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%I #14 Oct 06 2015 04:02:25
%S 1,0,-2,-3,2,15,19,-28,-134,-129,353,1254,791,-4238,-11818,-3123,
%T 49162,110007,-17783,-554458,-996323,690932,6096792,8624747,-12287153,
%U -65419110,-69285296,178655307,684550946,483569751,-2354830741,-6970706252,-2324044054,29195280375,68793790705
%N Row sums of triangle A104505, which is equal to the right-hand side of the triangle A084610 of coefficients in (1+x-x^2)^n.
%F G.f.: (x/((1-x))+1/((-sqrt(5*x^2-2*x+1)+x+1))*x*(1-(5*x-1)/(sqrt(5*x^2-2*x+1)))). - _Vladimir Kruchinin_, Oct 04 2015
%F a(n) = Sum_{j=0..n/2}((-1)^j*binomial(n,j)*binomial(n-j-1,n-2*j)). - _Vladimir Kruchinin_, Oct 04 2015
%t CoefficientList[Series[(x/((1 - x)) + 1/((-Sqrt[5 x^2 - 2 x + 1] + x + 1)) x (1 - (5 x - 1)/(Sqrt[5 x^2 - 2 x + 1]))), {x, 0, 40}], x] (* _Vincenzo Librandi_, Oct 05 2015 *)
%o (PARI) a(n)=sum(k=0,n,polcoeff((1+x-x^2)^n,n+k))
%o (Maxima)
%o a(n):=sum((-1)^j*binomial(n,j)*binomial(n-j-1,n-2*j),j,0,n/2); /* _Vladimir Kruchinin_, Oct 04 2015 */
%o (PARI) a(n) = sum(k=0, n/2, (-1)^k*binomial(n,k)*binomial(n-k-1,n-2*k));
%o vector(40, n, a(n-1)) \\ _Altug Alkan_, Oct 04 2015
%Y Cf. A104505.
%K sign
%O 0,3
%A _Paul D. Hanna_, Mar 11 2005
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