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A116156 a(n) = 5^n * n*(n + 1).

%I A116156 #28 Mar 10 2025 13:56:20
%S A116156 0,10,150,1500,12500,93750,656250,4375000,28125000,175781250,
%T A116156 1074218750,6445312500,38085937500,222167968750,1281738281250,
%U A116156 7324218750000,41503906250000,233459472656250,1304626464843750,7247924804687500
%N A116156 a(n) = 5^n * n*(n + 1).
%H A116156 Vincenzo Librandi, <a href="/A116156/b116156.txt">Table of n, a(n) for n = 0..1000</a>
%H A116156 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (15,-75,125).
%F A116156 G.f.: 10*x/(1-5*x)^3. - _Vincenzo Librandi_, Feb 28 2013
%F A116156 a(n) = 15*a(n-1) -75*a(n-2) +125*a(n-3). - _Vincenzo Librandi_, Feb 28 2013
%F A116156 a(n) = 10*A084902(n). - _Bruno Berselli_, Feb 28 2013
%F A116156 E.g.f.: 5*x*(2 + 5*x)*exp(5*x). - _G. C. Greubel_, May 10 2019
%F A116156 From _Amiram Eldar_, Jul 20 2020: (Start)
%F A116156 Sum_{n>=1} 1/a(n) = 1 - 4*log(5/4).
%F A116156 Sum_{n>=1} (-1)^(n+1)/a(n) = 6*log(6/5) - 1. (End)
%t A116156 Table[(n^2 + n) 5^n, {n, 0, 30}] (* or *) CoefficientList[Series[10 x/(1 - 5 x)^3, {x, 0, 30}], x](* _Vincenzo Librandi_, Feb 28 2013 *)
%o A116156 (Magma) [(n^2+n)*5^n: n in [0..30]]; // _Vincenzo Librandi_, Feb 28 2013
%o A116156 (Magma) I:=[0,10,150]; [n le 3 select I[n] else 15*Self(n-1)-75*Self(n-2)+125*Self(n-3): n in [1..30]]; // _Vincenzo Librandi_, Feb 28 2013
%o A116156 (PARI) a(n)=(n^2+n)*5^n \\ _Charles R Greathouse IV_, Feb 28 2013
%o A116156 (Sage) [5^n*n*(n+1) for n in (0..30)] # _G. C. Greubel_, May 10 2019
%o A116156 (GAP) List([0..30], n-> 5^n*n*(n+1)); # _G. C. Greubel_, May 10 2019
%Y A116156 Cf. A007758, A036289, A084902, A128796.
%K A116156 nonn,easy
%O A116156 0,2
%A A116156 _Mohammad K. Azarian_, Apr 08 2007