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%I #25 Mar 16 2017 22:45:06
%S 1,0,0,1,2,5,12,30,76,196,512,1353,3610,9713,26324,71799,196938,
%T 542895,1503312,4179603,11662902,32652735,91695540,258215664,
%U 728997192,2062967382,5850674704,16626415975,47337954326
%N Expansion of (1-x)*c(x/(1+x)), where c(x) is the g.f. of the Catalan numbers (A000108).
%C Apply the Riordan array (1-x,x/(1+x)) to C(n)=A000108(n).
%C Alternatively, this sequence corresponds to the number of positive walks with n steps {-1,0,1} starting at the origin, ending at altitude 2, and staying strictly above the x-axis. - D. Nguyen, December 1, 2016.
%H Vincenzo Librandi, <a href="/A105695/b105695.txt">Table of n, a(n) for n = 0..200</a>
%H C. Banderier, C. Krattenthaler, A. Krinik, D. Kruchinin, V. Kruchinin, D. Nguyen, and M. Wallner, <a href="https://arxiv.org/abs/1609.06473">Explicit formulas for enumeration of lattice paths: basketball and the kernel method</a>, arXiv:1609.06473 [math.CO], 2016.
%F G.f.: (1-x^2)*(1-sqrt((1-3*x)/(1+x)))/(2*x).
%F Let b(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*C(k) = A005043(n); then a(n) = b(n) - b(n-2).
%F Conjecture: (n+1)*a(n)+(2-3n)*a(n-1) +(1-n)*a(n-2)+3*(n-4)*a(n-3)=0. - _R. J. Mathar_, Dec 13 2011
%F a(n) ~ 3^(n-1/2) / (sqrt(Pi) * n^(3/2)). - _Vaclav Kotesovec_, Feb 01 2014
%F From _Peter Bala_, Oct 29 2015: (Start)
%F a(n) = Sum_{k = 1..floor((n-1)/2)} binomial(n-2,2*k-1)*Catalan(k) for n >= 1.
%F (n+1)*(n-3)*a(n) = (n-2)*(2*n-3)*a(n-1) + 3*(n-2)*(n-3)*a(n-2) with a(2) = 0, a(3) = 1. Mathar's 4-term recurrence above follows easily from this. (End)
%t CoefficientList[Series[(1-x^2)*(1-Sqrt[(1-3*x)/(1+x)])/(2*x), {x, 0, 20}], x] (* _Vaclav Kotesovec_, Feb 01 2014 *)
%o (PARI) x='x+O('x^50); Vec((1-x^2)*(1-sqrt((1-3*x)/(1+x)))/(2*x)) \\ _G. C. Greubel_, Mar 16 2017
%Y Cf. A002026, A005043.
%K easy,nonn
%O 0,5
%A _Paul Barry_, Apr 17 2005
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