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Expansion of 1/((1-4x)(1-5x)(1-11x)(1-12x)).
0

%I #15 Aug 31 2018 19:26:22

%S 1,32,665,11440,177441,2582592,36039145,488323280,6475959281,

%T 84497107552,1088651482425,13885593262320,175667802947521,

%U 2207466992246912,27583620276442505,343037771674508560

%N Expansion of 1/((1-4x)(1-5x)(1-11x)(1-12x)).

%H <a href="/index/Rec">Index entries for linear recurrences with constant coefficients</a>, signature (32,-359,1648,-2640)

%F From _Vincenzo Librandi_, Mar 17 2011: (Start)

%F a(n) = 32*a(n-1) - 359*a(n-2) + 1648*a(n-3) - 2640*a(n-4), n >= 4.

%F a(n) = 23*a(n-1) - 132*a(n-2) + 5^(n+1) - 4^(n+1), n >= 2. (End)

%F a(n) = 18*12^(n+1)/7 - 11^(n+3)/42 - 2*4^(n+1)/7 + 5^(n+3)/42. - _R. J. Mathar_, Mar 18 2011

%t CoefficientList[Series[1/((1-4x)(1-5x)(1-11x)(1-12x)),{x,0,30}],x] (* or *) LinearRecurrence[{32,-359,1648,-2640},{1,32,665,11440},30] (* _Harvey P. Dale_, Sep 24 2014 *)

%K nonn

%O 0,2

%A _N. J. A. Sloane_