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%I #23 Jun 27 2022 21:19:06
%S 1,35,790,14630,242011,3723545,54492880,769001660,10558544821,
%T 141919432655,1875649181770,24453637571090,315271616180431,
%U 4027179810711365,51043597496835460,642723844104282920
%N Expansion of 1/((1-3x)(1-9x)(1-11x)(1-12x)).
%H Vincenzo Librandi, <a href="/A028107/b028107.txt">Table of n, a(n) for n = 0..200</a>
%H <a href="/index/Rec">Index entries for linear recurrences with constant coefficients</a>, signature (35,-435,2205,-3564).
%F a(n) = (16*12^(n+3) - 27*11^(n+3) + 12*9^(n+3) - 3^(n+3))/432 = (1/16)*(-3^n + 3^n*4^(n+5) + 4*9^(n+2) - 11^(n+3)). - _Yahia Kahloune_, May 31 2013
%F a(n) = 35*a(n-1) - 435*a(n-2) + 2205*a(n-3) - 3564*a(n-4) for n > 3; a(0)=1, a(1)=35, a(2)=790, a(3)=14630. - _Vincenzo Librandi_, Jul 17 2013
%t CoefficientList[Series[1 / ((1 - 3 x) (1 - 9 x) (1 - 11 x)(1 - 12 x)), {x, 0, 20}], x] (* _Vincenzo Librandi_, Jul 17 2013 *)
%t LinearRecurrence[{35,-435,2205,-3564},{1,35,790,14630},20] (* _Harvey P. Dale_, Jul 31 2020 *)
%o (Magma) m:=25; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!(1/((1-3*x)*(1-9*x)*(1-11*x)*(1-12*x)))); /* or */ I:=[1,35,790,14630]; [n le 4 select I[n] else 35*Self(n-1)-435*Self(n-2)+2205*Self(n-3)-3564*Self(n-4): n in [1..25]]; // _Vincenzo Librandi_, Jul 17 2013
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_
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