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%I #17 Apr 11 2022 12:59:47
%S 3,1,-1,3,19,53,111,199,323,489,703,971,1299,1693,2159,2703,3331,4049,
%T 4863,5779,6803,7941,9199,10583,12099,13753,15551,17499,19603,21869,
%U 24303,26911,29699,32673,35839,39203,42771,46549,50543,54759,59203
%N a(n) = n^3 - 3n^2 + 3.
%C For n>2, fourth diagonal of A162611.
%H B. Berselli, <a href="/A177058/b177058.txt">Table of n, a(n) for n = 0..10000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F From _Bruno Berselli_, Jun 04 2010: (Start)
%F G.f.: (3-11*x+13*x^2+x^3)/(1-x)^4.
%F a(n)-4*a(n-1)+6*a(n-2)-4*a(n-3)+a(n-4) = 0, with n>3.
%F a(n)+a(n-1) = 2*A081438(n-3), with n>2. (End)
%F G.f.: 3+x+x^2*G(0) where G(k) = 1 - x*(k+1)*(k+1)*(k+4)/(1 - 1/(1 - (k+1)*(k+1)*(k+4)/G(k+1))); (continued fraction, 3-step). - _Sergei N. Gladkovskii_, Oct 16 2012
%t Table[n^3-3n^2+3,{n,0,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{3,1,-1,3},50] (* _Harvey P. Dale_, May 15 2020 *)
%o (PARI) a(n)=n^3-3*n^2+3 \\ _Charles R Greathouse IV_, Jan 11 2012
%Y Cf. A162611, A081438.
%K sign,easy
%O 0,1
%A _Vincenzo Librandi_, May 27 2010