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%I #18 May 20 2014 03:05:20
%S 1,2,-2,4,-12,40,-144,544,-2128,8544,-35008,145792,-615296,2625792,
%T -11311616,49124352,-214838528,945350144,-4182412288,18593224704,
%U -83015133184,372090122240,-1673660915712,7552262979584,-34178799378432,155096251351040,-705533929816064
%N G.f. A(x) satisfies: 2^n + 1 = Sum_{k=0..n} [x^k] A(x)^n for n>=1.
%F a(n) = (-2*(2*n-3)*a(n-1) + 4*(n-3)*a(n-2))/n for n>2, with a(0)=1, a(1)=2, a(2)=-2.
%F G.f.: A(x) = (1+2*x + sqrt(1+4*x-4*x^2))/2.
%F G.f. 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] F(x) denotes the coefficient of x^k in F(x).
%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) = g.f. of A025225, then B(x)=A(x)-1-x satisfies B(x)=x-C(x*B(x)). - _Michael Somos_, Sep 07 2005
%F G.f.: 4x^2/(1+2x - sqrt(1+4x-4x^2)). - _Michael Somos_, Sep 08 2005
%e From the table of powers of A(x), we see that
%e 2^n+1 = Sum of coefficients [x^0] through [x^n] in A(x)^n:
%e A^1: [1, 2], -2, 4, -12, 40, -144, 544, -2128, 8544, ...;
%e A^2: [1, 4, 0], 0, -4, 16, -64, 256, -1040, 4288, ...;
%e A^3: [1, 6, 6, -4], 0, 0, -8, 48, -240, 1120, -5088, ...;
%e A^4: [1, 8, 16, 0, -8], 0, 0, 0, -16, 128, -768, ...;
%e A^5: [1, 10, 30, 20, -20, -8], 0, 0, 0, 0, -32, ...;
%e A^6: [1, 12, 48, 64, -12, -48, 0], 0, 0, 0, 0, 0, ...;
%e A^7: [1, 14, 70, 140, 56, -112, -56, 16], 0, 0, 0, ...;
%e A^8: [1, 16, 96, 256, 240, -128, -256, 0, 32], 0, 0, ...; ...
%e In the above table of coefficients in A(x)^n, the main diagonal satisfies:
%e [x^n] A(x)^(n+1) = (n+1)*A009545(n+1) for n>=0.
%o (PARI) {a(n)=if(n==0,1,(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==0,1,if(n==1,2,if(n==2,-2,(-2*(2*n-3)*a(n-1)+4*(n-3)*a(n-2))/n)))}
%o (PARI) {a(n)=polcoeff( (1+2*x+sqrt(1+4*x-4*x^2+x^2*O(x^n)))/2,n)}
%o (PARI) a(n)=polcoeff((1+2*x+sqrt(1+4*x-4*x^2+x*O(x^n)))/2,n)
%Y Cf. A100223, A071356, A009545, A143045.
%Y a(n) = -(-1)^n * A025227(n), if n>1.
%K sign
%O 0,2
%A _Paul D. Hanna_, Nov 30 2004
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