira_, Sep 13 2024: (Start)

%F E.g.f.: exp(x)*(3*exp(8*x) + 1)/2.

%F a(n) = A199561(n)/2. (End)

%t LinearRecurrence[{10,-9},{2,14},30] (* or *) NestList[9#-4&,2,30] (* _Harvey P. Dale_, May 30 2012 *)

%o (Magma) [(3*9^n+1)/2: n in [0..30]];

%Y Cf. A066443, A199561.

%K nonn,easy

%O 0,1

%A _Vincenzo Librandi_, Nov 08 2011