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a(n) = n*(n-1)*5^n.
1

%I #25 Sep 08 2022 08:45:30

%S 0,0,50,750,7500,62500,468750,3281250,21875000,140625000,878906250,

%T 5371093750,32226562500,190429687500,1110839843750,6408691406250,

%U 36621093750000,207519531250000,1167297363281250,6523132324218750

%N a(n) = n*(n-1)*5^n.

%H Vincenzo Librandi, <a href="/A128799/b128799.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 (15,-75,125).

%F G.f.: 50*x^2/(1 - 5*x)^3. - _Vincenzo Librandi_, Feb 10 2013

%F a(n) = 50*A081135(n). - _R. J. Mathar_, Apr 26 2015

%F E.g.f.: 25*x^2*exp(5*x). - _G. C. Greubel_, May 17 2021

%F a(n) = 15*a(n-1) - 75*a(n-2) + 125*a(n-3). - _Wesley Ivan Hurt_, May 17 2021

%p seq(5^n*n*(n-1), n=0..30); # _G. C. Greubel_, May 17 2021

%t CoefficientList[Series[50x^2/(1-5x)^3, {x, 0, 30}] ,x] (* _Vincenzo Librandi_, Feb 10 2013 *)

%t Table[n(n-1)5^n,{n,0,30}] (* or *) LinearRecurrence[{15,-75,125},{0,0,50},30] (* _Harvey P. Dale_, Nov 05 2019 *)

%o (Magma) [(n^2-n)*5^n: n in [0..20]]; // _Vincenzo Librandi_, Feb 10 2013

%o (Sage) [2*5^n*binomial(n,2) for n in (0..30)] # _G. C. Greubel_, May 17 2021

%Y Cf. A007758, A036289, A081135.

%K nonn,easy

%O 0,3

%A _Mohammad K. Azarian_, Apr 07 2007