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%I #27 Sep 08 2022 08:45:40
%S 1,13,94,526,2551,11299,47020,186988,718429,2686729,9831658,35340826,
%T 125154355,437641663,1513809688,5187129880,17627632249,59469045061,
%U 199327841590,664232428390,2201904349231,7264715299483,23865295832644,78091766836996
%N Partial sums of A069996.
%C The first differences are in the third row of the square array of A072590.
%C The general formula for the partial sums of the sequence 1, 4*m, 9*m^2, 16*m^3, 25*m^4,...,n^2*m^(n-1),... is (n^2*m^(n+2)-(2*n*(n+1)-1)*m^(n+1)+(n+1)^2*m^n-m-1)/(m-1)^3 with m>1 (see also References).
%D "Supplemento al Periodico di Matematica", Raffaello Giusti Editore (Livorno) - Apr / May, 1913 - p. 99 (Problem 1277, case x=3).
%H Bruno Berselli, <a href="/A153703/b153703.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (10,-36,54,-27).
%F a(n) = (3^n*(n^2 - n + 1) - 1)/2.
%F G.f.: x*(1+3*x)/((1-x)*(1-3*x)^3).
%F a(n) = 10*a(n-1) - 36*a(n-2) + 54*a(n-3) - 27a(n-4) for n>4.
%F a(n) = 9*A027472(n+1) + A003462(n) for n>2.
%F E.g.f.: (1/2)*((1 + 9*x^2)*exp(x) - exp(-x))*exp(2*x). - _G. C. Greubel_, Aug 24 2016
%t CoefficientList[Series[(1 + 3 x) / ((1 - x) (1 - 3 x)^3), {x, 0, 25}], x] (* _Vincenzo Librandi_, Aug 19 2013 *)
%o (PARI) a(n) = (3^n*(n^2-n+1)-1)/2 \\ _Michel Marcus_, Jun 07 2013
%o (Magma) [(3^n*(n^2-n+1)-1)/2: n in [1..25]]; // _Vincenzo Librandi_, Aug 19 2013
%Y Cf. A069996, A072590, A027472, A003462.
%K nonn,easy
%O 1,2
%A _Bruno Berselli_, Dec 12 2010
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