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Expansion of 1/((1-x)^5*(x^3+x^2+x+1)^3).
4

%I #23 Jun 16 2016 23:27:44

%S 1,2,3,4,8,12,16,20,30,40,50,60,80,100,120,140,175,210,245,280,336,

%T 392,448,504,588,672,756,840,960,1080,1200,1320,1485,1650,1815,1980,

%U 2200,2420,2640,2860,3146,3432,3718,4004,4368

%N Expansion of 1/((1-x)^5*(x^3+x^2+x+1)^3).

%C The Gi1 triangle sums of A139600 lead to the sequence given above, see the formulas. For the definitions of the Gi1 and other triangle sums see A180662.

%H <a href="/index/Rec#order_14">Index entries for linear recurrences with constant coefficients</a>, signature (2, -1, 0, 3, -6, 3, 0, -3, 6, -3, 0, 1, -2, 1).

%F a(n) = sum(A056594(n-k)*A115269(k), k=0..n).

%F Gi1(n) = A189375(n-4) - A189375(n-5) - A189375(n-8) + 2*A189375(n-9) with A189375(n)=0 for n <= -1.

%F a(n) = (2*n^4+56*n^3+538*n^2+2044*n+2469+3*((2*n^2+28*n+89)*(-1)^n+(4*(-1)^((2*n-1+(-1)^n)/4)*(n^2+16*n+57-(n^2+12*n+29)*(-1)^n))))/3072. - _Luce ETIENNE_, Jun 25 2015

%p a:= n-> coeff(series(1/((1-x)^5*(x^3+x^2+x+1)^3), x, n+1), x, n):

%p seq(a(n), n=0..50);

%t CoefficientList[Series[1/((1-x)^5(x^3+x^2+x+1)^3),{x,0,50}],x] (* or *) LinearRecurrence[{2,-1,0,3,-6,3,0,-3,6,-3,0,1,-2,1},{1,2,3,4,8,12,16,20,30,40,50,60,80,100},50] (* _Harvey P. Dale_, Dec 05 2014 *)

%Y Cf. A139600, A189374, A189376.

%K easy,nonn

%O 0,2

%A _Johannes W. Meijer_, Apr 29 2011