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%I #36 Jan 28 2020 18:48:36
%S 1,1,0,1,0,2,0,5,0,14,0,42,0,132,0,429,0,1430,0,4862,0,16796,0,58786,
%T 0,208012,0,742900,0,2674440,0,9694845,0,35357670,0,129644790,0,
%U 477638700,0,1767263190,0,6564120420,0,24466267020,0,91482563640,0,343059613650,0
%N Expansion of 1 + 2x/(1 + sqrt(1 - 4x^2)).
%C Binomial transform is A097332. Second binomial transform is A014318.
%C Essentially the same as A126120. - _R. J. Mathar_, Jun 15 2008
%C Hankel transform is A087960(n) = (-1)^binomial(n+1,2). - _Paul Barry_, Aug 10 2009
%H Michael De Vlieger, <a href="/A097331/b097331.txt">Table of n, a(n) for n = 0..3340</a>
%H Jean-Luc Baril, Sergey Kirgizov, Armen Petrossian, <a href="http://math.colgate.edu/~integers/t46/t46.Abstract.html">Motzkin paths with a restricted first return decomposition</a>, Integers (2019) Vol. 19, A46.
%F a(n) = 0^n + Catalan((n-1)/2)(1-(-1)^n)/2.
%F Unsigned version of A090192, A105523. - _Philippe Deléham_, Sep 29 2006
%F From _Paul Barry_, Aug 10 2009: (Start)
%F G.f.: 1+xc(x^2), c(x) the g.f. of A000108;
%F G.f.: 1/(1-x/(1+x/(1+x/(1-x/(1-x/(1+x/(1+x/(1-x/(1-x/(1+... (continued fraction);
%F G.f.: 1+x/(1-x^2/(1-x^2/(1-x^2/(1-x^2/(1-... (continued fraction). (End)
%F G.f.: 1/(1-z/(1-z/(1-z/(...)))) where z=x/(1+2*x) (continued fraction); more generally g.f. C(x/(1+2*x)) where C(x) is the g.f. for the Catalan numbers (A000108). - _Joerg Arndt_, Mar 18 2011
%F Conjecture: (n+1)*a(n) + n*a(n-1) + 4*(-n+2)*a(n-2) + 4*(-n+3)*a(n-3)=0. - _R. J. Mathar_, Dec 02 2012
%F Recurrence: (n+3)*a(n+2) = 4*n*a(n), a(0)=a(1)=1. For nonzero terms, a(n) ~ 2^(n+1)/((n+1)^(3/2)*sqrt(2*Pi)). - _Fung Lam_, Mar 17 2014
%p A097331_list := proc(n) local j, a, w; a := array(0..n); a[0] := 1;
%p for w from 1 to n do a[w]:=a[w-1]-(-1)^w*add(a[j]*a[w-j-1],j=1..w-1) od; convert(a,list)end: A097331_list(48); # _Peter Luschny_, May 19 2011
%t a[0] = 1; a[n_?OddQ] := CatalanNumber[(n-1)/2]; a[_] = 0; Table[a[n], {n, 0, 48}] (* _Jean-François Alcover_, Jul 24 2013 *)
%o (Sage)
%o def A097331_list(n) :
%o D = [0]*(n+2); D[1] = 1
%o b = True; h = 1; R = []
%o for i in range(2*n-1) :
%o if b :
%o for k in range(h,0,-1) : D[k] -= D[k-1]
%o h += 1; R.append(abs(D[1]))
%o else :
%o for k in range(1,h, 1) : D[k] += D[k+1]
%o b = not b
%o return R
%o A097331_list(49) # _Peter Luschny_, Jun 03 2012
%K easy,nonn
%O 0,6
%A _Paul Barry_, Aug 05 2004
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