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a(n) = (5*9^n + 1)/2.
4

%I #28 Sep 08 2022 08:45:32

%S 3,23,203,1823,16403,147623,1328603,11957423,107616803,968551223,

%T 8716961003,78452649023,706073841203,6354664570823,57191981137403,

%U 514727830236623,4632550472129603,41692954249166423,375236588242497803

%N a(n) = (5*9^n + 1)/2.

%H Vincenzo Librandi, <a href="/A135423/b135423.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) = 9*a(n-1) - 4 for n>0, a(0)=3.

%F O.g.f.: (1/(1-x) - 5/(9*x-1))/2. - _R. J. Mathar_, Feb 19 2008

%F a(n) = 10*a(n-1) - 9*a(n-2). - _Vincenzo Librandi_, Nov 08 2011

%F E.g.f.: (1/2)*( 5*exp(9*x) + exp(x) ). - _G. C. Greubel_, Oct 14 2016

%t Table[ (5*9^n + 1)/2, {n,0,25}] (* or *) LinearRecurrence[{10, -9}, {3, 23}, 25] (* _G. C. Greubel_, Oct 14 2016 *)

%o (Magma) [(5*9^n+1)/2: n in [0..30]]; // _Vincenzo Librandi_, Nov 08 2011

%Y Bisection of A057198. Cf. A191450.

%K nonn,easy

%O 0,1

%A _Paul Curtz_, Feb 18 2008

%E More terms from _R. J. Mathar_, Feb 19 2008

%E Definition rewritten (with Mathar's formula) from _Bruno Berselli_, Nov 08 2011