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Expansion of 1/((1-6x)(1-10x)(1-12x)).
1

%I #18 Aug 20 2022 17:47:01

%S 1,28,532,8560,125776,1747648,23401792,305401600,3912299776,

%T 49432480768,618099069952,7666644643840,94496470552576,

%U 1158938055589888,14157139120832512,172384964172513280

%N Expansion of 1/((1-6x)(1-10x)(1-12x)).

%H Harvey P. Dale, <a href="/A020758/b020758.txt">Table of n, a(n) for n = 0..925</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (28,-252,720).

%F a(n) = 22*a(n-1) - 120*a(n-2) + 6^n for n>1, a(0)=1, a(1)=28. - _Vincenzo Librandi_, Mar 11 2011

%F a(n) = (2*12^(n+2) - 3*10^(n+2) + 6^(n+2))/24. [_Yahia Kahloune_, Jun 30 2013]

%t CoefficientList[Series[1/((1-6x)(1-10x)(1-12x)),{x,0,20}],x] (* or *) LinearRecurrence[ {28,-252,720},{1,28,532},20] (* _Harvey P. Dale_, Aug 20 2022 *)

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_.