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a(n) = (n+1)*(n+6)*3^n/6.
7

%I #27 Jan 11 2024 01:38:56

%S 1,7,36,162,675,2673,10206,37908,137781,492075,1732104,6022998,

%T 20726199,70681653,239148450,803538792,2683245609,8910671247,

%U 29443957164,96855122250,317297380491,1035574967097,3368233731366,10920608743932,35303692060125,113819103201843

%N a(n) = (n+1)*(n+6)*3^n/6.

%H Vincenzo Librandi, <a href="/A080420/b080420.txt">Table of n, a(n) for n = 0..1000</a>

%H Gregory Gerard Wojnar, Daniel Sz. Wojnar, and Leon Q. Brin, <a href="http://arxiv.org/abs/1706.08381">Universal peculiar linear mean relationships in all polynomials</a>, arXiv:1706.08381 [math.GM], 2017. See p. 4.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (9,-27,27).

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

%F From _G. C. Greubel_, Dec 22 2023: (Start)

%F a(n) = (n+6)*A288834(n)/2, for n >= 1.

%F a(n) = A136158(n+2, 2).

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

%F From _Amiram Eldar_, Jan 11 2024: (Start)

%F Sum_{n>=0} 1/a(n) = 17721/50 - 4356*log(3/2)/5.

%F Sum_{n>=0} (-1)^n/a(n) = 4392*log(4/3)/5 - 12591/50. (End)

%t CoefficientList[Series[(1 - 2 x) / (1 - 3 x)^3, {x, 0, 30}], x] (* _Vincenzo Librandi_, Aug 05 2013 *)

%t Table[(n+1)(n+6)3^n/6,{n,0,30}] (* or *) LinearRecurrence[{9,-27,27},{1,7,36},30] (* _Harvey P. Dale_, Apr 02 2019 *)

%o (Magma) [(n+1)*(n+6)*3^n/6: n in [0..30]]; // _Vincenzo Librandi_, Aug 05 2013

%o (SageMath) [(n+1)*(n+6)*3^n/6 for n in range(31)] # _G. C. Greubel_, Dec 22 2023

%Y T(n,2) in triangle A080419.

%Y Cf. A136158, A288834.

%K easy,nonn

%O 0,2

%A _Paul Barry_, Feb 19 2003