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%I #46 Sep 24 2021 13:40:47
%S 1,0,1,1,0,-2,-3,1,11,15,-13,-77,-86,144,595,495,-1520,-4810,-2485,
%T 15675,39560,6290,-159105,-324805,87075,1592843,2616757,-2136539,
%U -15726114,-20247800,32296693,152909577,145139491,-417959049,-1460704685,-885536173,4997618808,13658704994,3223741399
%N G.f. A(x) satisfies: 2^n - 1 = Sum_{k=0..n} [x^k]A(x)^n and also satisfies: (2+z)^n - (1+z)^n + z^n = Sum_{k=0..n} [x^k](A(x)+z*x)^n for all z, where [x^k]A(x)^n denotes the coefficient of x^k in A(x)^n.
%C The current sequence and A007440, A214649, A343773 form a cluster. The base of the cluster is A343773. We get the current sequence if we shift all the terms in A343773 to the right by 2 and take a(0) = 1. - _Gennady Eremin_, May 15 2021
%H Gennady Eremin, <a href="/A100223/b100223.txt">Table of n, a(n) for n = 0..800</a>
%H Gennady Eremin, <a href="https://arxiv.org/abs/2108.10676">Walking in the OEIS: From Motzkin numbers to Fibonacci numbers. The "shadows" of Motzkin numbers</a>, arXiv:2108.10676 [math.CO], 2021.
%F G.f.: A(x) = (1+x+sqrt(1-2*x+5*x^2))/2.
%F G.f.: A(x) = x/(series_reversion[x*(1-x)/(1-x-x^2)]).
%F D-finite with recurrence a(n) = ((2*n-3)*a(n-1) - 5*(n-3)*a(n-2))/n (for n>2), with a(0)=1, a(1)=0, a(2)=1.
%F Given g.f. A(x), then B(x) = A(x) - 1 + x series reversion is -B(-x). - _Michael Somos_, Sep 07 2005
%F Given g.f. A(x) and C(x) = reversion of x + x^2, then B(x) = A(x) - 1 + x satisfies B(x) = x + C(x * B(x)). - _Michael Somos_, Sep 07 2005
%F From _Paul Barry_, Nov 07 2010: (Start)
%F a(n+1) has Hankel transform (-1)^n*F(n).
%F a(n+2) has Hankel transform (-1)^comb(n+1,2).
%F a(n+3) has Hankel transform (-1)^comb(n+1,2)*F(n+2).
%F Hankel transform of a(n+4) is sum{k=0..n, (-1)^(n-k+1)*(n-k+1)(F(2k+2)-C(1,k)-C(0,k))}. (End)
%F G.f.: A(x) = x + G(0) = 1 + x^2/G(0) where G(k) = 1 - x + x^2/G(k+1); (continued fraction). - _Sergei N. Gladkovskii_, Dec 25 2011
%F Lim sup n->infinity |a(n)|^(1/n) = sqrt(5).
%F From _Gennady Eremin_, May 15 2021: (Start)
%F a(n+2) = A343773(n), n >= 0.
%F G.f.: A(x) = 1 + x^2*B(x), where B(x) is the g.f. of A343773.
%F Lim_{n->infinity} a(n)/A001006(n) = 0. (End)
%e G.f. = 1 + x^2 + x^3 - 2*x^5 - 3*x^6 + x^7 + 11*x^8 + 15*x^9 - 13*x^10 - 77*x^11 - 86*x^12 + 144*x^13 + ...
%e From the table of powers of A(x) (A100224), we see that
%e 2^n-1 = Sum of coefficients [x^0] through [x^n] in A(x)^n:
%e A^1=[1,0],1,1,0,-2,-3,1,11,...
%e A^2=[1,0,2],2,1,-2,-5,-2,12,...
%e A^3=[1,0,3,3],3,0,-5,-6,6,...
%e A^4=[1,0,4,4,6],4,-2,-8,-3,...
%e A^5=[1,0,5,5,10,10],5,-5,-10,...
%e A^6=[1,0,6,6,15,18,17],6,-9,...
%e A^7=[1,0,7,7,21,28,35,28],7,...
%e A^8=[1,0,8,8,28,40,60,64,46],...
%e the main diagonal of which is A001610 = [0,2,3,6,10,17,...],
%e where Sum_{n>=1} A001610(n-1)/n*x^n = log((1-x)/(1-x-x^2)).
%t CoefficientList[Series[(1+x+Sqrt[1-2*x+5*x^2])/2, {x, 0, 30}], x] (* _Vaclav Kotesovec_, Feb 11 2015 *)
%o (PARI) {a(n) = if( n<=0, n==0, (2^n - 1 - sum( k=0, n, polcoeff( sum( j=0, min(k, n-1), a(j) * x^j)^n + x * O(x^k), k))) / n)}
%o (PARI) {a(n) = if( n<=0, n==0, if( n==1, 0, if( n==2, 1, ((2*n-3) * a(n-1) - 5 * (n-3) * a(n-2)) / n)))}
%o (PARI) {a(n) = if( n<0, 0, polcoeff((1 + x + sqrt(1 - 2*x + 5*x^2 + x * O(x^n))) / 2, n))}
%o (Python)
%o A100223 = [1, 0, 1]
%o for n in range(3, 801):
%o A100223.append( ((2*n-3)*A100223[-1]
%o - 5*(n-3)*A100223[-2])//n ) # _Gennady Eremin_, Apr 24 2021
%Y Cf. A001006, A001610, A007440, A100224, A100225, A214649, A343773.
%K sign
%O 0,6
%A _Paul D. Hanna_, Nov 28 2004
%E Entry revised by _Paul D. Hanna_, Mar 19 2013
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