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%I #27 Jun 01 2020 20:48:57
%S 1,1,-1,2,-4,9,-21,51,-127,323,-835,2188,-5798,15511,-41835,113634,
%T -310572,853467,-2356779,6536382,-18199284,50852019,-142547559,
%U 400763223,-1129760415,3192727797,-9043402501,25669818476,-73007772802
%N A signed variant of the Motzkin numbers.
%C Hankel transform is A131713. Binomial transform is A166588.
%C [a(n+1)] = [1,-1,2,-4,9,...] is the inverse binomial transform of A126120. - _Philippe Deléham_, Nov 29 2009
%H G. C. Greubel, <a href="/A166587/b166587.txt">Table of n, a(n) for n = 0..500</a>
%F G.f.: (1+3x-sqrt(1+2x-3x^2))/(2x); (1+3x)/(1+2x-x^2/(1+x-x^2/(1+x-x^2/(1+x-x^2/(1+...))))) (continued fraction).
%F a(n) = 0^n + Sum_{k=0..n} binomial(n-1, k-1)*(-3)^(n-k)*A000108(k).
%F G.f.: (1+3*x-sqrt(1+2*x-3*x^2))/(2x) = (3-1/G(0))/2 ; G(k) = 1+2*x/(1-x/(1-x/(1+2*x/(1+x/(2+x/G(k+1)))))) ; (continued fraction). - Sergei N. Gladkovskii, Dec 11 2011
%F Conjecture: n*(n+1)*a(n) + n*(n+1)*a(n-1) - (5*n-3)*(n-2)*a(n-2) + 3*(n-2)*(n-3)*a(n-3) = 0. - _R. J. Mathar_, Nov 15 2012
%F G.f. G(x) satisfies (3 x^2 - 2 x^2 - x) G'(x) - (x+1) G(x) + 3 x + 1 = 0, from which follows 3*n*a(n) + (-3-2*n)*a(1+n) + (-3-n)*a(n+2) = 0 as well as Mathar's conjecture. - _Robert Israel_, May 17 2016
%F E.g.f.: 1 + Integral (exp(-x) * BesselI(1,2*x) / x) dx. - _Ilya Gutkovskiy_, Jun 01 2020
%e G.f. = 1 + x - x^2 + 2*x^3 - 4*x^4 + 9*x^5 - 21*x^6 + 51*x^7 - 127*x^8 + ...
%p f:= gfun:-rectoproc({3*n*a(n)+(-3-2*n)*a(1+n)+(-3-n)*a(n+2)=0,a(0) = 1, a(1) = 1}, a(n),remember):
%p map(f, [$0..100]); # _Robert Israel_, May 17 2016
%p with(PolynomialTools): CoefficientList(convert(taylor((1 + 3*x - sqrt(1 + 2*x - 3*x^2))/2/x, x = 0, 33), polynom), x); # _Taras Goy_, Aug 07 2017
%t CoefficientList[Series[(1 + 3*t - Sqrt[1 + 2*t - 3*t^2])/(2 t), {t, 0, 50}], t] (* _G. C. Greubel_, May 17 2016 *)
%K easy,sign
%O 0,4
%A _Paul Barry_, Oct 17 2009
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