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%I #33 Feb 19 2023 02:55:41
%S 20,182,506,992,1640,2450,3422,4556,5852,7310,8930,10712,12656,14762,
%T 17030,19460,22052,24806,27722,30800,34040,37442,41006,44732,48620,
%U 52670,56882,61256,65792,70490,75350,80372,85556,90902,96410,102080,107912,113906,120062
%N a(n) = (9*n+4)*(9*n+5).
%C Cf. comment of _Reinhard Zumkeller_ in A177059: in general, (h*n+h-k)*(h*n+k)=h^2*A002061(n+1)+(h-k)*k-h^2; therefore a(n)=81*A002061(n+1)-61. - _Bruno Berselli_, Aug 24 2010
%H Vincenzo Librandi, <a href="/A177073/b177073.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 (3,-3,1).
%F a(n) = 162*n+a(n-1) with n>0, a(0)=20.
%F From _Harvey P. Dale_, Jun 24 2011: (Start)
%F a(0)=20, a(1)=182, a(2)=506, a(n) = 3*a(n-1)-3*a(n-2)+a(n-3).
%F G.f.: -2*(x*(10*x+61)+10)/(x-1)^3. (End)
%F From _Amiram Eldar_, Feb 19 2023: (Start)
%F a(n) = A017209(n)*A017221(n).
%F Sum_{n>=0} 1/a(n) = tan(Pi/18)*Pi/9.
%F Product_{n>=0} (1 - 1/a(n)) = sec(Pi/18)*cos(sqrt(5)*Pi/18).
%F Product_{n>=0} (1 + 1/a(n)) = sec(Pi/18)*cosh(sqrt(3)*Pi/18). (End)
%t f[n_] := Module[{c = 9n}, (c+4)(c+5)]; Array[f, 40, 0] (* or *) LinearRecurrence[{3, -3, 1}, {20, 182, 506}, 40] (* _Harvey P. Dale_, Jun 24 2011 *)
%o (Magma) [(9*n+4)*(9*n+5): n in [0..50]]; // _Vincenzo Librandi_, Apr 08 2013
%o (PARI) a(n)=(9*n+4)*(9*n+5) \\ _Charles R Greathouse IV_, Jun 17 2017
%Y Cf. A002061, A017209, A017221, A177059.
%K nonn,easy
%O 0,1
%A _Vincenzo Librandi_, May 31 2010
%E Edited by _N. J. A. Sloane_, Jun 22 2010
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