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%I #30 Jan 30 2020 21:29:16
%S 1,0,-3,-3,-6,-12,-27,-63,-153,-381,-969,-2505,-6564,-17394,-46533,
%T -125505,-340902,-931716,-2560401,-7070337,-19609146,-54597852,
%U -152556057,-427642677,-1202289669,-3389281245,-9578183391,-27130207503,-77009455428,-219023318406
%N Expansion of 1 - 3*(1-x-sqrt(1-2*x-3*x^2))/2.
%C For n>0, a(n) = -3*A168049(n). Hankel transform is A168075. Another variant is A168073.
%H G. C. Greubel, <a href="/A168076/b168076.txt">Table of n, a(n) for n = 0..1000</a>
%F a(n) = 0^n - 3*Sum_{k=0..floor((n-2)/2), C(n-2,2k)*A000108(k)}.
%F D-finite with recurrence: n*a(n) + (-2*n+3)*a(n-1) + 3*(-n+3)*a(n-2) = 0. - _R. J. Mathar_, Dec 03 2014
%F Recurrence (for n >= 3) follows from the differential equation (3*x^2+2*x-1)*y' - (3*x+1)*y = 3*x-1 satisfied by the g.f. - _Robert Israel_, May 13 2018
%F a(n) ~ -3^(n+1/2) / (2*sqrt(Pi)*n^(3/2)). - _Vaclav Kotesovec_, Dec 03 2014
%F a(n) = -A168073(n) <= 0 for n >= 1. - _M. F. Hasler_, May 13 2018
%p f:= gfun:-rectoproc({(3*n-3)*a(n)+(1+2*n)*a(n+1)+(-n-2)*a(n+2), a(0) = 1, a(1) = 0, a(2) = -3},a(n),remember):
%p map(f, [$0..60]); # _Robert Israel_, May 13 2018
%t CoefficientList[Series[1 - 3*(1 - x - Sqrt[1 - 2*x - 3*x^2])/2, {x,0,50}] , x] (* _G. C. Greubel_, Jul 09 2016 *)
%o (PARI) x='x+O('x^99); Vec(1-3*(1-x-(1-2*x-3*x^2)^(1/2))/2) \\ _Altug Alkan_, May 13 2018
%o (PARI) A168076(n)=!n-3*sum(k=0,n\2-1, binomial(n-2,2*k)*binomial(2*k,k)/(k+1)) \\ _M. F. Hasler_, May 13 2018
%Y Cf. A168049, A168073, A168075.
%K easy,sign
%O 0,3
%A _Paul Barry_, Nov 18 2009
%E Comment corrected by _Vaclav Kotesovec_, Dec 03 2014
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