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a(n) = 9^n * n*(n+1).
1

%I #25 Sep 08 2022 08:45:24

%S 0,18,486,8748,131220,1771470,22320522,267846264,3099363912,

%T 34867844010,383546284110,4142299868388,44059007691036,

%U 462619580755878,4804126415541810,49413871702715760,504021491367700752

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

%H Vincenzo Librandi, <a href="/A116176/b116176.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (27,-243,729).

%F G.f.: 18*x/(1-9*x)^3. - _Vincenzo Librandi_, Feb 28 2013

%F a(n) = 27*a(n-1) - 243*a(n-2) + 729*a(n-3). - _Vincenzo Librandi_, Feb 28 2013

%F a(n) = 18*A081139(n+1). - _Bruno Berselli_, Mar 01 2013

%F E.g.f.: 9*x*(2 + 9*x)*exp(9*x). - _G. C. Greubel_, May 11 2019

%F From _Amiram Eldar_, Jul 20 2020: (Start)

%F Sum_{n>=1} 1/a(n) = 1 - 8*log(9/8).

%F Sum_{n>=1} (-1)^(n+1)/a(n) = 10*log(10/9) - 1. (End)

%t Table[(n^2 + n) 9^n, {n, 0, 20}] (* _Vincenzo Librandi_, Feb 28 2013 *)

%o (Magma) [(n^2+n)*9^n: n in [0..20]]; // _Vincenzo Librandi_, Feb 28 2013

%o (PARI) {a(n) = 9^n*n*(n+1)}; \\ _G. C. Greubel_, May 11 2019

%o (Sage) [9^n*n*(n+1) for n in (0..20)] # _G. C. Greubel_, May 11 2019

%o (GAP) List([0..20], n-> 9^n*n*(n+1)) # _G. C. Greubel_, May 11 2019

%Y Cf. A007758, A036289, A081139, A128796.

%K nonn,easy

%O 0,2

%A _Mohammad K. Azarian_, Apr 08 2007