%I #21 Sep 08 2022 08:44:45
%S 1,30,601,10050,151501,2135070,28702801,372684090,4712104501,
%T 58346365110,710428956601,8532288986130,101313313019101,
%U 1191569650755150,13901375026212001,161062105099480170
%N Expansion of 1/((1-9*x)*(1-10*x)*(1-11*x)).
%H G. C. Greubel, <a href="/A020982/b020982.txt">Table of n, a(n) for n = 0..955</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (30,-299,990)
%F If we define f(m,j,x) = Sum_{k=j..m} (binomial(m,k)*stirling2(k,j)*x^(m-k)) then a(n-2)=f(n,2,9), (n>=2). - _Milan Janjic_, Apr 26 2009
%F a(n) = 30*a(n-1) -299*a(n-2) +990*a(n-3), n>=3. - _Vincenzo Librandi_, Mar 18 2011
%F a(n) = 21*a(n-1) -110*a(n-2) +9^n, n>=2. - _Vincenzo Librandi_, Mar 18 2011
%F a(n) = 11^(n+2)/2+9^(n+2)/2-10^(n+2). - _R. J. Mathar_, Mar 20 2011
%t CoefficientList[Series[1/((1-9x)(1-10x)(1-11x)),{x,0,20}],x] (* or *) LinearRecurrence[{30,-299,990},{1,30,601},20] (* _Harvey P. Dale_, Jan 30 2013 *)
%o (PARI) x='x+O('x^30); Vec(1/((1-9*x)*(1-10*x)*(1-11*x))) \\ _G. C. Greubel_, Feb 09 2018
%o (Magma) R<x>:=PowerSeriesRing(Integers(), 30); Coefficients(R!(1/((1-9*x)*(1-10*x)*(1-11*x)))); // _G. C. Greubel_, Feb 09 2018
%K nonn
%O 0,2
%A _N. J. A. Sloane_
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