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Expansion of (1+x)^2/(1+x+x^2)^2.
2

%I #31 Jan 12 2016 10:58:05

%S 1,0,-2,2,1,-4,3,2,-6,4,3,-8,5,4,-10,6,5,-12,7,6,-14,8,7,-16,9,8,-18,

%T 10,9,-20,11,10,-22,12,11,-24,13,12,-26,14,13,-28,15,14,-30,16,15,-32,

%U 17,16,-34,18,17,-36,19,18,-38

%N Expansion of (1+x)^2/(1+x+x^2)^2.

%C Row sums of Riordan array (1/(1+x+x^2), x/(1+x)^2), A122917.

%C For n>=1, a(n) equals (-1)^(n+1) times the second immanant of the n X n matrix with 1's along the main diagonal, superdiagonal, and subdiagonal, and 0's everywhere else. The second immanant of an n X n matrix A is the immanant of A given by the partition (2, 1^(n-2)). - _John M. Campbell_, Apr 12 2014

%H Vincenzo Librandi, <a href="/A122918/b122918.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (-2,-3,-2,-1).

%F a(n) = 4 * sqrt(3) * cos(2*Pi*n/3 + Pi/6)/9 + 2(n+1) * sin(2*Pi*n/3 + Pi/6)/3. a(n) = sum{k=0..n} A057078(k) * A057078(n-k).

%F a(n) = (-1)^(n+1)*sum((-1)^i*binomial(n-i,i)*(n-2*i-1), i=0..[n/2]). - _John M. Campbell_, Jan 08 2016

%t CoefficientList[Series[(1 + x)^2/(1 + x + x^2)^2, {x, 0, 100}], x] (* _Vincenzo Librandi_, Apr 13 2014 *)

%t Print[Table[(-1)^(n+1)*Sum[Binomial[n-i, i]*(n-2*i-1)*(-1)^i, {i, 0, Floor[n/2]}], {n, 0, 100}]] ; (* _John M. Campbell_, Jan 08 2016 *)

%o (PARI) Vec((1+x)^2/(1+x+x^2)^2 + O(x^100)) \\ _Altug Alkan_, Jan 08 2015

%o (PARI) A122918(n)=(-1)^(n+1)*sum(i=0,n\2,(-1)^i*binomial(n-i,i)*(n-2*i-1)) \\ _M. F. Hasler_, Jan 12 2016

%Y Cf. A187430 (series reversion, with offset 1).

%K easy,sign

%O 0,3

%A _Paul Barry_, Sep 19 2006