%I #26 Jun 23 2022 18:55:27
%S 1,30,620,11160,188976,3108960,50434240,812507520,13044728576,
%T 209073047040,3348029967360,53591377582080,857645259698176,
%U 13723790036459520,219592368170516480,3513571713573027840,56217898008516427776,899492372901406310400
%N Expansion of 1/((1-2*x)*(1-4*x)(1-8*x)(1-16*x)).
%H T. D. Noe, <a href="/A028258/b028258.txt">Table of n, a(n) for n = 0..100</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (30,-280,960,-1024).
%F Difference of Gaussian binomial coefficients [ n+1, 4 ]-[ n, 4 ] (n >= 3).
%F a(n) = 30*a(n-1)-280*a(n-2)+960*a(n-3)-1024*a(n-4), with a(0)=1, a(1)=30, a(2)=620, a(3)=11160. - _Harvey P. Dale_, Jun 18 2011
%F a(n) = (2^n*(2^(n+1)-1)*(2^((n+1)+1)-1)*(2^(n+3)-1))/21. - _Harvey P. Dale_, Jun 18 2011; offset corrected by _Charles R Greathouse IV_, Feb 10 2017
%F E.g.f.: exp(2*x)*(64*exp(14*x) - 56*exp(6*x) + 14*exp(2*x) - 1)/21. - _Stefano Spezia_, Jun 23 2022
%t CoefficientList[Series[1/((1-2x)(1-4x)(1-8x)(1-16x)),{x,0,50}],x] (* or *) LinearRecurrence[{30,-280,960,-1024},{1,30,620,11160},50] (* or *) Table[(2^(n-1)(2^n-1)(2^(n+1)-1)(2^(n+2)-1))/21,{n,20}] (* _Harvey P. Dale_, Jun 18 2011 *)
%o (PARI) a(n)=(2^n*(2^(n+1)-1)*(2^((n+1)+1)-1)*(2^(n+3)-1))/21 \\ _Charles R Greathouse IV_, Feb 10 2017
%Y Cf. A006516, A016290, A006097.
%K nonn,easy,nice
%O 0,2
%A _N. J. A. Sloane_
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