login
a(n) = n*(n-1)^2*(n-2).
17

%I #43 Sep 08 2022 08:44:57

%S 0,12,72,240,600,1260,2352,4032,6480,9900,14520,20592,28392,38220,

%T 50400,65280,83232,104652,129960,159600,194040,233772,279312,331200,

%U 390000,456300,530712,613872,706440,809100,922560,1047552,1184832

%N a(n) = n*(n-1)^2*(n-2).

%H Vincenzo Librandi, <a href="/A047928/b047928.txt">Table of n, a(n) for n = 2..1000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).

%F a(n) = A002378(n)*A002378(n+1). - _Zerinvary Lajos_, Apr 11 2006

%F a(n) = 12*A002415(n+1) = 2*A083374(n) = 4*A006011(n+1) = 6*A008911(n+1). - _Zerinvary Lajos_, May 09 2007

%F a(n) = floor((n-1)^6/((n-1)^2+1)). - _Gary Detlefs_, Feb 11 2010

%F From _Amiram Eldar_, Nov 05 2020: (Start)

%F Sum_{n>=3} 1/a(n) = 7/4 - Pi^2/6.

%F Sum_{n>=3} (-1)^(n+1)/a(n) = Pi^2/12 - 3/4. (End)

%F G.f.: -12*x*(1+x)/(-1+x)^5. - _Harvey P. Dale_, Jul 31 2021

%F a(n) = (n-1)^4 - (n-1)^2. - _Katherine E. Stange_, Mar 31 2022

%p seq(floor(n^6/(n^2+1)),n=1..25); # _Gary Detlefs_, Feb 11 2010

%t f[n_]:=n*(n-1)^2*(n-2); f[Range[2,60]] (* _Vladimir Joseph Stephan Orlovsky_, Feb 10 2011 *)

%t LinearRecurrence[{5,-10,10,-5,1},{0,12,72,240,600},40] (* or *) CoefficientList[Series[-((12 x (1+x))/(-1+x)^5),{x,0,40}],x] (* _Harvey P. Dale_, Jul 31 2021 *)

%o (Magma) [ n*(n-1)^2*(n-2): n in [2..40]]; // _Vincenzo Librandi_, May 02 2011

%o (PARI) a(n)=n^4 - 4*n^3 + 5*n^2 - 2*n \\ _Charles R Greathouse IV_, May 02, 2011

%Y Cf. A002378, A002415, A006011, A008911, A083374.

%K nonn,easy

%O 2,2

%A _N. J. A. Sloane_